A233204 Number of ways to write n = k + m with 0 < k < m such that 2^k * prime(m) + 3 is prime.

0, 0, 0, 1, 2, 1, 3, 1, 4, 3, 2, 2, 2, 4, 3, 2, 6, 3, 2, 1, 8, 1, 2, 2, 4, 7, 2, 5, 6, 8, 5, 4, 4, 8, 3, 5, 2, 7, 5, 8, 5, 3, 4, 4, 4, 8, 6, 2, 4, 3, 7, 7, 3, 4, 7, 5, 3, 4, 6, 8, 4, 2, 6, 6, 4, 7, 7, 5, 7, 7, 6, 6, 2, 7, 8, 7, 7, 5, 11, 3, 4, 8, 2, 7, 8, 6, 9, 7, 6, 10, 11, 4, 5, 8, 4, 8, 8, 6, 7, 9

Offset: 1

Author

Zhi-Wei Sun, Dec 05 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 3. Also, any integer n > 2 can be written as k + m (0 < k <= m) with 2^k*prime(m) - 3 prime.

(ii) Any integer n > 6 can be written as k + m (0 < k < m) with prime(k) + 6 and prime(m) + 6 both prime. Each integer n > 4 can be written as k + m (0 < k < m) with prime(k) + 2 and prime(m) + 6 both prime. Also, for every integer n > 3 not among 11, 21, 32, 49, 171, there is a positive integer k < n with prime(k) + 2 and prime(n-k) + 2 both prime.

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..8000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000040, A000079, A231201, A231561, A233150, A233183.

Programs

Maple

a(6) = 1 since 6 = 2 + 4 with 2^2*prime(4) + 3 = 4*7 + 3 = 31 prime.
a(22) = 1 since 22 = 1 + 21 with 2^1*prime(21) + 3 = 2*73 + 3 = 149 prime.

Mathematica

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