

A236412


a(n) = {0 < k < n: m = phi(k)/2 + phi(nk)/8 is an integer with p(m)^2 + q(m)^2 prime}, where phi(.) is Euler's totient, p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).


5



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

1,14


COMMENTS

Conjecture: a(n) > 0 for all n > 17.
We have verified this for n up to 65000.
The conjecture implies that there are infinitely positive integers m with p(m)^2 + q(m)^2 prime. See A236413 for a list of such numbers m. See also A236414 for primes of the form p(m)^2 + q(m)^2.


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(15) = 1 since phi(2)/2 + phi(13)/8 = 1/2 + 12/8 = 2 with p(2)^2 + q(2)^2 = 2^2 + 1^2 = 5 prime.
a(69) = 1 since phi(5)/2 + phi(64)/8 = 2 + 4 = 6 with p(6)^2 + q(6)^2 = 11^2 + 4^2 = 137 prime.
a(89) = 1 since phi(73)/2 + phi(16)/8 = 36 + 1 = 37 with p(37)^2 + q(37)^2 = 21637^2 + 760^2 = 468737369 prime.


MATHEMATICA

p[n_]:=IntegerQ[n]&&PrimeQ[PartitionsP[n]^2+PartitionsQ[n]^2]
f[n_, k_]:=EulerPhi[k]/2+EulerPhi[nk]/8
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A000041, A233307, A236374, A236413, A236414.
Sequence in context: A303979 A301573 A061670 * A236263 A319572 A108063
Adjacent sequences: A236409 A236410 A236411 * A236413 A236414 A236415


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 24 2014


STATUS

approved



