A236483 - OEIS

%I #8 Apr 06 2014 10:49:02

%S 0,0,0,1,1,2,3,2,3,5,4,3,6,1,6,5,7,7,5,6,4,6,5,4,7,4,6,5,5,6,9,6,13,6,

%T 10,5,6,7,4,10,9,12,6,12,3,8,8,9,11,8,11,7,11,7,8,8,9,8,10,10,9,9,14,

%U 8,15,8,11,13,12,12,15,13,12,8,11,12,10,11,12,12,13,8,12,14,8,13,10,8,8,12,8,15,11,10,11,11,13,14,10,8

%N a(n) = |{0 < k < n-2: p(k) + 2^(phi(n-k)/2) is prime}|, where p(.) is the partition function (A000041) and phi(.) is Euler's totient function.

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

%C (ii) For any integer n > 1, there is a positive integer k < n such that p(k)*2^(phi(n-k)) + 1 is prime.

%C (iii) For any integer n > 1 not among 6, 23, 42, there is a positive integer k < n such that 2*p(k) + p(n-k) is prime.

%C Clearly, part (i) of the conjecture implies that there are infinitely many primes of the form p(k) + 2^m, where k and m are positive integers.

%H Zhi-Wei Sun, <a href="/A236483/b236483.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014

%e a(14) = 1 since p(7) + 2^(phi(7)/2) = 15 + 2^3 = 23 is prime.

%t a[n_]:=Sum[If[PrimeQ[PartitionsP[k]+2^(EulerPhi[n-k]/2)],1,0],{k,1,n-3}]

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

%Y Cf. A000010, A000040, A000041, A000079, A234337.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Jan 27 2014