login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) = |{0 < k < n: p(k)^2 + q(n-k)^2 is prime}|, where p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).
15

%I #19 Aug 05 2019 09:26:21

%S 0,1,2,2,1,1,4,2,3,2,2,4,4,3,2,2,5,3,1,5,3,5,6,3,3,2,2,1,1,2,2,5,3,4,

%T 3,5,3,1,6,4,7,10,3,5,4,2,4,5,3,4,2,3,7,9,5,6,8,2,5,3,3,5,4,3,5,4,6,7,

%U 6,3,2,9,8,6,1,6,7,7,6,2,5,8,4,6,2,6,4,8,7,3,5,3,3,5,4,5,8,5,6,2

%N a(n) = |{0 < k < n: p(k)^2 + q(n-k)^2 is prime}|, where p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).

%C Conjecture: (i) a(n) > 0 for all n > 1. Also, for any integer n > 4, p(k)*q(n-k) - 1 is prime for some 0 < k < n/2.

%C (ii) If n > 9, then prime(k)*p(n-k) + 1 is prime for some 0 < k < n. If n > 2, then prime(k)*q(n-k) - 1 is prime for some 0 < k < n, and also prime(k)*q(n-k) + 1 is prime for some 0 < k < n.

%C (iii) If n > 11, then prime(k) + p(n-k) is prime for some 0 < k < n. If n > 4, then prime(k) + q(n-k) is prime for some 0 < k < n, and also prime(k)^2 + q(n-k)^2 is prime for some 0 < k < n.

%H Zhi-Wei Sun, <a href="/A233307/b233307.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1312.1166">On a^n+ bn modulo m</a>, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.

%e a(5) = 1 since 5 = 1 + 4 with p(1)^2 + q(4)^2 = 1^2 + 2^2 = 5 prime.

%e a(6) = 1 since 6 = 3 + 3 with p(3)^2 + q(3)^2 = 3^2 + 2^2 = 13 prime.

%e a(19) = 1 since 19 = 3 + 16 with p(3)^2 + q(16)^2 = 3^2 + 32^2 = 1033 prime.

%e a(28) = 1 since 28 = 3 + 25 with p(3)^2 + q(25)^2 = 3^2 + 142^2 = 20173 prime.

%e a(29) = 1 since 29 = 6 + 23 with p(6)^2 + q(23)^2 = 11^2 + 104^2 = 10937 prime.

%e a(38) = 1 since 38 = 1 + 37 with p(1)^2 + q(37)^2 = 1^2 + 760^2 = 577601 prime.

%e a(75) = 1 since 75 = 13 + 62 with p(13)^2 + q(62)^2 = 101^2 + 13394^2 = 179409437 prime.

%e a(160) = 1 since 160 = 48 + 112 with p(48)^2 + q(112)^2 = 147273^2 + 1177438^2 = 1408049580373 prime.

%e a(210) = 1 since 210 = 71 + 139 with p(71)^2 + q(139)^2 = 4697205^2 + 8953856^2 = 102235272080761 prime.

%t a[n_]:=Sum[If[PrimeQ[PartitionsP[k]^2+PartitionsQ[n-k]^2],1,0],{k,1,n-1}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000009, A000040, A000041, A231201, A232504.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Dec 07 2013