OFFSET
1,1
COMMENTS
Any prime divisor of 4n^2+3 different from 3 is congruent to 1 modulo 6.
4n^2+3 is never a power of 3 for n > 0; hence a prime divisor congruent to 1 modulo 6 always exists.
a(n) = 7 if and only if n is congruent to 1 or -1 modulo 7.
REFERENCES
D. M. Burton, Elementary Number Theory, McGraw-Hill, Sixth Edition (2007), p. 191.
LINKS
Nick Hobson, Table of n, a(n) for n = 1..1000
EXAMPLE
The prime divisors of 4*3^2+3=39 are 3 and 13, so a(3) = 13.
PROG
(PARI) vector(60, n, factor(4*n^2+3)[2-(n^2)%3, 1])
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Nick Hobson, Nov 26 2006
STATUS
approved