A198411 a(n)= (4^(2^n) + 2^(2^n) + 1)/7.

1, 3, 39, 9399, 613576119, 2635249154000645559, 48611766702991209068831621643639680439, 16541727033902313631938712144098272550515752433223071786131565516477842550199

Offset

0,2

Comments

Let b(n) = 4^(2^n) + 2^(2^n) + 1, then b(n+1) = b(n)^2 - 2(8^(2^n) + 4^(2^n)+ 2^(2^n) ) == 1 + 4^(2^n)+ 2^(2^n)= b(n) == 0 (mod 7).

The next term (a(8)) has 154 digits. - Harvey P. Dale, Sep 13 2020

Links

Table of n, a(n) for n=0..7.

Example

a(2) = (4^(2^2) + 2^(2^2) + 1)/7 = 273/7 = 39.

Maple

for n from 0 to 9 do:x:=  (4^(2^n) + 2^(2^n) + 1)/7
:  printf(`%d, `, x):od:

Mathematica

Table[(4^(2^n)+2^(2^n)+1)/7, {n, 0, 8}] (* Harvey P. Dale, Sep 13 2020 *)

Crossrefs

Sequence in context: A188410 A188388 A076628 * A367033 A097421 A180418

Adjacent sequences: A198408 A198409 A198410 * A198412 A198413 A198414

Keyword

nonn

Author

Michel Lagneau, Oct 24 2011

Status

approved