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A236412 a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/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 (list; graph; refs; listen; history; text; internal format)
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

Zhi-Wei 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[n-k]/8

a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}]

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

Zhi-Wei Sun, Jan 24 2014

STATUS

approved

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Last modified October 21 14:10 EDT 2019. Contains 328301 sequences. (Running on oeis4.)