

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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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

ZhiWei Sun, Table of n, a(n) for n = 1..10000


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

Cf. A000010, A000040, A235592, A235728.
Sequence in context: A322587 A058973 A155520 * A277859 A308566 A288535
Adjacent sequences: A235909 A235910 A235911 * A235913 A235914 A235915


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 16 2014


STATUS

approved



