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A212156 a(n) = ((6*A023000(n))^3 + 1)/7^n, n >= 0.

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%I A212156 #15 Jan 13 2025 04:11:07
%S A212156 1,31,2257,116623,5757601,282424831,13840934257,678220602223,
%T A212156 33232913275201,1628413476849631,79792265450186257,
%U A212156 3909821042651007823,191581231339042552801,9387480337357087274431,459986536542705291758257
%N A212156 a(n) = ((6*A023000(n))^3 + 1)/7^n, n >= 0.
%C A212156 a(n) is an integer because 6*A023000(n) is one of three solution of X(n)^3+1 == 0 (mod 7^n), namely the one satisfying also X(n) == 6 (mod 7) == -1 (mod 7).
%C A212156 See the comments on A210852, and the Nagell reference given in A210848.
%F A212156 a(n) = (b(n)^3+1)/7^n, n>=0, with b(n):=6*A023000(n) given by a recurrence obtained from the one of A023000. There also programs for b(n)/6 are given.
%e A212156 a(0) = 1/1 = 1.
%e A212156 a(3) = ((6*57)^3 + 1)/7^3 = 40001689/343  = 116623,  (b(3) = 48^7 (mod 7^3) = 342 = 6*57).
%Y A212156 Cf.  A210848, A210849 (the p=5 case). A210853, A212154.
%K A212156 nonn,easy
%O A212156 0,2
%A A212156 _Wolfdieter Lang_, May 02 2012