A256119 Least number p that is zero or an odd prime, such that n - p is a generalized pentagonal number.

0, 0, 0, 3, 3, 0, 5, 0, 3, 7, 3, 11, 0, 11, 7, 0, 11, 5, 3, 7, 5, 19, 0, 11, 17, 3, 0, 5, 13, 3, 23, 5, 17, 7, 19, 0, 29, 11, 3, 13, 0, 19, 7, 3, 29, 5, 11, 7, 13, 23, 43, 0, 17, 13, 3, 29, 5, 0, 7, 19, 3, 59, 5, 23, 7, 43, 31, 41, 11, 29, 0, 31, 37, 3, 17, 5, 19, 0, 43, 53, 3, 11, 5, 13, 7, 59, 29, 17, 11, 19, 13, 79, 0, 23, 17, 3, 19, 5, 41, 7, 0

Offset: 0

Author

Zhi-Wei Sun, Mar 15 2015

Keywords

• nonn

Comments

By the conjecture in A256071, a(n) always exists.

Examples

			a(21) = 19 since 21 is not a generalized pentagonal number, and 19 is the least odd prime p with 21 - p a generalized pentagonal number.
a(26) = 0 since 26 = (-4)*(3*(-4)-1)/2 is a generalized pentagonal number.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000040, A001318, A256071.

Programs

Mathematica

Pen[n_]:=IntegerQ[Sqrt[24n+1]]
Do[If[Pen[n],Print[n," ",0];Goto[aa]];Do[If[Pen[n-Prime[k]],Print[n," ",Prime[k]];Goto[aa]],{k,2,PrimePi[n]}];Label[aa];Continue,{n,0,100}]