# [seqfan] Re: Discussion of A14960.

Olivier Gerard olivier.gerard at gmail.com
Sat May 22 22:13:51 CEST 2010

```Dear Neil and seqfans,

Thanks to Alexander for his interest in this sequence.

I do not know how to demonstrate that  A014960 is equivalent to n divides
24^n - 1, this was
one of the purpose of the sequence in the first place, along with the links
to modular functions.
I should have added that as a comment back then.  One point is that by using
modular arithmetic
you can compute terms n dividing  24^n - 1 quickly and efficiently and they
agree with the sequence
as far as I can compute with my current equipment today.

Perhaps it is evident. If Alexander has a nice proof, I would be happy to
see it.

Baruchel conjecture 1 is true if  the equivalence with 24^n -1  is true.

To me Baruchel conjecture 2 does not seem solid anyway.

Olivier

On Sat, May 22, 2010 at 18:52, N. J. A. Sloane <njas at research.att.com>wrote:

> The present entry is:
>
> %I A014960
> %S A014960
> 1,23,529,1081,12167,24863,50807,279841,571849,1168561,2387929,2870377
> %N A014960 Numbers n such that n divides s(n), where s(1)=1,
> s(k)=s(k-1)+k*24^(k-1) (A014942).
> %C A014960 Initial terms are 23^n, 23^(n-1)*47,
> 23^(n-2)*47^2,...23*47^(n-1),23^(n+1), etc. with som\
> etime a little "noise" between terms (eg.: for a(12)=23*124799 between
> a(11)=23*47^3 and maybe a(13)\
> =23^5). Maybe another sequence is interlaced, which would involve 23^n,
> 23^(n-1)*124799, etc., in wh\
> ich case an infinity of products of powers of 23 and powers of another
> prime factor may occur in the\
>  sequence. Conjecture: Next term, a(13), very probably is 23^5. Conjecture:
> All numbers in the seque\
> nce are multiple of 23. Conjecture: All numbers in the sequence have at
> most two different prime fac\
> tors. - Thomas Baruchel (baruchel(AT)users.sourceforge.net), Oct 10 2003
> %t A014960 s = 1; Do[ If[ Mod[ s, n ] == 0, Print[n]]; s = s + (n +
> 1)*24^n, {n, 1, 100000}]
> %Y A014960 Cf. A014942.
> %Y A014960 Sequence in context: A171328 A097778 A057193 this_sequence
> A171297 A009967 A147642
> %Y A014960 Adjacent sequences: A014957 A014958 A014959 this_sequence
> A014961 A014962 A014963
> %K A014960 nonn
> %O A014960 1,2
> %A A014960 Olivier Gerard (olivier.gerard(AT)gmail.com)
> %E A014960 More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 13
> 2000
> %E A014960 Four more terms from Thomas Baruchel (baruchel(AT)
> users.sourceforge.net), Oct 10 2003
>
> Alex Adamchuk has proposed some major edits to this, as follows:
>
> %S A014960
> 1,23,529,1081,12167,24863,50807,279841,571849,1168561,2387929,2870377,
> %T A014960
> 6436343,7009273,13152527,15954479,26876903,54922367,66018671,112232663,
> %U A014960
> 134907719,148035889,161213279,302508121,329435831,366953017,537539141
> %E A014960 a(13)-a(32) from Alexander Adamchuk (alex(AT)kolmogorov.com),
> May 16 2010
> %C A014960 Contribution from Alexander Adamchuk (alex(AT)kolmogorov.com),
> May 16 2010: (Start)
> %C A014960 Better definition: Numbers n such that n divides 24^n - 1.
> %C A014960 First two contrexamples for conjecture by Thomas Baruchel "...at
> most two different prime factors": 15954479 = 23*47*14759, 134907719 =
> 23*47*124799.
> %C A014960 Prime factors of a (n) in the order of their appearance are {23,
> 47, 124799, 304751, 14759, 497261, 49727, ...} = A087807. (End)
> %Y A014960 Contribution from Alexander Adamchuk (alex(AT)kolmogorov.com),
> May 16 2010: (Start)
> %Y A014960 Cf. A087807 = Prime factors of terms occurring in A014960.
> %Y A014960 Cf. A128356 = Least number k>1 (that is not the power of prime
> p) such that k divides (p+1)^k-1, where p = Prime[n]. (End)
>
> I don't have time to study this.  Are these changes correct?   I like to
> see proofs
> before I make such drastic changes.
>
> Thanks,  Neil
>
>
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>
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>

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