A123591 a(n) = ((2^n - 1)^(2^n) - 1)/(2^n)^2.

-1, 0, 5, 90075, 25657845139503479, 516742576223066713590751888575037849059948015

Offset

0,3

Comments

The next term is too large to include.

Last digit of a(n) is 5 or 9 for n>1. It appears that a(4k) == 4 mod 5 and a(4k+1) == a(4k+2) == a(4k+3) == 0 mod 5.

p divides a(p) for prime p>2. Composite numbers n such that n divides a(n) are listed in A127643 = {15,51,65,85,185,221,255,341,451,533,561,595,645,679,771,...}. - Alexander Adamchuk, Jan 22 2007

Links

G. C. Greubel, Table of n, a(n) for n = 0..8

Formula

a(n) = ((2^n - 1)^(2^n) - 1)/(2^n)^2.

a(n) = A085606(2^n)/(2^n)^2.

Mathematica

Table[((2^n-1)^(2^n)-1)/(2^n)^2, {n, 0, 7}]

Prog

(PARI) for(n=0, 7, print1(((2^n - 1)^(2^n) - 1)/(2^n)^2, ", ")) \\ G. C. Greubel, Oct 26 2017
(Magma) [((2^n - 1)^(2^n) - 1)/(2^n)^2: n in [0..7]]; // G. C. Greubel, Oct 26 2017

Crossrefs

Cf. A085606 (n-1)^n - 1.

Cf. A127643.

Sequence in context: A171981 A145232 A263174 * A133381 A366270 A236066

Adjacent sequences: A123588 A123589 A123590 * A123592 A123593 A123594

Keyword

sign

Author

Alexander Adamchuk, Nov 13 2006

Extensions

More terms from Alexander Adamchuk, Jan 22 2007

Status

approved