A060395 -id:A060395 - cp's OEIS frontend

A060396 Values of k associated with A060395.

1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1

Offset: 0

N. J. A. Sloane, Apr 04 2001

a(2n) = 0 for n >= 1. a(3n) = 0 for n >= 1. a(6n+1) = 1 for n >= 0. - Nathaniel Johnston, Apr 30 2011

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

a[n_] := Switch[n, 0, 1, 1, 1, , 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, 1]]]];
Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Aug 15 2022 *)

Name corrected and a(15)-a(97) from Nathaniel Johnston, Apr 30 2011

A060397 Smallest prime that divides k^2 + k + 2n + 1 for k = 0,1,2,....

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Apr 04 2001

Bisection of A060395.

			To obtain a(3), 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.

T. D. Noe, Table of n, a(n) for n = 0..10000
Carlos Rivera, Conjecture 17. The Ludovicus conjecture about the Euler trinomials, The Prime Puzzles & Problems Connection.

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

a[n_] := Switch[n, 0, 3, , Module[{f, kmax0 = 2}, f[kmax] := f[kmax] = MinimalBy[Table[{k, FactorInteger[k^2 + k + 2 n + 1][[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, 100}] (* Jean-François Alcover, Aug 15 2022 *)

a(n)=3 if n is equal to 0 or 1 mod 3.

More terms from Matthew Conroy, Apr 18 2001

cp's OEIS Frontend

A060396 Values of k associated with A060395.

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