A254713 All numbers k such that the number of distinct parts of all A045917(k) Goldbach partitions of 2k is prime.

4, 5, 6, 7, 11, 13, 17, 19, 23, 29, 31, 53, 59, 61, 67, 73, 83, 89, 97, 101, 103, 109, 113, 127, 131, 139, 151, 157, 163, 173, 179, 191, 193, 199, 223, 227, 229, 251, 263, 271, 307, 313, 337, 347, 353, 359, 367, 379, 389, 401, 449, 479, 521, 523, 577, 587, 599, 601, 607, 613, 631, 643

Offset: 1

Author

Ivan N. Ianakiev, Feb 06 2015

Keywords

• easy
• nonn

Comments

Conjecture: a(k) is prime for k > 3. Tested for k up to 3*10^4.

Examples

			For k = 4, 2k = 8. The number of the distinct Goldbach parts of 8 (3 and 5) is prime, therefore 4 is in the sequence.
5 is in the sequence because the 2 = A045917(5) Goldbach partitions of 10 are 5 + 5 and 3 + 7, and there are 3 distinct parts, namely 3, 5 and 7. - _Wolfdieter Lang_, Feb 23 2015
		

Links

• Eric Weisstein's World of Mathematics, Goldbach Partition.

Crossrefs

Cf. A035026, A002372, A045917.

Programs

Mathematica

lstIn={};lstFin={};
goldPart[x_]:=Module[{h=x/2},While[h>1,If[And[PrimeQ[h],PrimeQ[x-h]],AppendTo[lstIn,{h,x-h}]];h--];
lstFin=Length[Union[Flatten[lstIn]]];lstIn={};lstFin];
a254713=Flatten[Position[PrimeQ[goldPart/@Range[2,2002,2]],True]]

Extensions

Edited. Wolfdieter Lang, Feb 23 2015