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

1,3

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.

Links

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

Example

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.

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}]

Crossrefs

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

Sequence in context: A057536 A245574 A245573 * A345220 A265744 A331083

Adjacent sequences: A236965 A236966 A236967 * A236969 A236970 A236971

Keyword

nonn

Author

Zhi-Wei Sun, Feb 02 2014

Status

approved