A236968 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1 and k + phi(m) are all prime, where phi(.) is Euler's totient function.

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

Offset: 1

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Also, any n > 12 can be written as k + m (k > 0 and m > 2) with 6*k - 1, 6*k + 1 and k + phi(m)/2 all prime.

(ii) Each integer n > 34 can be written as p + q (q > 0) with p and p + phi(q) both prime. Also, any integer n > 14 can be written as p + q (q > 2) with p, p + 6 and p + phi(q)/2 all prime.

Clearly, part (i) of the conjecture implies that any integer n > 1 can be written as p + m - phi(m), where p is a prime and m is a positive integer.

Examples

			a(17) = 1 since 17 = 7 + 10 with 6*7 - 1 = 41, 6*7 + 1 = 43 and 7 + phi(10) = 7 + 4 = 11 all prime.
a(486) = 1 since 486 = 325 + 161 with 6*325 - 1 = 1949, 6*325 + 1 = 1951 and 325 + phi(161) = 325 + 132 = 457 all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A001359, A002822, A006512, A182662, A199920, A236531, A236831.

Programs

Mathematica

p[n_,k_]:=PrimeQ[6k-1]&&PrimeQ[6k+1]&&PrimeQ[k+EulerPhi[n-k]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]