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A119563
Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)+M(n)-1 = 2^(2^n) + 2^n - 1.
7
2, 5, 19, 263, 65551, 4294967327, 18446744073709551679, 340282366920938463463374607431768211583, 115792089237316195423570985008687907853269984665640564039457584007913129640191
OFFSET
0,1
COMMENTS
The first 5 entries are primes. Are there infinitely many primes in this sequence?
FORMULA
a(n) = A119561(n)-2=A000215(n)+A000225(n)-1. - R. J. Mathar, Apr 22 2007
EXAMPLE
F(2) = 2^(2^2)+1 = 17, M(2) = 2^2-1 = 3, F(2)+ M(2) - 1 = 19
PROG
(PARI) fm3(n) = for(x=0, n, y=2^(2^x)+2^x-1; print1(y", "))
CROSSREFS
Sequence in context: A218386 A055813 A119550 * A270398 A269997 A270547
KEYWORD
nonn
AUTHOR
Cino Hilliard, May 31 2006
EXTENSIONS
Edited by N. J. A. Sloane, Jun 03 2006
STATUS
approved