cp's OEIS Frontend

A386703 The residue of p(n) modulo q(n) in the interval (-q(n)/2, q(n)/2], where p(n) = A000041(n) and q(n) = A000009(n).

%I A386703 #21 Jul 31 2025 01:09:52
%S A386703 0,0,1,1,1,-1,0,-2,-2,2,-4,2,-7,3,-13,7,-7,17,4,-13,32,23,7,-11,-30,
%T A386703 -39,-62,-56,-43,-20,42,159,-161,22,258,-59,357,95,-239,-504,483,412,
%U A386703 471,719,-978,-426,434,-1137,533,-622,-1780,2087,2251,-2669,-1562,831,-3372,1772
%N A386703 The residue of p(n) modulo q(n) in the interval (-q(n)/2, q(n)/2], where p(n) = A000041(n) and q(n) = A000009(n).
%C A386703 Conjecture: |a(n)| > 1 for all n > 7.
%C A386703 This has been verified for all n = 8..10^5.
%C A386703 Verified for all n <= 2000000. - _Vaclav Kotesovec_, Jul 30 2025
%H A386703 Zhi-Wei Sun, <a href="/A386703/b386703.txt">Table of n, a(n) for n = 1..10000</a>
%H A386703 Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/498447">A conjecture involving the partition function and the strict partition function</a>, Question 498447 at MathOverflow, July 30, 2025.
%e A386703 a(6) = -1  since p(6) = 11 is congruent to -1 modulo q(6) = 4.
%e A386703 a(7) = 0 since p(7) = 15 is congruent to 0 modulo q(7) = 5.
%t A386703 rMod[m_,n_]:=Mod[Numerator[m]*PowerMod[Denominator[m],-1,n],n,(1-n)/2];
%t A386703 a[n_]=rMod[PartitionsP[n],PartitionsQ[n]];Table[a[n],{n,1,70}]
%Y A386703 Cf. A000009, A000041.
%K A386703 sign
%O A386703 1,8
%A A386703 _Zhi-Wei Sun_, Jul 30 2025