A060395 Smallest prime that divides k^2 + k + n for k = 0, 1, 2, ....

2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 11, 2, 3, 2, 3, 2, 17, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 41, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 11

Offset: 0

Author

N. J. A. Sloane, Apr 04 2001

Keywords

• nonn
• easy
• nice

Examples

			To obtain a(7), note that x^2+x+7 takes the values 7,9,13,19,... for k=0,1,2,... and the smallest prime dividing these numbers is 3.
		

Links

• Carlos Rivera, Conjecture 17. The Ludovicus conjecture about the Euler trinomials, The Prime Puzzles & Problems Connection.

Crossrefs

Cf. A060380, A060392-A060398. A060396 gives values of k.

Programs

Mathematica

a[n_] := Switch[n, 0, 2, 1, 3, , Module[{f, kmax0 = 2}, f[kmax] := f[kmax] = MinimalBy[Table[{k, FactorInteger[k^2 + k + n][[1, 1]]}, {k, 0, kmax}], Last, 1]; f[kmax = kmax0]; f[kmax = 2 kmax]; While[f[kmax] != f[kmax/2], kmax = 2 kmax]; f[kmax][[1, 2]]]];
Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Aug 15 2022 *)

Formula

a(n)=2 if n is equal to 0, 2 or 4 modulo 6; a(n)=3 if n is equal to 1 or 3 modulo 6.

Extensions

More terms from Matthew Conroy, Apr 18 2001