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A198411
a(n)= (4^(2^n) + 2^(2^n) + 1)/7.
0
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
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
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 24 2011
STATUS
approved