OFFSET
1,3
COMMENTS
0.66*n/log(n)^2 is a good approximation for a(n) as n increases
LINKS
Pierre CAMI, Table of n, a(n) for n = 1..10000
MATHEMATICA
Table[m = n*(n+1); Length[Select[Range[n+1], PrimeQ[m + 2*#-3] && PrimeQ[m + 2*#-1] &]], {n, 100}] (* T. D. Noe, Nov 07 2011 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Pierre CAMI, Nov 07 2011
STATUS
approved