A278448 -id:A278448

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A278808

Decimal expansion of b(1) in the sequence b(n+1) = c^(b(n)/n) A278448, where c=2 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

+20
7

2, 8, 7, 1, 8, 8, 0, 8, 2, 7, 0, 4, 5, 4, 5, 4, 6, 5, 8, 8, 9, 0, 5, 5, 1, 7, 5, 5, 0, 4, 5, 7, 5, 0, 4, 5, 8, 6, 5, 6, 5, 2, 5, 1, 1, 8, 4, 7, 9, 6, 5, 6, 5, 6, 7, 9, 2, 9, 9, 5, 4, 0, 1, 0, 8, 4, 0, 4, 5, 7, 9, 6, 8, 3, 0, 8, 9, 2, 7, 0, 3, 6, 0, 1, 8, 2, 8, 6, 3, 8, 1, 8, 6, 7, 6, 7, 8, 7, 5, 4, 8, 0, 8, 4, 3

(list; constant; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

For the given c there exists a unique b(1) for which the sequence b(n) does not converge to 1 and at the same time always satisfies b(n-1)b(n+1)/b(n)^2 < 1.

If b(1) were chosen smaller the sequence b(n) would approach 1, if it were chosen greater it would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

The value of b(1) is found through trial and error. Suppose one starts with b(1) = 2, the sequence b(n) would continue b(2) = 4, b(3) = 4, b(4) = 2.51..., b(5) = 1.54... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say b(1) = 3, which gives rise to b(2) = 8, b(3) = 16, b(4) = 40.31... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of b(1) is between 2 and 3.

No closed form expression is known. Probably transcendental but this is unproved. - Robert G. Wilson v, Dec 01 2016

LINKS

Rok Cestnik, Table of n, a(n) for n = 1..1000

Rok Cestnik, Plot of the dependence of b(1) on c

FORMULA

log_2(2*log_2(3*log_2(4*log_2(...)))). - Andrey Zabolotskiy, Nov 30 2016

EXAMPLE

2.87188082704545465889055175504575045865652511847965...

MATHEMATICA

c = 2;

n = 100;

acc = Round[n*1.2];

th = 1000000;

b1 = 0;

For[p = 0, p < acc, ++p,

For[d = 0, d < 9, ++d,

b1 = b1 + 1/10^p;

bn = b1;

For[i = 1, i < Round[n*1.2], ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

];

If[bn > th, {

b1 = b1 - 1/10^p;

Break[];

}];

];

N[b1, n]

RealDigits[Fold[Log2[#1*#2] &, 1, Reverse@Range[2, 144]], 10,

111][[1]] (* Robert G. Wilson v, Dec 01 2016 *)

CROSSREFS

For sequence round(b(n)) see A278448.

For different values of c see A278809, A278810, A278811, A278812.

For b(1)=0 see A278813.

KEYWORD

nonn,cons

AUTHOR

Rok Cestnik, Nov 28 2016

STATUS

approved

A278449

a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=3 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

+10
7

1, 3, 6, 9, 13, 16, 20, 24, 29, 33, 37, 42, 47, 51, 56, 61, 66, 71, 76, 81, 86, 92, 97, 102, 108, 113, 118, 124, 129, 135, 141, 146, 152, 158, 163, 169, 175, 181, 187, 193, 199, 205, 210, 216, 222, 229, 235, 241, 247, 253, 259, 265, 271, 278, 284, 290, 296, 303, 309, 315, 322, 328, 334, 341, 347, 354, 360, 367, 373, 379

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

For the given c there exists a unique b(1) for which the sequence b(n) does not converge to 1 and at the same time always satisfies b(n-1)b(n+1)/b(n)^2 < 1 (due to rounding to the nearest integer a(n-1)a(n+1)/a(n)^2 is not always less than 1).

In this case b(1) = 1.0828736095... A278809. If b(1) were chosen smaller the sequence would approach 1, if it were chosen greater the sequence would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

The value of b(1) is found through trial and error. Illustrative example for the case of c=2 (for c=3 similar): "Suppose one starts with b(1) = 2, the sequence would continue b(2) = 4, b(3) = 4, b(4) = 2.51..., b(5) = 1.54... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say b(1) = 3, which gives rise to b(2) = 8, b(3) = 16, b(4) = 40.31... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of b(1) is between 2 and 3."

b(n) = n*log_3((n+1)*log_3((n+2)*log_3(...))) ~ n*log_3(n). - Andrey Zabolotskiy, Dec 01 2016

LINKS

Rok Cestnik, Table of n, a(n) for n = 1..1000

Rok Cestnik, Plot of the dependence of b(1) on c

EXAMPLE

a(2) = round(3^1.08...) = round(3.28...) = 3.

a(3) = round(3^(3.28.../2)) = round(6.07...) = 6.

a(4) = round(3^(6.07.../3)) = round(9.26...) = 9.

MATHEMATICA

c = 3;

n = 100;

acc = Round[n*1.2];

th = 1000000;

b1 = 0;

For[p = 0, p < acc, ++p,

For[d = 0, d < 9, ++d,

b1 = b1 + 1/10^p;

bn = b1;

For[i = 1, i < Round[n*1.2], ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

];

If[bn > th, {

b1 = b1 - 1/10^p;

Break[];

}];

];

bnlist = {N[b1]};

bn = b1;

For[i = 1, i < n, ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

bnlist = Append[bnlist, N[bn]];

];

anlist = Map[Round[#] &, bnlist]

CROSSREFS

For decimal expansion of b(1) see A278809.

For different values of c see A278448, A278450, A278451, A278452.

For b(1)=0 see A278453.

KEYWORD

nonn

AUTHOR

Rok Cestnik, Nov 22 2016

STATUS

approved

A278450

a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=4 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

+10
7

0, 2, 4, 6, 9, 12, 14, 17, 21, 24, 27, 31, 34, 38, 41, 45, 49, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 101, 105, 109, 114, 118, 122, 127, 131, 135, 140, 144, 149, 153, 158, 162, 167, 172, 176, 181, 185, 190, 195, 200, 204, 209, 214, 218, 223, 228, 233, 238, 242, 247, 252, 257, 262, 267, 272, 277, 282, 287

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

In this case b(1) = 0.4970450000... A278810. If b(1) were chosen smaller the sequence would approach 1, if it were chosen greater the sequence would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

The value of b(1) is found through trial and error. Illustrative example for the case of c=2 (for c=4 similar): "Suppose one starts with b(1) = 2, the sequence would continue b(2) = 4, b(3) = 4, b(4) = 2.51..., b(5) = 1.54... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say b(1) = 3, which gives rise to b(2) = 8, b(3) = 16, b(4) = 40.31... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of b(1) is between 2 and 3."

b(n) = n*log_4((n+1)*log_4((n+2)*log_4(...))) ~ n*log_4(n). - Andrey Zabolotskiy, Dec 01 2016

LINKS

Rok Cestnik, Table of n, a(n) for n = 1..1000

Rok Cestnik, Plot of the dependence of b(1) on c

EXAMPLE

a(2) = round(4^0.49...) = round(1.99...) = 2.

a(3) = round(4^(1.99.../2)) = round(3.97...) = 4.

a(4) = round(4^(3.97.../3)) = round(6.28...) = 6.

MATHEMATICA

c = 4;

n = 100;

acc = Round[n*1.2];

th = 1000000;

b1 = 0;

For[p = 0, p < acc, ++p,

For[d = 0, d < 9, ++d,

b1 = b1 + 1/10^p;

bn = b1;

For[i = 1, i < Round[n*1.2], ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

];

If[bn > th, {

b1 = b1 - 1/10^p;

Break[];

}];

];

bnlist = {N[b1]};

bn = b1;

For[i = 1, i < n, ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

bnlist = Append[bnlist, N[bn]];

];

anlist = Map[Round[#] &, bnlist]

CROSSREFS

For decimal expansion of b(1) see A278810.

For different values of c see A278448, A278449, A278451, A278452.

For b(1)=0 see A278453.

KEYWORD

nonn

AUTHOR

Rok Cestnik, Nov 22 2016

STATUS

approved

A278451

a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=5 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

+10
7

0, 1, 3, 5, 7, 9, 11, 14, 17, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 56, 59, 62, 66, 69, 73, 76, 80, 83, 87, 90, 94, 98, 101, 105, 109, 112, 116, 120, 123, 127, 131, 135, 139, 143, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 190, 194, 198, 202, 206, 210, 214, 218, 222, 226, 231, 235, 239

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

In this case b(1) = 0.1775819188... A278811. If b(1) were chosen smaller the sequence would approach 1, if it were chosen greater the sequence would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

The value of b(1) is found through trial and error. Illustrative example for the case of c=2 (for c=5 similar): "Suppose one starts with b(1) = 2, the sequence would continue b(2) = 4, b(3) = 4, b(4) = 2.51..., b(5) = 1.54... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say b(1) = 3, which gives rise to b(2) = 8, b(3) = 16, b(4) = 40.31... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of b(1) is between 2 and 3."

b(n) = n*log_5((n+1)*log_5((n+2)*log_5(...))) ~ n*log_5(n). - Andrey Zabolotskiy, Dec 01 2016

LINKS

Rok Cestnik, Table of n, a(n) for n = 1..1000

Rok Cestnik, Plot of the dependence of b(1) on c

EXAMPLE

a(2) = round(5^0.17...) = round(1.33...) = 1.

a(3) = round(5^(1.33.../2)) = round(2.91...) = 3.

a(4) = round(5^(2.91.../3)) = round(4.78...) = 5.

MATHEMATICA

c = 5;

n = 100;

acc = Round[n*1.2];

th = 1000000;

b1 = 0;

For[p = 0, p < acc, ++p,

For[d = 0, d < 9, ++d,

b1 = b1 + 1/10^p;

bn = b1;

For[i = 1, i < Round[n*1.2], ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

];

If[bn > th, {

b1 = b1 - 1/10^p;

Break[];

}];

];

bnlist = {N[b1]};

bn = b1;

For[i = 1, i < n, ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

bnlist = Append[bnlist, N[bn]];

];

anlist = Map[Round[#] &, bnlist]

CROSSREFS

For decimal expansion of b(1) see A278811.

For different values of c see A278448, A278449, A278450, A278452.

For b(1)=0 see A278453.

KEYWORD

nonn

AUTHOR

Rok Cestnik, Nov 22 2016

STATUS

approved

A278452

a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c = e = 2.71828... and b(1) is chosen such that the sequence neither explodes nor goes to 1.

+10
7

1, 4, 7, 11, 15, 19, 23, 28, 33, 37, 42, 48, 53, 58, 64, 69, 75, 80, 86, 92, 97, 103, 109, 115, 121, 127, 133, 139, 146, 152, 158, 165, 171, 177, 184, 190, 197, 203, 210, 216, 223, 230, 236, 243, 250, 256, 263, 270, 277, 284, 290, 297, 304, 311, 318, 325, 332, 339, 346, 353, 360, 367, 375, 382, 389, 396, 403, 410, 418, 425

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

In this case b(1) = 1.3679012617... A278812. If b(1) were chosen smaller the sequence would approach 1, if it were chosen greater the sequence would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

The value of b(1) is found through trial and error. Illustrative example for the case of c=2 (for c=e similar): "Suppose one starts with b(1) = 2, the sequence would continue b(2) = 4, b(3) = 4, b(4) = 2.51..., b(5) = 1.54... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say b(1) = 3, which gives rise to b(2) = 8, b(3) = 16, b(4) = 40.31... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of b(1) is between 2 and 3."

b(n) = n*log((n+1)*log((n+2)*log(...))) ~ n*log(n). - Andrey Zabolotskiy, Dec 01 2016

LINKS

Rok Cestnik, Table of n, a(n) for n = 1..1000

Rok Cestnik, Plot of the dependence of b(1) on c

EXAMPLE

a(2) = round(e^1.36...) = round(3.92...) = 4.

a(3) = round(e^(3.92.../2)) = round(7.12...) = 7.

a(4) = round(e^(7.12.../3)) = round(10.74...) = 11.

MATHEMATICA

c = E;

n = 100;

acc = Round[n*1.2];

th = 1000000;

b1 = 0;

For[p = 0, p < acc, ++p,

For[d = 0, d < 9, ++d,

b1 = b1 + 1/10^p;

bn = b1;

For[i = 1, i < Round[n*1.2], ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

];

If[bn > th, {

b1 = b1 - 1/10^p;

Break[];

}];

];

bnlist = {N[b1]};

bn = b1;

For[i = 1, i < n, ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

bnlist = Append[bnlist, N[bn]];

];

anlist = Map[Round[#] &, bnlist]

CROSSREFS

For decimal expansion of b(1) see A278812.

For different values of c see A278448, A278449, A278450, A278451.

For b(1)=0 see A278453.

KEYWORD

nonn

AUTHOR

Rok Cestnik, Nov 22 2016

STATUS

approved

A278453

a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where b(1)=0 and c is chosen such that the sequence neither explodes nor goes to 1.

+10
7

0, 1, 2, 4, 6, 8, 10, 12, 15, 17, 19, 22, 25, 27, 30, 33, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 75, 78, 81, 84, 88, 91, 94, 98, 101, 104, 108, 111, 114, 118, 121, 125, 128, 132, 135, 139, 142, 146, 149, 153, 157, 160, 164, 167, 171, 175, 178, 182, 186, 189, 193, 197, 201, 204, 208, 212, 216

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

There exists a unique value of c for which the sequence b(n) does not converge to 1 and at the same time always satisfies b(n-1)b(n+1)/b(n)^2 < 1 (due to rounding to the nearest integer a(n-1)a(n+1)/a(n)^2 is not always less than 1).

Its value: c = 5.7581959391... A278813. If c were chosen smaller the sequence would approach 1, if it were chosen greater the sequence would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

The value of c is found through trial and error. Suppose one starts with c = 5, the sequence would continue b(2) = 1, b(3) = 2.23..., b(4) = 3.31..., b(5) = 3.80..., b(6) = 3.39..., b(7) = 2.48..., b(8) = 1.77... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say c = 6, which gives rise to b(2) = 1, b(3) = 2.44, b(4) = 4.31..., b(5) = 6.92..., b(6) = 11.94..., b(7) = 35.38... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of c is between 5 and 6.

b(n) = n*log_c((n+1)*log_c((n+2)*log_c(...))). At n=1 this gives the relation between c and b(1). It follows that b(n) ~ n*log_c(n). - Andrey Zabolotskiy, Nov 30 2016

LINKS

Rok Cestnik, Table of n, a(n) for n = 1..1000

Rok Cestnik, Plot of the dependence of b(1) on c

FORMULA

a(n) = round(n*log_c((n+1)*log_c((n+2)*log_c(...)))). - Andrey Zabolotskiy, Nov 30 2016

EXAMPLE

a(2) = round(5.75...^0) = round(1) = 1.

a(3) = round(5.75...^(1/2)) = round(2.39...) = 2.

a(4) = round(5.75...^(2.39.../3)) = round(4.05...) = 4.

MATHEMATICA

b1 = 0;

n = 100;

acc = Round[n*1.2];

th = 1000000;

c = 0;

For[p = 0, p < acc, ++p,

For[d = 0, d < 9, ++d,

c = c + 1/10^p;

bn = b1;

For[i = 1, i < Round[n*1.2], ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

];

If[bn > th, {

c = c - 1/10^p;

Break[];

}];

];

bnlist = {N[b1]};

bn = b1;

For[i = 1, i < n, ++i,

bn = N[c^(bn/i), acc];

If[bn > th, Break[]];

bnlist = Append[bnlist, N[bn]];

];

anlist = Map[Round[#] &, bnlist]

CROSSREFS

For decimal expansion of c see A278813.

For different values of b(1) see A278448, A278449, A278450, A278451, A278452.

KEYWORD

nonn

AUTHOR

Rok Cestnik, Nov 22 2016

STATUS

approved

page 1

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