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A230230
Number of ways to write 2*n = p + q with p, q, 3*p - 10, 3*q + 10 all prime.
5
0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 3, 4, 3, 3, 5, 1, 5, 5, 3, 4, 5, 3, 2, 6, 4, 3, 6, 3, 3, 6, 3, 5, 6, 3, 6, 5, 4, 4, 9, 5, 4, 9, 5, 3, 9, 4, 4, 6, 4, 5, 6, 5, 5, 10, 4, 8, 10, 3, 7, 12, 3, 6, 11, 5, 7, 8, 3, 4, 6, 6, 4, 7, 2, 7, 9, 2, 10, 9, 3, 9, 8, 3, 5, 14, 8, 4, 12, 5, 5, 11, 5, 6, 8, 3, 8, 7, 4, 9, 11, 3
OFFSET
1,7
COMMENTS
Conjecture: a(n) > 0 for all n > 3.
This is stronger than Goldbach's conjecture for even numbers. If 2*n = p + q with p, q, 3*p - 10, 3*q + 10 all prime, then 6*n is the sum of the two primes 3*p - 10 and 3*q + 10.
Conjecture verified for 2*n up to 10^9. - Mauro Fiorentini, Jul 08 2023
LINKS
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588 [math.NT], 2012-2017.
EXAMPLE
a(5) = 1 since 2*5 = 7 + 3 with 3*7 - 10 = 11 and 3*3 + 10 = 19 both prime.
a(16) = 1 since 2*16 = 13 + 19 with 3*13 - 10 = 29 and 3*19 + 10 = 67 both prime.
MATHEMATICA
PQ[n_]:=n>3&&PrimeQ[3n-10]
SQ[n_]:=PrimeQ[n]&&PrimeQ[3n+10]
a[n_]:=Sum[If[PQ[Prime[i]]&&SQ[2n-Prime[i]], 1, 0], {i, 1, PrimePi[2n-2]}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 12 2013
STATUS
approved