A254713 - OEIS

%I #16 Feb 25 2015 12:34:14

%S 4,5,6,7,11,13,17,19,23,29,31,53,59,61,67,73,83,89,97,101,103,109,113,

%T 127,131,139,151,157,163,173,179,191,193,199,223,227,229,251,263,271,

%U 307,313,337,347,353,359,367,379,389,401,449,479,521,523,577,587,599,601,607,613,631,643

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

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

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GoldbachPartition.html ">Goldbach Partition</a>.

%e 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.

%e 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

%t lstIn={};lstFin={};

%t goldPart[x_]:=Module[{h=x/2},While[h>1,If[And[PrimeQ[h],PrimeQ[x-h]],AppendTo[lstIn,{h,x-h}]];h--];

%t lstFin=Length[Union[Flatten[lstIn]]];lstIn={};lstFin];

%t a254713=Flatten[Position[PrimeQ[goldPart/@Range[2,2002,2]],True]]

%Y Cf. A035026, A002372, A045917.

%K easy,nonn

%O 1,1

%A _Ivan N. Ianakiev_, Feb 06 2015

%E Edited. _Wolfdieter Lang_, Feb 23 2015