A240713 Number of decompositions of 2n=p1+p2 (prime p1 <= p2), where at least one other such pair 2n=p3+p4 (prime p3 <= p4) exists such that |p1-p3|= 6 or 12.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 0, 3, 4, 2, 2, 4, 2, 3, 5, 3, 3, 5, 2, 4, 6, 2, 4, 5, 0, 4, 5, 4, 4, 7, 3, 4, 6, 3, 4, 9, 3, 5, 7, 3, 5, 7, 3, 5, 7, 5, 6, 10, 5, 6, 12, 2, 5, 10, 2, 6, 7, 4, 4, 4, 5, 7, 9, 6, 5, 11, 0, 6, 10, 3, 7, 8, 4, 4, 13, 8
Offset: 1
Examples
For n = 11, 2n=22. 22 = 3 + 19 = 5 + 17 = 11 + 11. |5-11|=6 so pair 5+17 and 11+11 are counted. So a(11)=2. ... For n = 17, 2n=34. 34 = 3 + 31 = 5 + 29 = 11 + 23 = 17 + 17. |5-11|=6, so 5+29 and 11+23 are counted. Also since |11-17|=6, 17+17 is also counted (where 11+23 is already counted). In case |5-17|=12, both instances are already counted. So overall three instances are found. a(17)=3.
Links
- Lei Zhou, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
Table[s = 2*n; ct = 0; p = 1; While[p = NextPrime[p]; p <= n, If[PrimeQ[s - p], ok = 0; a1 = p - 12; b1 = s - a1; a2 = p - 6; b2 = s - a2; a3 = p + 6; b3 = s - a3; a4 = p + 12; b4 = s - a4; If[a1 > 0, If[PrimeQ[a1] && PrimeQ[b1], ok = 1]]; If[a2 > 0, If[PrimeQ[a2] && PrimeQ[b2], ok = 1]]; If[a3 <= n, If[PrimeQ[a3] && PrimeQ[b3], ok = 1]]; If[a4 <= n, If[PrimeQ[a4] && PrimeQ[b4], ok = 1]]; If[ok == 1, ct++]]]; ct, {n, 1, 85}]
Comments