login
Zero terms of A276520.
0

%I #6 Nov 22 2016 21:24:46

%S 1,2,3,4,5,7,32,52,55,61,128,194,214,244,292,334,388,782,902,992,1414,

%T 1571,1712,1916,2551

%N Zero terms of A276520.

%C It is conjectured that this sequence is finite and all terms are found.

%e A276520(1,2,3,4,5)=0, so a(1)=1, a(2)=2, a(3)=3, a(4)=4, and a(5)=5.

%e The next zero: A276520(7)=0, so a(6)=7.

%t p = 3; sp = {p}; m = 0; Table[While[m++; l = Length[sp]; While[sp[[l]] < m, While[p = NextPrime[p]; cp = 2*3^(Floor[Log[3, 2*p - 1]]) - p; ! PrimeQ[cp]]; AppendTo[sp, p]; l++]; c = 2 - Mod[m + 1, 2]; ct = 0; Do[If[MemberQ[sp, m - c*sp[[i]]], If[c == 1, If[(2*sp[[i]]) <= m, ct++], ct++]], {i, 1, l}]; ct != 0]; m, {n, 1, 25}]

%Y Cf. A276520.

%K nonn,easy,fini,full

%O 1,2

%A _Lei Zhou_, Nov 15 2016