Sum : 47 (49) for 101 and for 100 47-7+6=41 or 49-7+6=48
II. Minus after death of Nostr and grave .
Adds : 1,1,1, 1, 6/60×13,1,1,2 - 20 numbers .
To deduct 6/60×13,1,1,2 need from 1418 , because it ecu. As will to deduct 4 ? From another 4, multipliers 4 no go, if for heirs less of money , what changer in the line of "money" no haven"t, for example, 36=4,9.
14188/10 -1-1-2-13/10=1412,5 , need "almost" 1419
1412=4,353 , this number Nostr show us numbers in first number.
400+(1412...+1+1+2+13/10 )+1200+(36,100,1,79,126,4,2,1,1,8,10,17,4, 12,12,12) =3444,10 with debt oк without 1100, 23 numbers or 22 (for101)
2,3,4,16,25,353+425=2,3,4,5,7,8,9,11?,16,17,25,79,353- 12(13) multipliers, sum 528(539)
2,3,4,16,25,353+425=2,3,4,5,7,8,9,16,17,25,79,101,353 - 13 multipliers, sum 639
2,3,4,5,7,8,9,11?,16,17,25,79,101,353 - 13(14) multipliers, sum 639 (650)
Sum : 47(49) for 101 , for pair of binomial coefficients (28,35) 41(43)для 100 for pair (21,35) or (35,21) ? binomial coefficients ( no threes of Pifagor!)
III. All the distribution of the inheritance.
Has 4 variables.
1200=10+500+100+100+490
1200=10+500+100+100+100+100 +290
1200=10+500+400+100+100 +90
Number 500 from testament need to dispense, because children of Anna Ponsard no to born, their no.
For example we no dispensed on 500: 1200=500+500+100+100+10 - no, it more 1200.
Numbers 100 and 600 takes 2 and 1 numbers as enumeration "and , and". Numner 400 uses 2 numbers : accrued and paid. Nostr - chief accountant.
For 100 have 353, but 288 did for 101, so I took 2 variables. In case 353 the line of "money" no array , perhaps it key or key + the line of "money" for 353.
2,4,5,9,16,25,53,125+425=2,3,4,5,7,8,9,11?,16,17,25,53,79 ,125 - 14 (13) m, sum364 (353)
Lines of money 288 and 353 go to Gorapollon 1 the book,1 the part on 11 and 2 the part on 14 . Under the array two the line no go. Why? No hasn"t of key .
All calculations I did without key. Remained will do key and to calculate all years !
It the calculation rightly, he new another than for hronologies. Lines find each other in algorithm of Evclid.
The line of objects:
Summ "before": 2 ,2, 2,2,1,(1?),2,6,4,12,6,6,6,1,1,1,1,3,1, 1,1- 21 numbers, sum 62(61)
Sum "after" : 2 ,2,1/3, 2,2,1,(1?),2,6,4,12,6,6,6,1,1,1,1,3,1, 1,2/3 - 22 numbers, sum (61)62
Array from testament NO USE ! Their need to sort on threes or binomial coefficients. For us need 1001-1002, 300, 353,288.
4. Arrays of 1,2,3 hronologies
Assavoir mon : 2,3,4,5,7,9,11,13,16,25,49,61 - sum 205, 12 multipliers
The line assavoir mon: 4,11,14,25,36,48,49,52,61,65 - sum 365, 10 numbers
First the hronolohie:
515=5×103
516=3×4×43
570=2×3×5×19
600=3×8×25
1350=2×25×27
1080=5×8×27
1242=2×27×23
For sort of years for 516 ( 600 with or without)+ assavoir mon:
2,3,4,5,7,8,9,11,13,16,19,23,25,27,43,49,61- sum 325 , 17 multipliers
Building of testament with minus of the legacy on 288,300,353 right and remained.
Now I did beginning building. 1001 or 1002 of prophets can got , if will do factorization under all minus or all factorization with partial minus.
1200+435(currency)+400+1418...=3444,1
Currency:
2,3,4,5,7,8,9,17,79,101 - sum 235(232) , multipliers 10
2,3,4,5,7,8,9,17,25,79 - sum 159(156), multipliers 10
1200=100+100+400+500+90+10=2,4,5,9,16,25,125 or 1200=800+400=16,32,25
1418...=600+808+8-11...
To mistake can only in minus 1418... , another defined clear. 600 deducted only one number (and you ,and you ) + details.
Sum for 101:
2,3,4,5,7,8,9,16,17,25,79,101,125,601 - 14 multipliers, sum 1002 - No
Sum for 100:
2,3,4,5,7,8,9,16,17,25,79 ,125,701- 13 multipliers, sum 1001
2,3,4,5,7,8,9, 17,25,32,79,809(811) - 13 multipliers, sum 1000-1002, 8 power numbers
2,3,4,5,7,8,9,16,17,25,79,101,125,601 - 14 multipliers, sum 1002
Sum: 59
2,3,4,5,7,8,9,16,17,25,79 ,125,701- 13 multipliers, sum 1001
Sum: 57
7. The secret of letter "L".
Need to take, what " the secret " this letter deprecated on 500 years and stitched white thread.
This letter very important and open the sort on catrens. In the concept of codes it name point of shift. For example : (l+L)mod?=?
"L" is 11 letter of Roman ABC. Сам "text" of catrens similar to simple text , which need to decrypt. So first letter can to be another, but to add her need only as 11. (o+ABC with L)mod11(22 or 23?). In more difficult the variable "L" no one , she change, it our case. She change with help of the line ABC .
ABC beside numbers 1, 2,1,4,1,11,22,1,1,1,1 have letters a,b,f,g, h,l,n,t,v,y,z . Following letter after "L" go "N=22".
As the variable can use key words .
8. ARRAY OF THE MINI HRONOLOGIE
To this the building I return again as to continue.
We has key 120×5=600; 3 hronologies : 1,2, (1+2) or 3 hronologies 61×6=366; array assavoir mon on 58 numbers and need also 154 or 165 numbers of catrens of almanacs .
1610+621+1566=3797, by this, as we remember, 1610 - year astrological end of key
4092-4173=81 - even 2 the hronologie , while her formally no haven"t, takes invisible part, multipliers 81 go to 1566 year.
2,5,7,23,27,29,81- sum 171
The mini hronologie + assavoir mon:
2,3,4,5,7,9,11,13,16,23,25,27,29,49,61,81 - 16 numbers, 11 power, sum 365
3/2=1+1/2
3/4=1+1/3
4/5=1+1/4
7/5=1+2/5 5/2=2+1/2
9/7=1+2/7 7/2=3+1/2
11/9=1+2/9 9/2=4+1/2
13/11=1+2/11 11/2=5=1/2
16/13=1+3/13 13/3=4+1/3
23/16=1+7/16 16/7=2+2/7 7/2=3+1/2
25/23=1+2/23 23/2=11+1/2
27/25=1+2/25 25/2=12+1/2
29/27=1+2/27 27/2=13+1/2
49/29=1+20/29 29/20=1+9/20 20/9=2+2/9 9/2=4+1/2
61/49=1+12/49 49/12=4=1/12
81/61=1+20/61 61/20=3+1/20
Sum: 61
So, array for rings of Evclid need uses with 1 the part first the book of Gorapollon and lines of objects.
Numbers for almanacs must gave from (1+2) hronologies ? or from 3 the hronologie , or any from these 2 arrays would need gone no whole, or array 1002-1001 of prophets.
2) а)First the hronologie. For the sort of years for 516 ( with 600 or without)+ assavoir mon:
2,3,4,5,7,8,9,11,13,16,19,23,25,27,43,49,61- sum 325 , 17 numbers , 11 power numbers, go for 1 the part of Gorapollon.
Array 61 calculate catrens from centuries.
б) Two the hronologie. For 476 without 600 +assavoir mon:
2,3,4,5,7,9,11,13,16,17,25,43,49,59,61,251 - sum 575, 16 numbers for Gorapollon,I
the book , 3 the part.
Array 61 calculate catrens from centuries.
в) Together (1+2)=4173,816
Array 61, calculate catrens from centuries and Galen .
г) Assavoir mon calculate estetiche, array 58.
д) Key , array 120×5=600 calculate catrens from centuries. He give majority of catrens.
3) Three the hronologie no haven"t 61 and 60 . Or her no haven"t and another array calculate of almanacs, for example, (1+2), or catrens no 154 and number another . Number 14 from Gorapollon have only in following the variable of the hronologie, but array have number 79, as minimum on 158 catrens, if gone array to the end. Until I no see array for almanacs and number 14. If 3 the hronologie participates as the part of mini the hronologie , then the beginning (1+2) need revision and need would gave 14 numbers for 2 the part 1 the book of Gorapollon.Can almanacs calculate array 1002-1001 from testament.
Key have 120 numbers for 2 the book of Gorapollon.
Three the hronologie with 600 and 1566: 2,3, 5,8, 11, 25,27,29,49,53, 79,163,257,613 - 14 multipliers
If no calculate 3 the hronologie gets 6 arrays and 3-4 array dates, which must be equivalent to arrays of years by the sort of years on Evclid and in parallel to they dates.
4) About date. The point of counting from 27.06.1558 year can in one case: if adds 7.73.
As need calculate dates ? Selection the variable of the calculation.
1) Dates:
1555.3.01(331 days or 11 months.1day)+(177).3.11(101 days or 3 m.11 d) - from letter Cezar ,59(three 3,4,5)
(73).7(210 days or 7 m.)+1557.3.14(344 days or 11 m.14 d) - from letter of Henry, three 7(4,3,5), (48,55,73)
10+3+25-yy.
6.27=87 days or 2 m.27 days
All arrays from my another fails use for calculate years, , for example, if array 120 go 5 numbers with threes of Pifagor , got 600 years, if go on 3 arrays of hronologies , got 61×6=366, sum 966 catrens, 58 give array assavoir mon, but for 11 from Galen and 154 for almanac it can be array 1+2 hronolohies.
We have 3 questions until not solved.
First, as to calculate dates.
Two ,as to combine years (or parts years which got from the sort on arrays) with got centuries and catrens. As the line of catrens be astrologically " empty " years, their need add to whole years. In our case can to multiply, because the line of catrens be components of " empty " years.
Three, have also encrypted 58 catrens . They no have not cipher of Cezar, I wrote about the line. The part I did , but need my the calculate to add.
About letter of the part of the line catrens .De get numbers of catrens and century , as I you wrote, use (the line of catrens+ the line АВС)mod 11 with view point of shift "L". Can change available the line of ABC code on 11 letters under whole Roman on mod 22 or 23 letters. In this case you table of triangle numbers from triangle of Pascal will appropriate with amend "k" if her to delete , and "U" to add or no to delete. It a kind of cipher G.Y.Cezar or Weiner , what almost equivalent, because the code of Weiner consist from a set of ciphers Gay Yuly Cezar. For example , (2+6)mod11≡3, shift begin from "0" on 3 letters forward and .... But thus years and catrens no connect .I don"t know as.
Can С(n,m)×years/the line of catrens, also my common the calculation on samples ( (4) ,(11),(22) more in detail.
Until about dates. Suppose what years from arrays did calculated with help of algorithm of Evclid.
Day from letter of Cezar 1555 year.
3.01(331 days or 11 months.1day)+(177).3.11(101 days or 3 m.11 days) - from letter of Cezar ,59(threes 3,4,5). Dates to take on days already passed!
I did different variables of the calculation:
Years and dates must combined between themselves . Dates 177 and 73 years with months and day Nostr give in collected view, as and hronologies , us need take on threes.
For example ,1 mart 1555 year - 331 days. In array go, for example, mod 3,4,5. Go on dates:
331=1mod3 , (331+1)=0mod4 332=2mod5 334=1mod3 и т.д.
In more difficult the variable the line of objects (after death of Nostr ) multiply on remains 1,0,2 and .
The sort similar the sort on letters for chipher .With one the side can to calculate, until go to the date 3 m. and 11 days, and to calculate again on three . With another the side to dates adds even dates . Can to take one common sum , but their need expand. Because I got apparently half the calculate of dates , need did difficult his.
In books nave and another variables of the sort on dates.
For example months and days I did one number , but point of beginning and add dates now shown separately. Also can months and days. As the variable can to calculate across suitable fractals. This the calculation I did as example in another my file.
For 1.03 1555: 331х+101у=1
For 14.03.1557 : 344х+210y=1
If add only three of Pifagor , the increase of dates will go very quickly.
Across array for 1.03.1555:
331/101=3+28/101 101/28=3+17/28 28/17=1+11/17 17/11=1+6/11 11/6=1+5/6 6/5=1+1/5 (12) can 14
For 14.3.1557:
344/210=1+134/210 210/134=1+76/134 134/76=1+68/76 76/68=1+8/68 68/8=8+4/8 8/2=2+0(11 with out 0)
Conditionally this the variable suitable for us. In this the variable Me no like uncertainty .
Have very good the variable building of array the line assavoir mon, common sum 365 . Figaro play role in the calculation of dates ?! In another case need express dates and months in years. Probably numbers of Gorapollon uses for calculate dates.
I might the variable have stopped. Me very like this the variable. The conditions of equivalence from module mathematic necessarily for any case , if you added with the right +3 or to multiply , or..., need add also with the left +3 or ... . Also I see here numbers of Nostr assavoir mon and numbers of Gorapollon.
So years of future or lines of testament 300,353,288, need to calculate with 3 arrays of hronologies+ the line of objects 1(before death of Nostr); numbers of Gorapollon first the book to calculate to dates, but without years of future. To the key go 2 the book of Gorapollon and the line of objects (after death of Nostr). Threes of Pifagor for all. Foreword of Gorapollon go to 58 catrens or to array assavoir mon.
Rings of Evclid would can be field ,if mod - simple numbers.(3,4,5), what unlikely, because have three of Antichrists (48,55,73) and another threes change . Rind should be closed.
Astrological ring 360 № equivalent to rings of Evclid all or partially!
V). Can also dates and years connect and add the line of catrens× years/ the line of identification. Numbers will very bulky.
The line of catrens now have "L" , so can done catrens and century in one a line , but need add their to years. It one the calculation or need calculate two numbers ? I don"t know. Numbers from triangle of Pascal and the line of АВС give us numbers catrens and century , 3797/46=3772+25, 28×46=966. Combinatorial analysis need. I did only common combinatorial the calculation, but need also connection with years. , need use the line of catrens (digital the part) and the line of identification. Can use catrens ×years the line of identification, digital the part consistent with letter the part .But the sort on frames different from direct the sort of "L"ю Now I no finished. Have a discrepancy.
I did no all, but few to figured out with dates. I gave to you enough to calculate on arrays yeas and dates.
10. THE CALCULATION "BIG" YEARS FROM LETTER OF HENRY
If the calculation no ended on 2797 y., can to calculate again across arrays.
Nostr gave only years without dates.
10(11)x+3y+25z=2797
In case across suitable fractals we had in 2 numbers less numbers.
Can do separate no big array for algorithm of Evclid : 10or 11,3,25,2797:
(11/3=3+2/3 3/2=1+1/3)
10/3=3+1/3
25/3=8+1/3
2797/25=111+22/25 25/22=1+3/22 22/3=7+1/3
Sum: 11(13) for 1 the part 1 of book Gorapollon for catrens from Galen.
From this no big the calculation sow , what years near of future and very far future can was coexist .
Sum: with 0 without reduce12+4+4+8+6+6+8+40+4+24=116 ; without 0 without reduce 9+3+3+8+5+6+7+40+3+24=112 ; with 0 with reduce 8+2+4+2+12+12+40+4+24=106 ; without 0 with reduce 6+2+3+2+12+12+40+3+24=114
For binomial coefficients . For another winter Ме and Ме with 2 number summer.
b) For multipliers : 365 days: from 1.03.1555 for 1.01 :
For 91: 2,3,5,7,8,13,16,18,25,31,37,41,47,113, - 14 multipliers, sum 366
S: with 0 without reduce 14+10+38+34+14+8+10or8+12or8+6=140-146 ; without 0 without reduce14+14+38+34+14+7+10or8+12or8+6; with 0 with reduce 14 14+14+38+34+12+8+10or8+12or 8+6; without 0 with reduce 14+14+38+34+12+7+10or8+12or8+6=143...
Sum: with 0 without reduce 10+10+68+10+24+8or12+4or8+8=142-150 ; without 0 without reduce 10+9+68+9+8or12+4or8+8=112-120 ; with 0 with reduce 10+8+68+8+8or12+4or8+8=110-118 ; without 0 with reduce 10+8+68+8+8or12+4or8+8=114-122 ;
For 365 days, sum120: 1) For 1.03.1555 from 1.04
Without 0 without reduce 10+9+68+9+8or12+4or8+8=112-124 for Ме 3-24 winter, Ме 2-24 summer or Ме 3-27 winter,Ме 1-24 summer
Sum: with 0 without reduce 4+84+8+10or14+12+16=134-138; without 0 without reduce 3+84+7+10or14+12+16=132-136 ; with 0 with reduce 4+84+6+10or14+12+16=140-144 ; without 0 with reduce 3+84+6+10or14+12+16=141-145
Sum : с 0 без сокр. 14+12+12+12+12+10+12+16+28+10+8+14or12+8=168-166 ; без 0 без сокр14+11+12+11+12+9+11+15+26+9+7or8+14or11+7=158-156 ; с 0 с сокр 14+10+6+12+12+8+10+14+24+6+6+14or10+6=142-138 ;с 0 без сокр 14+10+6+12+12+8+10+14+24+6+6+14or10+6=142-138
Sum : with 0 reduce 6+8+12+12+8+14+12+16+8+6or12+8+8or10=118-126 for Ме 3-27wimter , Ме 2-24 summer ; without 0 without reduce 6+7+11+12+8+13+10+14+8+6or11+8or9+7= ; with 0 without reduce 6+6+8+12+20+10+14+8+6or8+8+6=102-107 ; without out 0 with reduce 6+6+8+12+20+10+14+8+5or8+8+6=103-108
For 360 days, sum 120:
From 1.04. , all, 360 days,120:
For 120: with 0 without reduce 6+8+12+12+8+14+12+16+8+6+8+10=120 for Ме 3-24 winter, Ме 2-24 summer
Sum : with 0 without reduce 12+9+8+26+10+8+14+10+8+10+6+10+12=143 ; without 0 without reduce 11+8+7+26+9+7+13+9+7+9+5+10or9+11=132-131 ; with 0 with reduce 12+6+6+26+14+24+8+6+4+10or8 +6=122-120 ; with 0 with reduce 11+6+6+26+14+24+7+6+4+10or8+6=118-120
For sum 120, 360 days:
For 1.03.1555 from 1.04, sum 120:
For 120: with 0 with reduce 12+6+6+26+14+24+8+6+4+8 +6=120 for Ме 3-24 (or 3-27) winter, Ме 2-24 summer; without 0 with reduce 11+6+6+26+14+24+7+6+4+10+6=120 for Ме 3-24 (or 3-27) winter, Ме 1(or 2)-24 summer
with 0 without reduce 4+8+10+12+12+10+12+8+14+6+12or8+14or12+10+8=134-140 ; without 0 without reduce -3+7+9+11+11+9+11+7+13+5+12or 8+14 or 12+10+8=130(126)-124 for Ме 3-27 winter, summer Ме 1-24; with 0 with reduce 2+6+8+10+10+8+10+6+12+4+12(or 8)+ 12+ 10+8= ; without 0 with reduce; 2+6+8+10+10+8+10+6+12+4+12(or 8)+12+10+8=114-118
Sum: with 0 without reduce 4+8+10+14+10+12+8+14+14or 12+16=110-108; without 0 without reduce 3+7+10+13+9+11+7+13+13or12+16=104-103 ; with reduce 0 with reduce 2+6+10+8+10+10+6+8+12+16=89 ; 0 with reduce 2+6+10+8+10+10+6+8+12+16=89
For whole asssavoir mon:
4,11,14,25,36,48,49,52,61,65 - 10 multipliers, sum 365
Sum: with 0 without reduce 4+10+10+4+22+4+6+10=70 ; without 0 without reduce 3+10+9+3+22+4+6+9=66 ; with 0 with reduce 2+10+2+6+22+6+6+6=62 ; without 0 with reduce 2+10+2+6+22+6+6+5=61
The variables for 365:
4,11,14,25,36,48,49,52,61,65 - 10 multipliers, sum 365
Sum : with 0 without reduce 8+8+32+12=60; without 0 without reduce 8+7+32+11=58; with 0 with reduce8+6+32+2+2=50 ; without 0 with reduce 8+6+32+2=48
For ABC ?
How many days in years ? In Gorapollon 360, common year 360+5 (6, 0) ...
4 array did for catrens for almanac, centuries, 11 of Galen, estetiche calculate assavoir mon-Nostradams. If not enough 4 arrays have "spare" it lines 3 hronologie .
Lines of Nostr with "big" numbers founded their place .
Need with hronologies ,which have 3 arroys+3 lines of "money", later array key+the line of identification. Nostr mixed year past and future !
SUM: for 360 days ,126:
1) From 1.04. for all, 360 days, for Ме 3-27 winter, Ме 2-24 summer 126 numbers
2) From 14.03.1557 from 1.04 , for Ме 3-27 winter, summer Ме 1-24 126 numbers
For 360 days, sum 120:
From 1.04. , all, 360 days,120:
For 120: with 0 without reduce 6+8+12+12+8+14+12+16+8+6+8+10=120 for Ме 3-24 winter, Ме 2-24 summer
For 1.03.1555 from 1.04, sum 120:
For 120: with 0 with reduce 12+6+6+26+14+24+8+6+4+8 +6=120 for Ме 3-24 (or 3-27) winter, Ме 2-24 summer; without 0 with reduce 11+6+6+26+14+24+7+6+4+10+6=120 for Ме 3-24 (or 3-27) winter, Ме 1(or 2)-24 summer
For 365 days :
1) From 1.04. for all, 365 days, 126 :
For Ме another winter 3-24 or 3-27 or Ме summer another ; or Ме 3-24 with Ме 1-24 ; or Ме3-27 with Ме 2-24 126 numbers
For 365 days, sum120: 1) For 1.03.1555 from 1.04
Without 0 without reduce 10+9+68+9+8or12+4or8+8=112-124 for Ме 3-24 winter, Ме 2-24 summer or Ме 3-27 winter,Ме 1-24 summer
We see , what key it no key, this is array. Whole lines and arrays interchangeable. Also I did one the variables of assavoir mon for whole numbers without decomposition on multipliers. With key all o"kay.
If to take 4 arrays ,to add to their remains=fractals, also to change arrays and lines , we got 1123 .
If no to change arrays and lines, we got more 900 numbers. For example , for remains: 3/2=1+1/2. 1/2=0+1/2 2/1=1+0 or 3/2=1+1/(1+1/1).
If Nostr took from 1.04 , then 1558 to calculate separately no need.
I did all calculation for summer Ме 1(2)-24 , for Me 1(2)-14 I no sow.
What are the criteria of sort on years. First it building of a generalized triangle Pascal (?) or divisibility (?). I don"t know.But also to search as move array. I see two variables, as mod across algorithm of Evclid and two as the line for suitable fractals, for example (" ...the chain of years..." as wrote Nostr). With this building I finish preparatory the part of the code.
From 2 variables can to take one , it suitable fractals (the chain ), it no mod. It would give for us more numbers of catrens. Approximately ∑▒〖suitable fractals×〗(nm)=.... . This method give only a approximate to first number with step , for example 8/5=3+2/3 ... , the approximate 3,... .
So algorithm of Evclid better. To calculate can from top to bottom and Vice versa. The code allow improvisation . Сriterion is MCD and remains for algorithm of Evclid. Have 3 the variable it CTR, but in this case lines of Nostr go out.
Sum : with 0 without reduce 46 , without reduce without 0 45, with reduce with 0 44 , with reduce without 0 44
It"s for the sort of estetiche. We know , as to calculate years for century, Galen, almanac , but we no know as theirs to sort. We know , as to sort estetiche , but we no know as their to calculate. Have also 2 variablles assavoir mon=Nostradamvs , but I no calculated it.
Can it the calculate and the sort together.
We have only numbers 126 ? No. Can 120, it have in Gorapollon 120×3=360№ as numbers days in years , also 120 numbers have 2 the book of Gorapollon , also 84 . 84 numbers I whole no sow.
As need to sort on years. Some thoughts have at Me.
To calculate across mod. To do equations or system of equations.
For example 3х+y≡6m+27n , x,y, m,n - change .
Also have simple the variable. We take algorithm of Evclid a=b×(q+n)+(r+m), n, m change.
All depend from numbers of Gorapollon, it or Cnm or 6 and 1, 1and 2 , and ... .
Criterion CommonMaxDivider for 2 numbers and theirs CMD.
Then which have difficulty ? Need also to calculate triangle of Pascal extended or generalized. Because add numbers change.
11. Arrays key and assavoir mon (compact the variable)
Sum: with 0 without reduce 10+10+68+10+24+8or12+4or8+8=142-150 ; without 0 without reduce 10+9+68+9+8or12+4or8+8=112-120 ; with 0 with reduce 10+8+68+8+8or12+4or8+8=110-118 ; without 0 with reduce 10+8+68+8+8or12+4or8+8=114-122 ;
For 365 days, sum120: 1) For 1.03.1555 from 1.04
Without 0 without reduce 10+9+68+9+8or12+4or8+8=112-124 for Ме 3-24 winter, Ме 2-24 summer or Ме 3-27 winter,Ме 1-24 summer
Sum : with 0 reduce 6+8+12+12+8+14+12+16+8+6or12+8+8or10=118-126 for Ме 3-27wimter , Ме 2-24 summer ; without 0 without reduce 6+7+11+12+8+13+10+14+8+6or11+8or9+7= ; with 0 without reduce 6+6+8+12+20+10+14+8+6or8+8+6=102-107 ; without out 0 with reduce 6+6+8+12+20+10+14+8+5or8+8+6=103-108
For 360 days, sum 120:
From 1.04. , all, 360 days,120:
For 120: with 0 without reduce 6+8+12+12+8+14+12+16+8+6+8+10=120 for Ме 3-24 winter, Ме 2-24 summer
Sum : with 0 without reduce 12+9+8+26+10+8+14+10+8+10+6+10+12=143 ; without 0 without reduce 11+8+7+26+9+7+13+9+7+9+5+10or9+11=132-131 ; with 0 with reduce 12+6+6+26+14+24+8+6+4+10or8 +6=122-120 ; with 0 with reduce 11+6+6+26+14+24+7+6+4+10or8+6=118-120
For sum 120, 360 days:
For 1.03.1555 from 1.04, sum 120:
For 120: with 0 with reduce 12+6+6+26+14+24+8+6+4+8 +6=120 for Ме 3-24 (or 3-27) winter, Ме 2-24 summer; without 0 with reduce 11+6+6+26+14+24+7+6+4+10+6=120 for Ме 3-24 (or 3-27) winter, Ме 1(or 2)-24 summer
II. For whole asssavoir mon:
4,11,14,25,36,48,49,52,61,65 - 10 multipliers, sum 365
Sum: with 0 without reduce 4+10+10+4+22+4+6+10=70 ; without 0 without reduce 3+10+9+3+22+4+6+9=66 ; with 0 with reduce 2+10+2+6+22+6+6+6=62 ; without 0 with reduce 2+10+2+6+22+6+6+5=61
The variables for 365:
4,11,14,25,36,48,49,52,61,65 - 10 multipliers, sum 365
Another 1,2,3 only multipliers Chronologies+assavoir mon (Patris Gunar) = "kappa, teta, lambda" or 8,10,11
14. CALCULATE (1+2) hronologies 4173,8 for algorithm of Evclid
I did 3 variables (1+2) + "fita".
Can two change : only for 515 and 516 , 112 ... and 113... Two for 4 and 480 . For 480 and 1350 =789 . For 782 have 4173, 816 , another 4174 ... .
For Gorapollon , 2 the part =14 have one the line of three the hronologie , can assavoir mon and the change places. The line (1+2) no haven"t length 14.
I did all the calculate for algorithm of Evclid or for years of past. For years of future I few no finished. We can to begin calculate years in the side more with help threes of Pifagor .
15. Calendare numeration in the code of Nostrt
"All lie calendars ..."
A. Griboedov "Woe from wit", madam Hlestacova
Something in the code can be explained in terms of calendar.
In letters has years of future is threes and dates. We see double of year future , with the side it array of key, with another the side years and dates it threes of Pifagor with dates already has. Key can go to one hrohologie , threes to another .It no very clear. At that 5 arrays need to calculate as array separately . Can variables.
We have lines , which we calculated, but no explained. For example the line of objects . It the line of tableware and cushions ?
Calendar on 365 days, every 4 years leap year, his name is Julian. Middle continuance of Julian year few more of Tropical year on 0,0078 middle solar day. Approximately across 128 years different accumulated and comprise 1 day. Lenght of moon month in the code not specified.
Now further on Nostr gone . Threes of Pifagor (28,21,35)=7;(4,3,5)=84 . Why this was awarded this threes? At once need to talk about number 488, it sum of you the line catrens, two the part of this the line is cipher.
Now return to calendar.
84-year cycle Tropical years , Avgust, III-V century, 84 years=30680,365 day=1039months , in which 551 months full ,31 months insert) and 488 blank ( it sum you the line of catrens). In the end of cycle the full moon moved forward and got number 30682 days . This calendar match with Julian ,it all facilities.
You understand to Me , what identification blank ?It identification, what has winter the season when nature sleep.
As variable .
The line assavoir mon has sum 365 , it days , key has days and months. Muslim calendar Hijra calculate days ( go to first hronolohie + assavoir mon) , Christian calendar (Julian) days and months, it (key and 2 hronologie) . Different solar years and lunar 354 years (Muslim calendar comprise 11 days. Combination first and two hronologies it division one numbers on another (lambda of Pifagor).
Lines of "money" is parts years future, which we no know , but Nostr calculated.
Would be all good and logic in the code , but 3 lines of "money" (288,300,353 sum 941 ) go to future , but they want to attached to 3 hronolohies , that is to past ?
Arrays need to calculate all separately , key very big on 120 years, array assavoir mon has 58 numbers of length .About all years . If on Jotish, 120 years it whole cycle solar years, after all repeats .
Next. What is it the line of objects? It ADD MOON days to solar, for example multiply on remains in rings of Evclid . But role their one it filling scarce of days . Something from the code Nostr put in catrens , it "damp" letters and completion missing lunar days.
Tropical year is 365 days. As 12 months in year conditional for us , lunar got almost 13 months from the calculate approximately 29,5 days in year , in testament it 12-13 hears.
Century III, catren 4:
"Quand seront proches le defaut des lunaires,
De l'un a l'autre ne distant grandement,
Froid, siccite,danger vers les frontieres,
Mesmes ou l'oracle a prins commencement".
Why Nostr took 22 or 23 numbers ? I think , what it"s cycles of the new moon for 2 years for 22 and 1 for one. That is for year (12 months ) the new moon comes on 11 days before. This numeration do to cycle of Meton.
353 as number of catrens, or as sum one from lines of "money" go to simple in Yuda calendar hronologie 353,354,353 , separately go leap years 383,384,385, at that 353 and 383 calculates " insufficient ", 355 and 385 " excessive ".
Julian calendar entered Julian Caesar in 46 year before our era , it 46 letters from 11 letters "ABC" cipher Nostr .In vintage Paschal books age of Moon for year with row of Moon L=1 takes 11 , but cycle of Meton was incorrect. With help of row Moon calculated phase of Moon. So Nostr did the point of shift L=11 .
Cycle of Meton it 19-year cycle , uses in China , Vavilon, open Greek scientists again in 432 y. before our era .
Tentatively lunar month of Nostr was 29 or 29,5 days, last number 29,5;2=59 taken from ancient Greek calendar.
About 3797. If divided number 3797 we see ,what Nostr symbolically indicates, what 97 days go to Gregorian calendar , but 37 means 3 solar years or 37 lunar months, when Tropical year 365, 2422 divided on 29,53059, fractal in 2 approximation 1/3 +12 months. In Gregorian calendar за 400 years add day 366 adds 97 numbers, on compare with Julian here 3 days aborts . Nostr reminds us number , except text, that different lunar and solar days in time of year cycle was posted, also about , what he know about Gregorian reform, which been.
What mean latitude of the earth in astronomy ( the line in identification).
???? I don"t know calculates in mundane astrologie .
Lines of "money" (parths of years future) have similar numbers, so their sort with help threes of Pifagor. After retiring from hronologies their we cannot collect with the same numbers, it right of mathematic. So if you see such "decryption" of the code , you must to know, what you are deceived.
Numbers of Gorapollon it combination C(n,m), it MCD (or q?) ,need to see by the calculation of array. I think ,what Numbers of Gorapollon more need for calculate of rings Evclid .Asross them ( or threes ) go connection years with catrens.
So when we to took numbers of Nostr for the calculation , we been know , what meant Nostr, why and for what him liked numbers , which we see in the code. All lines of Nostr got rationale beside numbers of Gorapollon.
16. The choice of the variable calculation years .
I took one example for one array .
For 516 ( 600 with or without)+ assavoir mon:
2,3,4,5,7,8,9,11,13,16,19,23,25,27,43,49,61- sum 325 , 17 numbers , 11 powre numbers , for 1 the part of Gorapollon
You see, what the line closes on itself, so he NO NEED.
2) We took classic the variable algorithm of Evclid , it our array.
For example we calculate algorithm across coefficients of Bezu. We would got numbers "+" and "-", so which numbers also NO NEED. Nosrt begin from IX. Need to calculate or to sort on years with help mod in the side of ascending.
Array consist from rings of Evclid. It is possible to take only one mod or to change mod.
Main for algorithm of Evclid : if a×b/c and one from numbers mutually simple with "c", so two number divided on "c".
What we have? We have 5 array ; separately 3 lines of "money", the line of identification and of catrens (number the part); separately years with dates and threes of Pifagor with dates .
Once there 2 variables. First array it only years of past (array) and years of future (letters) to calculate separately. Two the variables it array add to years , this pieces "find" other with help, for example, numbers of Gorapollon or another common the line . Possible what numbers of Gorapollon go only to dates ? Need to select. The sort with help mod ,it separately and big the part of the calculated algorithm of Evclid , used theory of comparisons and deductions.
It need to emanate from properties of numbers.
Hint for their: to use factorization in rings.
3) Also have the variable across CTR for dates. I until it no sow . Why CTR? Because building for assavoir mon and key similar on CTR (theorem about remains) in the beginning.
P.S. Also can decision across equation of Fibonacci equation х2-х-1=0.
It 2 method of decision.
17. AS TO CALCULATE YEARS OF NOSTR AND CHECK OF THE CORRECTNESS OF THE CALCULATE ON YEARS
We already know , what Nostr used algorithm of Evclid.
Variables :
The variable Љ1. It most simple , which has name as algorithm of Evclid or CTR (Chinese theorem about remains or the calculate of calendar ).
Need got equality of remains , for example, on 3, then по 4 and ....
Hronologies and key no mixed! it no right , key independently calculate years , hronologies would catrens 6×61=366 .
The variable Љ2, for example, cyclic group for whole numbers Z , which by division on k, for example, k=5, gave remains 0,1,2,3,4,5≡0. It the calculation applies to commutative algebra.
The variable Љ3, complicated the variable of first. Calculate on module mathematics from accounting equality of classes deduction on module. It an Abelian group additive and commutative. For example, class deductions on module m=4 equal 0,1,2,3,4.
We have as whole numbers and also with fractals.
For examples of the calculation.
To collect more difficult , then to calculate numbers in smaller a side.
Different in this variables small. I will give example of the calculate , numbers are similar to those, which have in the code.
In this the variable important , that 2 numbers was mutually simple , as private a case CTR.
If mod - always simple number , it"s box of rings. Need talk, what mutual a simple takes for mod.
Threes of Pifagor: 7(4,3,5) , 59(4,3,5), (48,55,73). Threes need add , for example , to lines of testament 288,353,300, which close to each other on numbers , I was took example for all 3 variables. Can add and to the line of " prophets " on 1001 or 1002.
Example Љ1 with of classes deduction: 5/3=3×1+2/3 3/2=2×1+1/2
{█(х≡2(mod3)@х≡1(mod2))┤
х=1+2y Substitute one in another х=1+2y≡2(mod3), 2y≡1(mod3) НОД(2;3)=1
the decision the only thing , y=2, х=5 , the decision got accounting of classes deduction.
This can got also with help formula of Eyler.
This the variable would can use for at the same time the calculate of years and dates (months and days from letter of Henry) as equivalent .
Main criterions: (a-b)/mod, ac=bc(modm) if (c,m)=d; a=bx+r, D(a,b)=d=D(r,b) - for ring
Example Љ2 on CTR: be 2 rings , also two ring longer (here I have in view only years), for example , {2,5,8 }
Then the system of equations: х=с1(mod m1).... х=сn(mod mn)
Solution : х=х0(mod M)
On place 4 can be (4,3,5) remains or another numbers , which no contrary to rights add and multiplication Abelian rings. So to calculate for every group of ring.
Need got remains: (4,3,5). It can be : objects×remain of array.
It"s classic the variable of the calculation on CTR.
Main criterions for CTR : mutual simple 2 of neighbouring numbers NOD(A,B)=1
Example Љ3 on formula of Eyler and suitable fractals : 5/2=2×2+1 ; 8/5=5×1+3 5/3=3×1+2 3/2=2×1+1
Suitable fractals:
k1=1 k2=1 ,2,3
q1= 2 q2=1,1,1
P1=1, 5 P2 =1,2,5,8
х1≡(-1)2×2×remain(mod5) х1≡2×remain(mod5) ; х2≡(-1)4×5×remain(mod8) х2≡5×remain(mod8) - for every ring remain will set.
This the variable Nostr also would can know. If mod, for example, big numbers , the class of deductions to sort long, because would can take formula of Eyler. For more details see rights for formula Eyler. As to calculate suitable fractals I you wrote.
Main criterion for formula of Eyler: ap-1=1(modp); ap=a(modp)
Tail of ring MULTYPLYS on the line of objects , which already no objects, but no missing days to lunar years, and as total - solar the calculate by lunar dates.
For algorithm of Evclid does not matter whether number fractals or whole .
So can calculate key 5 once with 1 three or with 2 , need select. Array can begin with the end or with the begining. As ?
Tip for their:
If took from less to big (array in my the building ), so years been increase quickly, if took with the end , so slower. If distance between numbers great , so the density is less .
For date need took same threes, to the calculating was equivalent.
Nostr taken care of about us and give their numbers from book of Gorapollon.
Example for cyclic group.
Mod 4 1 ̅,2 ̅,4 ̅ , dividers of number 4. For example , number 26≡2mod4 right, because in remain 2. The calculation go on specific whole remains.
Cyclic group consist from set (a0,a2,a3...an) , n×a also belongs to a group.
Can need number to express the degree. 3=31, 4=30×4, 5=30×5, because group have in their consist 3. If in the code threes 7(4,3,5) calculates no add consistently their numbers , as takes on 7 раз every (3,6,9,12,15,18,21) , we have also cyclic group.
For LDE (linear Diophantine equations) : a=bx+r cyclic group not suitable, because have remain. It the variable need delete.
Last the variable with classes of deductions on module .
Here it is important to determine, whole system classes of deductions or given.
For example, for mod 4 0 ̅,1 ̅,2 ̅,3 ̅,4 ̅ has 5 classes whole of deductions natural numbers , given for mod 4 1 ̅, 3 ̅ only 2 classes , because they mutually simple with 4. Example Љ1 see higher. For more details see rights adds and multiply for modules.
Mod no can be all time 3,4,5 , numbers must grow. Example higher with deductions I cited , he no difficult. This the method is error-free. Imagine only , class deductions for 1000, it 1001 of classes deductions ! Something has to limit which this scope . It remains or MCD (max common divider ) and rights of module mathematics.
In the lookup adds numbers of Nostr the calculation few gets rough. Its examples until only semis of the calculation code.
So from our set I would have stopped CTR and classes of deductions, gave need pick up already with numbers of Nostr. Nostr nothing simple us not bequeathed , where is possible not thought. For algorithm of Evclid need programm.
It the calculation on years , as to years need add equality them dates and to calculate in parallel or simultaneously as system of equation. More precisely, the choice of the variable sort Evclid for numbers of Gorapollon and testament . By another methods of the calculate , result will been the same.
No less interesting, how to check the correctness of the calculation years in application to the code of Nostr. . Planetary combinations it"s secondary control . They can have error before month and more.
Here again on help will go come mathematics , need date can calculate across coefficients of Bezu , instead of one date we would have 2 , once from which has minus ,she is counted in reverse the side . The calculation also continuous for all years, but would cam take one separately. In this check the calculation can put another catren . Example of catrens:
Сentury VIII, catren 47:
"Lac Transmenien portera tesmoignage,
Des coniurez sarez dedans Perouse
Vn despolle contrefera le sage:
Tuant Tedesque sterne & minuse".
(about Transmenien lake , Two Punic War in Italy, 217 y. before our era.)
It catren with double the date. A translate another , because the date is determined primarily the calculation .
Catrens with the mention of past , Transmenien lake , Hannibal and ... need to calculate also in minusс. To past applies all before 1555 y. . Also all catrens with names the ancient generals, legacy name of country and ... .
Example: century I ,catren 9
"De l'Orient viendra le cueur Punique
Facher Hadrie & les hoirs Romulides,
Acompaigne de la classe Libycque,
Trembler Mellites:& proches isles vuides".
Here on the contrary, can search the date of past ! The Temple of the goddess of fertility Melitte standing in Vavilon , women had to come in and for a fistful of money surrender a first comer strangers in the Temple. Next under patronage of Melitte women were all life. In a roundabout way it can applies to religious laxity.
Can I early refused from suitable fractals, across their possible connect years and the line of catrens, about need think.
18. AS TO CALCULATE YEARS OF NOSTR IN RINGS OF EVCLID , adds
Adds for CTR and algorithm of Evclid. MCD different. What to do ?
I. For algorithm of Evclid in rings. For LDE (linear Diophantine equation) of view a=bx+r can been , what а/в no haven"t common MCD (Chinese theorem on remains) or he lost in rings . For this introduced the concept of main ideal, that is multiple а={all dividers of а}, if b⊂а,then "b" also came in multiple of main ideal. This includes n×b ; (b-c), where b∈a and c∈a . It simple . But if "a" and "b" no haven"t common MCD, then (а-b)∈I by this a∈I and b∈I , where I - ideal in ring R . For (a+b) the same thing. For remains, I took from Vilenkin , a∈I ⋀ r∈R→ r×a∈I , ⋀ - cursor of logical konukcu , cursor → - a sign that shows of multiple , R - ring.
b) If MCD (a,b)=1 , then we could used formula of arithmetic progression on formula Dirichlet. I very to doubt, what always in arrays have MCD=1, for example MCD(67, 68)=1 , because numbers very much. Another ring of main ideals is MCD-ring. Another ring of main ideals is factorial. MCD(a,b)=1
a, a+b....a+(n-1)b,+... - only simple numbers
It no our the variable.
II. For the sort on CTR .
Is 2 compares : а≡b(modc) , a≡y(modz). For another Z 2 mutual simple numbers MCD(c, z)=1 the system of equations is equivalent to modc×z , b+с×t≡ y(modz), then c×t≡y-b(modz) , resolve x≡x1 modcz. It simple the variable of CTR. I have very undoubtedly, what always in rings of Evclid meets 2 mutual simple numbers , because in arrays we sow very much of numbers. Remains DIFFERENT by TWO DIFFERENT modules, which MUTUAL SIMPLE .
III. Compare.
а) Compare can been on module (mod I - main ideal ) by equal remains. For example, by COMMON module 32 need got COMMON remains , by this a≡b(modm).
a=mq1+r
b=mq2+r
(a-b)=m(q1-q2), (a-b)=m×t
b) Compare for different modules. Modules DIFFERENT and NO MUTUAL SIMPLE .
a=m1 q1+r
b=m2 q2+r2
a≡b(modm) and a≡b(modn), with m⋮n, then (a-b)/n on
a≡b(modn). For example m=16 , n=4 it property of transitivity.
In this case uses all properties of equivalence: add (a+b) , minus (a-b) and another for equivalence .
IV. Module m no divides n , but they have common multiplier. For example : MCD(9,15)=3
Modules DIFFERENT and NO MUTUAL SIMPLE .
a=m1 q1+r
b=m2 q2+r2
For example: х=4(mod15), х=2(mod15) 4-2=2 2 no divides on 3 , so system of equations no haven"t solution.
Only to calculate years it rings of Evclid, if years and date it system of compares.
Next need whould think as connect years and cipher (digital the patr) , also to finish formula of the sort catrens (letter the part). Our the code finally found the point of shift "L" . Two the code is assavoir mon for 58 estetica. It"s already another area of mathematics , it combinatorics.I did only beginning of code part. I don"t know as it to calculated. Need to search . For algorithm of Evclid (all arrays) or of the calculation of years you need search specialist in the theory of numbers to make the program and to calculate the arrays and pick them digits, for manually of the calculation time do not need to spend. In our time this algorithm well known and
are not complex.
I think, what you also understood my the calculation .
19. Utilities for the code
I. Years connect with catrens across binomial coefficients.
binomial coefficients Length the line of binomial coefficients different with the line of objects only on one.
Lines of money .
2,3,4,5,7,8,9,17,53,79,101 - sum 288
Array: 39
2,3,4,5,7,8,9,16,17,25,53,79,125- sum 353
Array : 47
2,3,4,5,7,8,9,16,17,25, 79,125- sum 300
Array: 43
Assavoir mon:
2,3,4,5,7, 9,11,13,16,25,49,61 - 205
Array: 38
It binomial coefficients in different variables his of the calculation.
First the variable with the calculation "ff, gg" or all pairs of letters .
All numbers 0f catrens.
Сm/n=n!/m!×(n-m- 1)! for pairs and Сm/n=n!/(n-m)! for another
Take combinations for each of multitude .
(h,f,g,l,N,l,g,g,f,f,l), ONLY for 11-ff=9, 11-gg=9, TWO PAIRS
С9/11=11!/(9!×(11-9-1)!=110
110×2=220
С10/11=11!/10!=11
11×7=77
(T,y,y,y) , yyy=3
С1/4=4!/(1!×(4-1-1)!=12
С3/4=4!/3!=4
(F,T)
C1/2=2
2×2=4
Sum for (F,T) and (T,y,y,y):
( 12+4)+4 =20
(h,A,T,h,g,f,z,h,A,v,g,v,b,g,y,f,y,v,y,T,h,v)
С=22×22=484 -for multitude from 22 letters
С21/22=22!/21!×0!=22
(T), (V),(l), (f), (L),(T), (z)
Sum 46×7=322
All sum : 16+4+110×2+484+322+77=1123
Sum: 958+154+11=1123 , Galen - no ?.
In multitude for 11 have 5 letters:
h,g, f,l,N
or
1, 3,3,3,1
In multitude with 22 letters have 9 letters.
A,T,h,g,f,z,y,v,b
All have 9 letters in multitude for 22 letters , they .
2,2,4,3,2,1,3,4.1
2)Two variamle . Сm/n=P/m!=n!/m!(n-m)! for common numbers of letters , yyy, AA, TT, hhhh... . P=n!
For multitude (Tyyy) 24+4=28 .
C1/4=4!/1!=24
C3/4=4!/3!=4
(F,T)
C1/2=2
2×2=4
For multitude from 11 letters ABC : (h,f,g,l,N,l,g,g,f,f,l)
С4/5 =5!/4!=5
5×2=10
С2/5=5!/2!=60
60×3=180
Sum :190
For multitude 22 from 9 letters ABC , (h,A,T,h,g,f,z,h,A,v,g,v,b,g,y,f,y,v,y,T,h,v)
С5/9=9!/5!4!=126
126×2=252
С6/9=9!/6!×3!=84
84×2=168
С7/9=9!/7!×2!=36
36×3=108
С8/9=9!/8! ×1!=9
9×2=18
Sum : 168+252+18+108+536
7 letters of multitudes from 11 . It"s T,V,l, f,L ,T, z - 5 letters of ABC
Or
2,1,2,1,1,
С2/5=5!/2!=60
60×3=180
С3/5=5!/3!=20
20×3=60
Sum :
180+60=240
All sum: 536+190+240+38+4=966+32=998
We have two of calculations from one the line for 46 letters or 11 letters of ABC.
I hope ,that you very attentively read my calculations and all understand . I find rightly the decision whole with mini mistakes and more we no get lost.
III. Cipher for estetica.
Assavoir mon . For 10 letters+space bar has 11 numbers.
ABC Nn L/l a s s a v o i ? r - m о ? n
2 3 4 5 7 9 11 13 16 25 49 61
12 1 N 61
13 2 o 9
17 3 s 3
18 4 t 11?
16 5 r 13
1 6 a 2
4 7 d ?
1 8 a 5
11 9 m 25
20 or 19 10 v 7
17 11 s 4
"i" go in "t" , "o" go in "d", i=t, 0=d
In ABC j, k, w no haven"t :
a b c d e f g h i l m n o p q r s t
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
u v x y z
19 20 21 22 23
Or U=V:
v x y z
19 20 21 22
In this the cipher no haven"t selections with repetitions and without , but has space bar .
21. ENCRYPTING THE PART OF CODE NOSTRADAMUS, common
Sum from calculation for years : CTR and progression also will no suited, because have single solution , our MCD more 1. Need use compares and rings of Evclid.
Also we know , what key and dates on multipliers no decompose.
It finishing the part of code , which concerns cipher of Nostr.
As to connect years of arrays Evclid with cipher of Nostr (digital of the part )? To connect whole years with "empty" can different variables. Need to connect theory of numbers with kombinatorika.
To connect need years with years , bur no letters with years . In simple view it can be a sort of threes on the line catrens by the calculation of dates if on their place no been numbers of Gorapollon . "Empty" years and letters of cipher connected , but whole years no connected with letters, to add or to multiple . As it to done ?
For example , years calculated , a the line of catrens stayed intact , first the variable binding across dates no used , then need would think , what has in stock. We has different variables. How else can bind полные годы к "empty" , so connect with letters ? To common combinatorial the calculation I no return , this the calculation on frames with repetition для ff,ggg, ll and... , and the calculation on frames without repetition for l, g, N, f, g and ... .
It our "empty" years , digital the part cipher you of the line catrens . Every number matches letter. Until letters no need us , the part formula of sort on letters catrens I did . In this the fail I try decide question accession letters across to arrays of Evclid. "Empty" years : 2,3,4,5,9,16,19,23, 27,31,47,49,61 - sum 296, 13 multipliers
In whole view : 2,36,38,45,46,47,48,49,54,61,62 - sum 488, 11 numbers
It numbers need bind to arrays of Evclid with help unknown until the calculation .
Also there are numbers can mean numbers of catrens , but can and no mean. In another case need bind of catrens to arrays , this problem no go out. It the line is Janus-faced.
(the line of catrens from 0+the line АВС (with L- beginning )≡ ?mod(22 or 23)
or
(letter of the line ABC +L)≡?(mod11) - ?????
Number before module need would calculate.
No finished formula gave ORDER OF POSITION letters catrens .
Another variables. With the side of arrays years can help formula of Euler , suitable fractals (it"s unlikely ), binomial coefficients (only if uses instead threes of Pifagor ), can the line of objects . With the side of cipher the part has common combinatorial the calculation , the line of catrens , also letters and the line of identification .
Also we have the line of identification or latitudes : 2,3,4,5,7,9,13,16,25,37,41 - sum 162, 11 multipliers .
The line of identification need used in cipher .
Also you table of "triangular" numbers , which snatched from triangle of Pascal with Romanic ABC , can she no deprived of sense.
Common combinatorial calculate gave us NUMBERS OF CATRENS 966, 154, 11 and 58 for assavoir mon, and take into account frames in view NUMBERS OF LETTERS in cipher on frames and without their , he not tied to array , but he can unite years and catrens .
In sum we have ordered between themselves years and calculated their number , and aparted in order catrens without account their number ( because combinatorial the part no tied until to modules). So catrens and years "no see " other other, a sorted catrens "no know " , how much their . Need do help them meets. Common combanotorial the calculation I no launched (it reshuffles). The solution until has been no found, because his nobody no searched .
Suddenly I found new adds array for cipher :
The line of latitudes: 2,3,4,5,7,9,13,16,25,37,41 - sum 162, 11 numbers . In astrology and astronomy 1min.=60№, so all sum 162№=2,7 minutes .
Sum: with 0 : 18+12+4+6+12=52 , without : 18+10+4+5+11=48, with reduction with 0: 18+2+4+2+4+10=40, with reduction without 0: 18+2+4+2+4+10=40
It the variable no need.
Rightly it? I think, what yes. It array play very important role in the sort years and catrens.
So array of identification =46 play very important role in the sort years and catrens.
Until years , letters of cipher and numbers of catrens no met.
So algorithm of Evclid participates in the calculations of years and also in encryption of the code . Degrees in astrology need converted in minutes, hours and days. The line of identification is common the line and perfect to all lines . Without of latitude impossible to calculate As in a horoscope.
The line of catrens no haven"t of array . With help of the line of identification and new already array of years (or old ?) can build system of compares equations ), remains connect their with the line of catrens (digital the part).
For example for one array of years and array of the line identification:
{█(а1≡r1(modb1)@а2≡r2(modb2))┤ old or new array of years
Got number go in formula of the sort on catrens, in left the part.
For assavoir mon the sort another.
Also need search , need would common combinatorial the calculation converted in specific with help already no combinations, but transpositions. Encrypting the part is no simple for XVI century. Can have another variables of the sort.
Literature:
Vilenkin N.Y. "Algebra and theory of numbers", М. " Education ", 1984
Ore "Invitationin theory of numbers", translate with of English, M., 2003
Buhshtab A.A. "Theory of numbers", M, "Education", 1966
O.Zarissky, P. Samuel "Commutative algebra", M., translate with of English ,1966
5. К. Aierland, М.Rouzen " Classic introduction in modern theory of numbers ", translate with English,М. "Peace", 1987
6. I.A. Climishin "Calendar and hrohologie", Мoskow , Science ,1985
7. Magazine "World of mathematic", т.38, " Calendars, a measure of length and mathamatic", Мoskow, edition "De Agostini",2014.
8. Sait of history : http://www.istorya.ru/hronos.php, also hronologie on Europe in maps, http://www.istorya.ru/hronos/chronmap.php
9. Site of library Moshkov www.lib.ru, "Samizdat" Прохорова Наталья Григорьевна, my page about Nostradamus, poetry and another : http://samlib.ru/editors/p/prohorowa_n_g/