

A236439


a(n) = {0 < k < n2: A000009(m)^2 + A047967(m)^2 is prime with m = k + phi(nk)/2}, where phi(.) is Euler's totient function.


2



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

1,5


COMMENTS

Conjecture: a(n) > 0 for all n > 3.
We have verified this for n up to 50000.
The conjecture implies that there are infinitely many positive integers m with A000009(m)^2 + A047967(m)^2 prime. See A236440 for such numbers m.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
Z.W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014


EXAMPLE

a(14) = 1 since 2 + phi(12)/2 = 4 with A000009(4)^2 + A047967(4)^2 = 2^2 + 3^2 = 13 prime.
a(17) = 1 since 10 + phi(7)/2 = 13 with A000009(13)^2 + A047967(13)^2 = 18^2 + 83^2 = 7213 prime.


MATHEMATICA

p[n_]:=PrimeQ[PartitionsQ[n]^2+(PartitionsP[n]PartitionsQ[n])^2]
a[n_]:=Sum[If[p[k+EulerPhi[nk]/2], 1, 0], {k, 1, n3}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000009, A000010, A000040, A047967, A236412, A236417, A236419, A236440.
Sequence in context: A154840 A205102 A247108 * A011154 A048466 A096838
Adjacent sequences: A236436 A236437 A236438 * A236440 A236441 A236442


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 25 2014


STATUS

approved



