

A060392


Let f(m) = smallest prime that divides k^2 + k + m for k = 0,1,2,...; sequence gives smallest m >= 0 such that f(m) is the nth prime.


5



0, 1, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

Chris Nash (see the Prime Puzzles link) has shown that such an m always exists.
For n>1, least odd number d such that the Legendre symbol (14d/prime(k)) = 1 for k = 2,...,n, but not for n+1.  T. D. Noe, Apr 19 2004


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, 113139. Math. Rev. 96m:11082


LINKS



EXAMPLE

k^2 + k takes the values 0, 2, 6, 12, ... for k = 0,1,2,...; the smallest prime divisor of these numbers is 2, so f(0) = 2.


MATHEMATICA

nn=20; a=Table[0, {nn}]; d=1; While[Length[Select[a, # == 0&]] != 1, d=d+2; i=2; While[JacobiSymbol[14d, Prime[i]]==1, i++ ]; If[i<=nn && a[[i]]==0, a[[i]]=d]]; a (* corrected by JeanFrançois Alcover, Feb 06 2019 *)


PROG

(PARI) lista(nn) = {va = vector(nn); d = 1; while (#select(x>(x==0), va) != 1, d += 2; i = 2; while(kronecker(14*d, prime(i)) == 1, i++); if ((i <= nn) && (va[i] == 0), va[i] = d); ); va; } \\ Michel Marcus, Feb 05 2019


CROSSREFS



KEYWORD

nice,nonn


AUTHOR

Luis RodriguezTorres (ludovicusmagister(AT)yahoo.com), Apr 03 2001


EXTENSIONS



STATUS

approved



