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A238458
Number of primes p < n with 2*P(n-p) + 1 prime, where P(.) is the partition function (A000041).
3
0, 0, 1, 2, 2, 3, 2, 3, 4, 2, 3, 3, 3, 5, 2, 4, 4, 5, 4, 5, 4, 4, 3, 3, 3, 4, 4, 4, 2, 4, 2, 5, 4, 4, 5, 3, 3, 6, 3, 4, 1, 3, 4, 7, 6, 4, 4, 4, 4, 4, 4, 5, 3, 5, 5, 7, 3, 3, 4, 6, 5, 8, 5, 5, 4, 4, 2, 7, 5, 7, 3, 6, 5, 7, 6, 7, 5, 5, 4, 7, 4, 5, 3, 5, 6, 8, 5, 3, 4, 6, 3, 5, 4, 5, 4, 5, 2, 6, 4, 5
OFFSET
1,4
COMMENTS
Conjecture: a(n) > 0 for all n > 2. Also, for each n > 3 there is a prime p < n with 2*P(n-p) - 1 prime.
We have verified the conjecture for n up to 10^5.
See also A238459 for a similar conjecture involving the strict partition function.
LINKS
EXAMPLE
a(3) = 1 since 2 and 2*P(3-2) + 1 = 2*1 + 1 = 3 are both prime.
a(41) = 1 since 37 and 2*P(41-37) + 1 = 2*5 + 1 = 11 are both prime.
MATHEMATICA
p[n_, k_]:=PrimeQ[2*PartitionsP[n-Prime[k]]+1]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, PrimePi[n-1]}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 27 2014
STATUS
approved