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%I #11 Feb 23 2017 23:28:57
%S 1,4,608,100082,1033865,147695,363432817,493771113103,2362056468993,
%T 408352474516087,11132773648769182,1051698129414636470,
%U 55996715400581424222,4972138747809482684591,29726859239716779753649
%N (A210852(n)^3 + 1)/7^n, n >= 0.
%C a(n) is integer because A210852(n) is one of the three solutions of X(n)^3 + 1 == 0 (mod 7^n), namely the one satisfying also X(n) == 3 (mod 7).
%C See the comments on A210852, and the Nagell reference given in A210848.
%F a(n) = (b(n)^3 + 1)/7^n, n>=0, with b(n):=A210852(n) given by a recurrence. See also a Maple program for b(n) there.
%e a(0) = 1/1 = 1.
%e a(3) = (325^3 + 1)/7^3 = 34328126/343 = 100082, (b(3) = 31^7 (mod 7^3) = 325).
%Y Cf. A210848, A210849 (the p=5 case).
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, May 02 2012
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