cp's OEIS Frontend

A245525 Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that p(n) == r (mod prime(n)), where p(.) is the partition function given by A000041.

%I A245525 #22 Jul 28 2014 14:46:31
%S A245525 1,-1,-2,-2,-4,-2,-2,3,7,13,-6,3,19,6,-12,19,2,19,21,-12,-11,-25,10,
%T A245525 -27,18,12,23,-27,-13,-46,-16,-35,5,-61,-17,8,-29,-65,-44,-30,12,-40,
%U A245525 40,-95,90,88,53,93,97,-42,-47,47,2,117,-16,34,27,51,-11,108,-24,115,-29,30,-32,-90,-87,141,24,131,-166,-115,-96,-111,84,-191,163,-156,115,78
%N A245525 Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that p(n) == r (mod prime(n)), where p(.) is the partition function given by A000041.
%C A245525 Conjecture: a(n) is always nonzero, i.e., prime(n) never divides the partition number p(n).
%C A245525 This conjecture does not hold with the smallest counterexample being n=1119414 (cf. A245662). - _Max Alekseyev_, Jul 27 2014
%H A245525 Zhi-Wei Sun, <a href="/A245525/b245525.txt">Table of n, a(n) for n = 1..10000</a>
%F A245525 a(n) = A094252(n) or A094252(n)-A000040(n), depending on whether A094252(n) <= A000040(n)/2.
%e A245525 a(20) = -12 since p(20) = 627 == -12 (mod prime(20)=71).
%t A245525 rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
%t A245525 a[n_]:=rMod[PartitionsP[n],Prime[n]]
%t A245525 Table[a[n],{n,1,80}]
%Y A245525 Cf. A000040, A000041, A094252, A245526.
%K A245525 sign
%O A245525 1,3
%A A245525 _Zhi-Wei Sun_, Jul 25 2014