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%I #15 Dec 10 2013 13:31:13
%S 24,240,168,480,264,21840,24,16320,3192,2640,552,43680,24,6960,57288,
%T 32640,24,15353520,24,216480,7224,5520,1128,1485120,264,12720,3192,
%U 13920,1416,454293840,24,65280,258888,240,18744,2241613920,24,240,13272,7360320,1992
%N Largest integer m such that m divides (sigma_(2n+1)(2k-1)-sigma(2k-1)) for all k>=1.
%C a(n)==0 mod 24. It seems that a(n)==0 (mod 2n+1) if and only if 2n+1 is an odd prime.
%C It appears that a(n)=24 for n in A045979, a(n)=168 for n in A051227, a(n)=264 for n in A051229, and a(n)=240 or 480 if n is in A051225. - _Michel Marcus_, Dec 07 2013
%o (PARI) ds(n, k) = sigma(2*k-1, 2*n+1) - sigma(2*k-1);
%o a(n) = {my(m = ds(n, 1)); for (k=2, 100, m = gcd(m, ds(n, k));); m;} \\ Script computes gcd of 100 terms; for current data, 10 terms are actually sufficient; is there a better way? - _Michel Marcus_, Dec 07 2013
%Y Cf. A000203.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Apr 12 2003
%E a(12) corrected and more terms from _Michel Marcus_, Dec 07 2013
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