A233439 a(n) = |{0 < k < n: prime(k)^2 + 4*prime(n-k)^2 is prime}|

0, 0, 0, 1, 2, 1, 2, 1, 3, 4, 4, 8, 4, 6, 3, 1, 7, 3, 8, 5, 2, 9, 2, 11, 8, 7, 5, 4, 8, 7, 8, 8, 8, 7, 5, 9, 5, 10, 9, 7, 13, 9, 11, 10, 14, 5, 11, 10, 10, 11, 12, 7, 13, 10, 10, 8, 15, 11, 12, 11, 13, 14, 6, 12, 11, 22, 21, 5, 15, 7, 13, 15, 17, 15, 10, 16, 11, 13, 14, 12, 17, 12, 16, 16, 19, 22, 17, 12, 19, 17, 19, 17, 16, 17, 18, 20, 19, 17, 10, 16

Offset

1,5

Comments

Conjecture: (i) a(n) > 0 for all n > 3.

(ii) For any integer n > 10, prime(j)^3 + 2*prime(n-j)^2 is prime for some 0 < j < n, and prime(k)^3 + 2*prime(n-k)^3 is prime for some 0 < k < n.

(iii) If n > 5, then prime(k)^3 + 2*p(n-k)^3 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n > 2, then prime(k)^3 + 2*q(n-k)^3 is prime for some 0 < k < n, where q(.) is the strict partition function (A000009).

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014-2016.

Example

a(4) = 1 since prime(3)^2 + 4*prime(1)^2 = 5^2 + 4*2^2 = 41 is prime.
a(6) = 1 since prime(5)^2 + 4*prime(1)^2 = 11^2 + 4*2^2 = 137 is prime.
a(8) = 1 since prime(3)^2 + 4*prime(5)^2 = 5^2 + 4*11^2 = 509 is prime.
a(16) = 1 since prime(6)^2 + 4*prime(10)^2 = 13^2 + 4*29^2 = 3533 is prime.

Mathematica

a[n_]:=Sum[If[PrimeQ[Prime[k]^2+4*Prime[n-k]^2], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000040, A002313, A220413, A232174, A232269, A232465, A232502, A233150, A233204, A233206, A233296, A233307, A233346.

Sequence in context: A210850 A051276 A226212 * A256600 A137752 A331917

Adjacent sequences: A233436 A233437 A233438 * A233440 A233441 A233442

Keyword

nonn

Author

Zhi-Wei Sun, Dec 09 2013

Status

approved