# [seqfan] Re: Discussion of A14960.

Max Alekseyev maxale at gmail.com
Mon May 24 17:54:10 CEST 2010

```Using explicit formula for A014942, we have:

If n divides A014942(n) then n divides 1+24^n*(23*n-1) and hence 24^n - 1;

Vice versa, for n that divides 24^n - 1, let n=23^k*m where
gcd(m,23)=1. Clearly, m divides A014942(n).
On the other hand, we have 24^n == 1 + 23^(k+1)*m (mod 23^(k+2)) and thus
A014942(n) == (1 + (1 + 23^(k+1)*m)*(23^(k+1)*m - 1))/23^2 == 0 (mod 23^k).
That is, n divides A014942(n).

Therefore, Alex Adamchuk's observation is correct.

Max

On Sat, May 22, 2010 at 12:52 PM, 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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