

A235912


a(n) = {0 < k < n  2: 2*m + 1, m*(m1)  prime(m) and m*(m+1)  prime(m) are all prime with m = phi(k) + phi(nk)/2}, where phi(.) is Euler's totient function.


3



0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 3, 2, 4, 2, 6, 5, 6, 7, 4, 8, 7, 8, 8, 11, 7, 12, 9, 9, 12, 5, 14, 10, 9, 9, 9, 9, 7, 8, 11, 9, 8, 7, 14, 8, 6, 9, 5, 5, 9, 11, 3, 9, 6, 13, 8, 8, 6, 7, 6, 5, 4, 3, 1, 5, 5, 5, 6, 5, 7, 7, 4, 7, 11, 8, 3, 5, 3, 10, 4, 4, 3, 9, 2, 4, 4, 5, 8, 12, 13, 4, 9, 5, 11, 5, 12, 7, 4, 4
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OFFSET

1,12


COMMENTS

Conjecture: a(n) > 0 for all n > 11.
This implies that there are infinitely many odd primes p = 2*m + 1 with q = m*(m1)  prime(m) and r = m*(m+1)  prime(m) both prime. Note that r  q = 2*m.


LINKS



EXAMPLE

a(10) = 1 since phi(5) + phi(5)/2 = 6 with 2*6 + 1 = 13, 5*6  prime(6) = 30  13 = 17 and 6*7  prime(6) = 42  13 = 29 all prime.


MATHEMATICA

PQ[n_]:=n>0&&PrimeQ[n]
p[n_]:=PrimeQ[2n+1]&&PQ[n(n1)Prime[n]]&&PQ[n(n+1)Prime[n]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[nk]/2
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n3}]
Table[a[n], {n, 1, 100}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



