A060380 Let f(m) = smallest prime that divides k^2 + k + m for k = 0,1,2,...; sequence gives smallest m >= 2 such that f(m) is the n-th prime, or -1 if no such m exists.

2, 3, 5, 47, 11, 221, 17, 1217, 941, 2747, 8081, 9281, 41, 55661, 19421, 333491, 1262201, 601037, 5237651, 9063641, 12899891, 26149427, 24073871, 28537121, 352031501, 398878547, 160834691, 67374467, 146452961, 24169417397

Offset: 1

Author

Luis Rodriguez-Torres (ludovicusmagister(AT)yahoo.com), Apr 03 2001

Keywords

• hard
• nice
• nonn

Comments

Chris Nash (see the Prime Puzzles link) has shown that such an m always exists.

For n>2, least odd number d such that the Legendre symbol (1-4d/prime(k)) = -1 for k = 2,...,n, but not for n+1. See A060392. - T. D. Noe, Apr 19 2004

Examples

			k^2 + k + 2 takes the values 2, 4, 8, 14, ... for k = 0,1,2,...; the smallest prime divisor of these numbers is 2, so f(2) = 2.
		

References

• R. F. Lukes, C. D. Patterson and H. C. Williams, Numerical sieving devices: their history and some applications. Nieuw Arch. Wisk. (4) 13 (1995), no. 1, 113-139. Math. Rev. 96m:11082

Links

• G. W. Fung and H. C. Williams, Quadratic polynomials with high density of primes, Mathematics of Computation, Vol. 55, 1990.
• C. Rivera, www.primepuzzles.net, Conjecture 17

Crossrefs

Cf. A060392-A060398. A060393 gives associated values of k.

Programs

Mathematica

(* This program is not convenient beyond a(24) *) a[1] = 2; a[2] = 3; a[n_] := For[d = 1, True, d = d+2, If[And @@ (# == -1 & /@ Table[JacobiSymbol[1 - 4d, Prime[k]], {k, 2, n}]) && JacobiSymbol[1 - 4d, Prime[n+1]] != -1, Return[d]]]; Table[Print[an = a[n]]; an, {n, 1, 24}] (* Jean-François Alcover, Oct 14 2013, after T. D. Noe *)

Extensions

Corrected by T. D. Noe, Apr 19 2004