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A174711
Composites of the form 2*n^n + 1 = A216147(n).
0
9, 55, 513, 6251, 93313, 1647087, 33554433, 774840979, 20000000001, 570623341223, 605750213184507, 22224013651116033, 875787780761718751, 36893488147419103233, 1654480523772673528355, 3956839311320627178247959
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.
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
Complement of A216148 in A216147. - M. F. Hasler, Sep 02 2012
Sequence in context: A096191 A362088 A281454 * A231857 A041148 A307844
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 27 2010
STATUS
approved