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 A198411 a(n)= (4^(2^n) + 2^(2^n) + 1)/7. 0

%I

%S 1,3,39,9399,613576119,2635249154000645559,

%T 48611766702991209068831621643639680439,

%U 16541727033902313631938712144098272550515752433223071786131565516477842550199

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

%C 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).

%C The next term (a(8)) has 154 digits. - _Harvey P. Dale_, Sep 13 2020

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

%p for n from 0 to 9 do:x:= (4^(2^n) + 2^(2^n) + 1)/7

%p : printf(`%d, `, x):od:

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

%K nonn

%O 0,2

%A _Michel Lagneau_, Oct 24 2011

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Last modified May 14 16:48 EDT 2021. Contains 343898 sequences. (Running on oeis4.)