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Revision History for A034884

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Showing entries 1-10 | older changes
Numbers k such that k < d(k)^2, where d(k) = A000005(k).
(history; published version)
#38 by Sean A. Irvine at Mon May 08 23:32:46 EDT 2023
STATUS

proposed

approved

#37 by Jon E. Schoenfield at Sat May 06 10:44:51 EDT 2023
STATUS

editing

proposed

Discussion
Sat May 06
20:22
Kevin Ryde: A276734 is where floor(sqrt(k)) arises.  That the better place?  (If these numbers are not interesting enough to make a sequence.)
20:40
Bill McEachen: @Kevin As you wish. If change is reverted I can relocate it to there.
Sun May 07
04:15
Kevin Ryde: Or if this sequence is floor(sqrt(k)) < numdiv(k) then say that (if that's right).
09:00
Bill McEachen: @Kevin yeah, I don't see the particular use in that
#36 by Jon E. Schoenfield at Sat May 06 10:44:04 EDT 2023
COMMENTS

The union of this sequence and A276734 has 74 total terms which are all k with floor(sqrt(k)) <= d(k). - Bill McEachen, Apr 07 2023

STATUS

proposed

editing

#35 by Bill McEachen at Sat May 06 07:34:25 EDT 2023
STATUS

editing

proposed

#34 by Bill McEachen at Fri May 05 22:27:46 EDT 2023
COMMENTS

The union of this sequence and A276734 has 74 total terms which are all k with floor(sqrt(k ))<= d(k)^2. - Bill McEachen, Apr 07 2023

STATUS

proposed

editing

Discussion
Fri May 05
22:28
Bill McEachen: @Kevin thanks.
#33 by Kevin Ryde at Tue Apr 11 19:54:46 EDT 2023
STATUS

editing

proposed

Discussion
Tue Apr 11
20:12
Bill McEachen: Yes, the 2 taken together. Didn't feel any new sequence was deserved, thus the comment
Fri May 05
14:20
Amiram Eldar: Can you please check: I see that 5 is in A276734, but d(5)^2 = 2^2 = 4 < 5, so k =5 is not a solution to k <= d(k)^2. In fact by changing k < d(k)^2 to k <= d(k)^2 I get only 2 more terms, 1 and 9.
15:00
Bill McEachen: @Amiram here's the script I used:
genit(nterms=75,maxx=17000)={my(arr=List(),comp=List(),dbg=0,cnt=0,combo=List());listput(combo,1);listput(combo,2);listput(combo,3);listput(combo,5);listput(combo,7);forcomposite(x=4,maxx,if(#combo>=nterms,break);q=1.0* #divisors(x)/sqrtint(x);listput(comp,x);listput(arr,q);if(q>=1,listput(combo,x));if(dbg>0,print(x,"  ",q));cnt+=1);print("see SET of list=combo  ",#combo);Set(combo)}
15:18
Bill McEachen: OR this script just for the A276734 part, yields 22 terms:
genit2(nterms=75,maxx=17000)={my(arr=List(),comp=List(),dbg=0,cnt=0,combo=List());forcomposite(x=4,maxx,if(#combo>=nterms,break);q=1.0* #divisors(x)/sqrtint(x);listput(comp,x);listput(arr,q);if(q==1,listput(combo,x));if(dbg>0,print(x,"  ",q));cnt+=1);x=1;q=1.0* #divisors(x)/sqrtint(x);listput(comp,x);listput(arr,q);if(q==1,listput(combo,x));x=5;q=1.0* #divisors(x)/sqrtint(x);listput(comp,x);listput(arr,q);if(q==1,listput(combo,x));x=7;q=1.0* #divisors(x)/sqrtint(x);listput(comp,x);listput(arr,q);if(q==1,listput(combo,x));print("see SET of list=combo  ",#combo);Set(combo)}

Maybe I missed something in my characterization (?)
22:20
Kevin Ryde: Ahh, "floor(sqrt(k)) <= d(k)" is not the same as "k <= d(k)^2".  Presume the former you started is intention.
22:23
Kevin Ryde: (If the numbers of this condition are interesting then a sequence would ensure they're found by a search.)
22:25
Bill McEachen: @Kevin I have no interest in any new sequence, I will revert the comment wording as it was originally intended.
#32 by Kevin Ryde at Tue Apr 11 19:52:45 EDT 2023
COMMENTS

This The union of this sequence along with and A276734 has 74 total terms, representing where which are all k with k <= d(k)^2. - Bill McEachen, Apr 07 2023

STATUS

proposed

editing

Discussion
Tue Apr 11
19:54
Kevin Ryde: Massaged the wording, as it read to me like "together with" meant both this sequence and A276734 had 74 terms, or something like that.  But if k <= d(k)^2 is interesting then could be very inclined to have it as its own sequence instead of a comment.
#31 by Bill McEachen at Tue Apr 11 18:12:41 EDT 2023
STATUS

editing

proposed

#30 by Bill McEachen at Tue Apr 11 18:11:56 EDT 2023
COMMENTS

This sequence along with A276734 has 74 total terms, representing where d(k) / floor(sqrt <= d(k)) >= 1^2. - Bill McEachen, Apr 07 2023

STATUS

proposed

editing

Discussion
Tue Apr 11
18:12
Bill McEachen: @Michel ok, done, just the type of stuff I don't think about
#29 by Wesley Ivan Hurt at Tue Apr 11 13:58:25 EDT 2023
STATUS

editing

proposed

Discussion
Tue Apr 11
17:33
Michel Marcus: maybe use k <= d(k)^2  (to be like name here)  ?