A236483 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.

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, 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, 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

Offset

1,6

Comments

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

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

(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.

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.

Links

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

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Example

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

Mathematica

a[n_]:=Sum[If[PrimeQ[PartitionsP[k]+2^(EulerPhi[n-k]/2)], 1, 0], {k, 1, n-3}]
Table[a[n], {n, 1, 100}]

Crossrefs

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

Sequence in context: A052369 A110976 A318271 * A266714 A151570 A341520

Adjacent sequences: A236480 A236481 A236482 * A236484 A236485 A236486

Keyword

nonn

Author

Zhi-Wei Sun, Jan 27 2014

Status

approved