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%I #5 Jun 08 2021 02:23:55
%S 7,703,387400807,58149737003032434092905183,
%T 196627050475552913618075908526912116282660024455971729157367165907347241304007
%N a(n) = 1 - 3^(3^n) + 9^(3^n).
%C a(n) is the ratio of two consecutive base-3 Fermat numbers A129290(n) = 3^(3^n) + 1 = {4, 28, 19684, 7625597484988, ...}.
%F a(n) = A002061(3^(3^n)). a(n) = A129290(n+1) / A129290(n).
%t Table[1 - 3^3^n + 9^3^n, {n,0,5}]
%Y Cf. A129290 (3^(3^n) + 1).
%Y Cf. A055777 (3^(3^n)).
%Y Cf. A002061 (central polygonal numbers: n^2 - n + 1).
%K nonn
%O 0,1
%A _Alexander Adamchuk_, Apr 08 2007
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