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%I #8 Feb 23 2017 23:29:20
%S 1,18,140,20,479393,219600095,4804461081,686351583,6679631931865,
%T 82080661415031,8898622841908566,174149720118385232,
%U 7290250572352382182,65315972853762054047,98713213404986046050649
%N (A212153(n)^3 + 1)/7^n, n >= 0.
%C a(n) is integer because A212153(n) is one of the three solutions of X(n)^3+1 == 0 (mod 7^n), namely the one satisfying also X(n) == 5 (mod 7).
%C See the comments on A210853, and the Nagell reference given in A210848.
%F a(n) = (b(n)^3+1)/7^n, n>=0, with b(n):=A212153(n) given by a recurrence. See also a Maple program for b(n) there.
%e a(0) = 1/1 = 1.
%e a(3) = (19^3 + 1)/7^3 = 6860/343 = 20, (b(3) = 19^7 (mod 7^3) = 19).
%Y Cf. A210848, A210849 (the p=5 case). A210853, A212156.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, May 02 2012
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