login
A236358
a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m + 1 prime}|, where phi(.) is Euler's totient function.
4
0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 2, 1, 0, 2, 3, 1, 1, 0, 2, 1, 0, 2, 3, 3, 3, 2, 3, 2, 1, 4, 1, 4, 1, 4, 5, 3, 5, 7, 7, 8, 5, 5, 4, 4, 7, 7, 4, 7, 3, 6, 4, 5, 5, 6, 7, 6, 4, 5, 7, 6, 9, 5, 8, 7, 7, 4, 6, 5, 4, 6, 9, 8, 3, 6, 8, 9, 8, 8, 7, 8, 8, 9, 8, 4, 7, 4, 7, 7, 5, 4, 8, 6, 6, 7, 11
OFFSET
1,16
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 26.
(ii) For any integer n > 37, there is a positive integer k < n such that m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m - 1 prime.
We have verified both parts for n up to 50000. Clearly, part (i) implies that there are infinitely many positive integers m with 2*3^m + 1 prime, while part (ii) implies that there are infinitely many positive integers m with 2*3^m - 1 prime.
EXAMPLE
a(36) = 1 since phi(15)/2 + phi(21)/12 = 4 + 1 = 5 with 2*3^5 + 1 = 487 prime.
MATHEMATICA
p[n_]:=IntegerQ[n]&&PrimeQ[2*3^n+1]
f[n_, k_]:=EulerPhi[k]/2+EulerPhi[n-k]/12
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 23 2014
STATUS
approved