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A236419
a(n) = |{0 < k < n: r = phi(k) + phi(n-k)/6 + 1 and p(r) + q(r) are both prime}|, where phi(.) is Euler's totient function, p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).
4
0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 2, 1, 0, 1, 0, 2, 0, 1, 1, 0, 0, 4, 1, 2, 1, 0, 2, 3
OFFSET
1,68
COMMENTS
Conjecture: a(n) > 0 for all n > 127.
We have verified this for n up to 30000.
The conjecture implies that there are infinitely many primes r with p(r) + q(r) also prime.
LINKS
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.
EXAMPLE
a(15) = 1 since phi(1) + phi(14)/6 + 1 = 3 with p(3) + q(3) = 3 + 2 = 5 prime.
a(54) = 1 since phi(41) + phi(13)/6 + 1 = 43 with p(43) + q(43) = 63261 + 1610 = 64871 prime.
MATHEMATICA
pq[n_]:=PrimeQ[n]&&PrimeQ[PartitionsP[n]+PartitionsQ[n]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/6+1
a[n_]:=Sum[If[pq[f[n, k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 25 2014
STATUS
approved