

A230040


Number of ways to write n = x + y + z with y <= z such that all the five numbers 6*x1, 6*y1, 6*z1, 6*x*y1 and 6*x*z1 are Sophie Germain primes.


6



0, 0, 1, 2, 2, 3, 3, 1, 3, 4, 5, 2, 1, 1, 3, 4, 4, 3, 4, 6, 5, 2, 2, 6, 5, 1, 2, 4, 2, 2, 3, 6, 5, 7, 6, 2, 3, 4, 4, 2, 3, 5, 1, 4, 7, 4, 6, 3, 9, 4, 2, 5, 4, 3, 9, 2, 4, 3, 6, 3, 5, 8, 8, 5, 8, 6, 2, 4, 3, 4, 1, 6, 4, 3, 8, 8, 6, 6, 9, 11, 2, 4, 2, 8, 3, 4, 6, 10, 5, 11, 7, 8, 6, 10, 4, 1, 3, 1, 3, 3
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OFFSET

1,4


COMMENTS

Conjecture: a(n) > 0 for all n > 2.
This implies that 6*n3 with n > 2 can be expressed as a sum of three Sophie Germain primes (i.e., those primes p with 2*p+1 also prime).
We have verified the conjecture for n up to 10^8. Note that any Sophie Germain prime p > 3 has the form 6*k1.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588.


EXAMPLE

a(4) = 2, since 4 = 1 + 1 + 2 = 2 + 1 + 1, and 6*11=5 and 6*21=11 are Sophie Germain primes.
a(26) = 1, since 26 = 15 + 2 + 9, and all the five numbers 6*151=89, 6*21=11, 6*91=53, 6*15*21=179 and 6*15*9=809 are Sophie Germain primes.


MATHEMATICA

SQ[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
a[n_]:=Sum[If[SQ[6i1]&&SQ[6j1]&&SQ[6(nij)1]&&SQ[6i*j1]&&SQ[6*i(nij)1], 1, 0], {i, 1, n2}, {j, 1, (ni)/2}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A005384, A219842, A219864, A227938, A229969, A229974, A230037.
Sequence in context: A227314 A128924 A239957 * A242361 A116464 A284532
Adjacent sequences: A230037 A230038 A230039 * A230041 A230042 A230043


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 06 2013


STATUS

approved



