OFFSET
1,1
COMMENTS
If p = n+2 is prime, then p divides 2*n^n + 1. Proof: Let p = n+2 prime. Then, according to Fermat's theorem, n^(p-1) == 1 (mod p). Because p-1 = n+1, n^(n+1) == 1 (mod p), and with n = p-2 == -2 (mod p), we obtain successively: n*n^n == 1 (mod p), -2*n^n == 1 (mod p), 2*n^n == -1 (mod p) => p divides 2*n^n + 1.
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
J. M. De Koninck, A. Mercier, 1001 problemes en theorie classique des nombres, Ellipses 2004, p. 52.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 2.
LINKS
C. K. Caldwell, Composite Numbers
EXAMPLE
a(2) = 9 = 3^2, a(3) = 55 = 5*11, a(4) = 513 = 3 ^ 3 * 19.
MAPLE
with(numtheory):for n from 0 to 50 do: x:=2*n^n + 1 : if type(x, prime)=false then print (x):else fi:od:
MATHEMATICA
Select[Table[2n^n+1, {n, 20}], CompositeQ] (* Harvey P. Dale, Jun 21 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 27 2010
STATUS
approved