

A235728


a(n) = {0 < k < n  2: 2*m + 1, m*(m+1)  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.


4



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

1,6


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 191.
(ii) If n > 8 is not equal to 32, then there is a positive integer k < n  2 such that 2*m + 1, m*(m+1)  prime(m) and m*(m+1) + prime(m) are all prime, where m = sigma(k) + phi(nk)/2, and sigma(k) is the sum of all positive divisors of k.
(iii) If n > 444, then there is a positive integer k < n such that 2*m + 1, m^2  prime(m) and m^2 + prime(m) are all prime, where m = sigma(k) + phi(nk).
Clearly, part (i) of the conjecture implies that there are infinitely many odd primes p = 2*m + 1 with m*(m+1)  prime(m) = (p^21)/4  prime((p1)/2) and m*(m+1) + prime(m) = (p^21)/4 + prime((p1)/2) both prime.


LINKS

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


EXAMPLE

a(6) = 2 since phi(1) + phi(5)/2 = phi(3) + phi(3)/2 = 3 with 2*3 + 1 = 7, 3*4  prime(3) = 7 and 3*4 + prime(3) = 17 all prime.
a(191) = 1 since phi(153) + phi(38)/2 = 105 with 2*105 + 1 = 211, 105*106  prime(105) = 11130  571 = 10559 and 105*106 + prime(105) = 11130 + 571 = 11701 all prime.


MATHEMATICA

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


CROSSREFS

Cf. A000010, A000040, A235508, A235592, A235682, A235703, A235727.
Sequence in context: A318048 A235805 A168231 * A205546 A081315 A035662
Adjacent sequences: A235725 A235726 A235727 * A235729 A235730 A235731


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 15 2014


STATUS

approved



