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a(n) = (A212153(n)^3 + 1)/7^n, n >= 0.
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%I #21 Jan 14 2025 08:40:54

%S 1,18,140,20,479393,219600095,4804461081,686351583,6679631931865,

%T 82080661415031,8898622841908566,174149720118385232,

%U 7290250572352382182,65315972853762054047,98713213404986046050649,11532114009920222592500432,356054521382275298405890644,28999349909865958163356878647

%N a(n) = (A212153(n)^3 + 1)/7^n, n >= 0.

%C a(n) is an 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.

%H Paolo Xausa, <a href="/A212154/b212154.txt">Table of n, a(n) for n = 0..500</a>

%F a(n) = (A212153(n)^3 + 1)/7^n.

%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).

%t Join[{1}, MapIndexed[(#^3 + 1)/7^#2[[1]] &, FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 20]]]] (* _Paolo Xausa_, Jan 14 2025 *)

%Y Cf. A210848, A210849 (the p=5 case), A210853, A212153, A212156.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, May 02 2012