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.

1, -1, -2, -2, -4, -2, -2, 3, 7, 13, -6, 3, 19, 6, -12, 19, 2, 19, 21, -12, -11, -25, 10, -27, 18, 12, 23, -27, -13, -46, -16, -35, 5, -61, -17, 8, -29, -65, -44, -30, 12, -40, 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

Offset: 1

Author

Zhi-Wei Sun, Jul 25 2014

Keywords

• sign

Comments

Conjecture: a(n) is always nonzero, i.e., prime(n) never divides the partition number p(n).

This conjecture does not hold with the smallest counterexample being n=1119414 (cf. A245662). - Max Alekseyev, Jul 27 2014

Examples

			a(20) = -12 since p(20) = 627 == -12 (mod prime(20)=71).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A000041, A094252, A245526.

Programs

Mathematica

rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
a[n_]:=rMod[PartitionsP[n],Prime[n]]
Table[a[n],{n,1,80}]

Formula

a(n) = A094252(n) or A094252(n)-A000040(n), depending on whether A094252(n) <= A000040(n)/2.