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Number of ways to write n = p + p(k) + p(m) with 0 < k <= m, where p is a prime and p(.) is the partition function (A000041).
2

%I #38 Aug 05 2019 09:26:45

%S 0,0,0,1,2,3,4,4,6,5,7,5,7,5,10,6,10,5,12,7,13,5,13,6,15,6,15,6,15,6,

%T 13,7,15,8,17,10,14,8,14,11,12,9,13,11,14,14,16,13,16,14,15,12,12,14,

%U 16,14,13,10,14,16,15,14,18,17,15,17,14,14,15,16,14,13,15,19,18,18,16,15,13,17,18,14,19,17,19,18,18,15,21,17,22,13,17,14,20,15,19,13,15,15

%N Number of ways to write n = p + p(k) + p(m) with 0 < k <= m, where p is a prime and p(.) is the partition function (A000041).

%C Conjecture: (i) a(n) > 0 for all n > 3.

%C (ii) For any integer n > 2, |n - p(k)| is prime for some k = 1,...,n. Also, for any positive integer n not equal to 7, n + p(k) is prime for some k = 1,...,n.

%C We have verified part (i) of the conjecture for all n = 4, 5, ..., 2*10^7.

%H Zhi-Wei Sun, <a href="/A202650/b202650.txt">Table of n, a(n) for n = 1..3000</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(6) = 3 since 6 = 3 + p(1) + p(2) = 2 + p(1) + p(3) = 2 + p(2) + p(2) with 2 and 3 prime.

%t PQ[n_]:=n>1&&PrimeQ[n]

%t a[n_]:=Sum[If[PQ[n-PartitionsP[m]-PartitionsP[k]],1,0],{m,1,n},{k,1,m}]

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

%Y Cf. A000040, A000041, A232398.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Nov 24 2013