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a(n) is the number of p cases such that p, p - 6, p + 6 and 2n - q are primes.

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

1,8

a(n) = 0 only when n = 1, 2, 3, 4, 5, 6, 19, 31, 331, 499

for n = 7, it is found when p = 11, p - 6, p, p + 6, and 2n - p are 5, 11, 17, 3, all primes. So a(7) = 1;

for n = 12, it is found,

when p = 11, {p - 6, p, p + 6, 2n - p} = {5, 11, 17, 13}, all primes;

when p = 13, {p - 6, p, p + 6, 2n - p} = {7, 13, 19, 11}, all primes;

when p = 17, {p - 6, p, p + 6, 2n - p} = {11, 17, 23, 7}, all primes;

So a(12) = 3;

for n = 19; 2n = 38 = 7 + 31 = 19 + 19 = 31 + 7, none of p = 7, 19, 31 can make p -6 and p + 6 both prime. Thus a(19) = 0

m = 200; ps = {}; p = 7; While[p = NextPrime[p]; If[PrimeQ[p - 6] && PrimeQ[p + 6], AppendTo[ps, p]]; p < 2*m]; a = {}; Do[ct = 0; k = 0; While[k++; ps[[k]] < n, q = n - ps[[k]]; If[PrimeQ[q], ct++]]; AppendTo[a, ct]; If[ct == 0, AppendTo[b, n]], {n, 2, m, 2}]; a

Ref. A006489; Cf. A045917.

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nonn,easy

Lei Zhou, Oct 12 2024

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allocated for Lei Zhou

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allocated for Lei Zhou

Starting from Goldbach decomposition of 10 = p + q = 5 + 5, 12 = 7 + 5, and 14 = 7 + 7, a(n) is the first number in A001057 such that if 2n - 6 = p + q, 2n = p' + q', where p' = p + 6 * a(n) and q' = 2n - q' are both primes.

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, -1, 1, 1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 2, 0, 0, 0, 0, 1, 1, 0, -1, 1, 1, 1, -2, -1, -1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 1, 1, 0, 1, -1, 0, -2, 0, 1, 0, 0, 2, 0, 1, 0, 1, 2, 1, 1, 0, 1, -2, 1, -2, 1, 1, 0, 1, 1, 1, -1, 2, 1, 1

8,27

By definition, this sequence starts from n = 8.

Hypothesis: a(n) is defined for all n >= 8 and for all n >=8, the corresponding Goldbach decomposition 2n = p + q has positive primes p and q.

When n = 8, 2n = 16. 2n - 6 = 10. 10 = p + q = 5 + 5 (by definition). a(8) = 0, p' = p + a(8) = 5, q' = 2n - p' = 16 - 5 = 11. P' and q' are both primes.

...

When n = 10, 2n = 20. 2n - 6 = 14. 14 = p + q = 7 + 7 (by definition). a(10) = 0, p' = p + a(10) = 7, q' = 2n - p' = 20 - 7 = 13. P' and q' are both primes.

...

When n = 13, 2n = 26. 2n - 6 = 20. 20 = p + q = 7 + 13 (per above evaluation). a(13) = 0, p' = p + a(13) = 7, q' = 2n - p' = 26 - 7 = 19. P' and q' are both primes.

When n = 16, 2n = 32. 2n - 6 = 26. 26 = p + q = 7 + 19 (per above evaluation). a(16) = 1, p' = p + a(16) = 13, q' = 2n - p' = 32 - 13 = 19. P' and q' are both primes. It is tested when a(16) is 0, q' = 25 is not a prime, thus a(16) = 1 is the first number in A001057 that makes both p' and q' primes.

a = {}; p = {5, 7, 7}; Do[Do[n = 6*k - 4 + 2*j; i = 0; While[i++; m = 1/4 + (i - 1/2)*(-1)^i/2; pr = p[[j]] + 6*m; q = n - pr; ! (PrimeQ[pr] && PrimeQ[q])]; p[[j]] = pr; AppendTo[a, m], {j, 1, 3}], {k, 3, 30}]; Print[a]

Ref. A001057; Cf. A045917

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sign,easy

Lei Zhou, Oct 09 2024

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allocated for Lei Zhou

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proposed

a(n) is the number of pair pairs of primes p+q=2*(n+4) with 5 <= p <= n such that either p+6 or q+6 is also prime.

a(n) is the number of prime pairs pair of primes p+q=2*(n+4) with 5 <= p <= n such that either p+6 or q+6 is also prime.

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