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A322931 Digits of the 8-adic integer 3^(1/3).

%I A322931 #17 Aug 14 2019 08:33:01
%S A322931 3,7,5,0,7,3,0,1,7,1,7,6,4,2,3,6,7,7,7,0,3,1,2,0,1,7,2,6,1,2,5,4,1,1,
%T A322931 1,2,3,5,5,2,3,5,4,7,3,6,0,0,3,4,7,1,3,3,6,4,6,0,0,4,4,6,0,5,6,4,1,5,
%U A322931 5,6,0,0,0,6,2,6,1,0,7,1,6,0,0,4,6,5,0,7,0,1,3,7,3,7,0,0,2,0,6,1
%N A322931 Digits of the 8-adic integer 3^(1/3).
%C A322931 The octal version of A225404.
%H A322931 Seiichi Manyama, <a href="/A322931/b322931.txt">Table of n, a(n) for n = 0..10000</a>
%H A322931 Wikipedia, <a href="https://en.wikipedia.org/wiki/Hensel%27s_lemma">Hensel's Lemma</a>.
%F A322931 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 5 * (b(n-1)^3 - 3) mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n. - _Seiichi Manyama_, Aug 14 2019
%e A322931 4671710370573^3 == 3 (mod 8^13) in octal.
%o A322931 (PARI) N=100; Vecrev(digits(lift((3+O(2^(3*N)))^(1/3)), 8), N) \\ _Seiichi Manyama_, Aug 14 2019
%o A322931 (Ruby)
%o A322931 def A322931(n)
%o A322931   ary = [3]
%o A322931   a = 3
%o A322931   n.times{|i|
%o A322931     b = (a + 5 * (a ** 3 - 3)) % (8 ** (i + 2))
%o A322931     ary << (b - a) / (8 ** (i + 1))
%o A322931     a = b
%o A322931   }
%o A322931   ary
%o A322931 end
%o A322931 p A322931(100) # _Seiichi Manyama_, Aug 14 2019
%Y A322931 Cf. A225404 (decimal version), A290563, A309698, A322932, A322933.
%K A322931 nonn,base,easy
%O A322931 0,1
%A A322931 _Patrick A. Thomas_, Dec 31 2018