OFFSET
1,1
COMMENTS
All terms are divisible by 5, hence conjecture: there is no such n that f[n],f[n+1],f[n+2] and f[n+3] are primes.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
15 is a term because f[15]=1861, f[16]=2113 and f[17]=2381 are all primes.
MATHEMATICA
Flatten[Position[Partition[Table[PrimeQ[8n^2+4n+1], {n, 14000}], 3, 1], {True, True, True}]] (* Harvey P. Dale, Oct 08 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Feb 15 2005
STATUS
approved