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A210852 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

%I A210852 #40 Aug 26 2022 23:40:41
%S A210852 0,3,31,325,1354,1354,34968,740862,2387948,25447152,146507973,
%T A210852 1276408969,9185715941,78392151946,272170172760,950393245609,
%U A210852 10445516265495,43678446835096,974200502783924,10744682090246618,22143577275619761
%N A210852 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).
%C A210852 The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 3^(7^(n-1)) (mod 7^n), n >= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n >= 1. a(0) = 0 satisfies this congruence as well. The proof can be done by showing that each term in the binomial expansion of (3^(7^(n-1)))^3 +1 = (28 -1)^(7^(n-1)) + 1 has a factor 7^n.
%C A210852 a(n) == 3 (mod 7), n >= 1. This follows from the formula given above, and 3^(7^(n-1)) == 3 (mod 7), n >= 1 (proof by induction).
%C A210852 The digit t(n), n >= 0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to this sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n >= 1, with b(n):= (a(n)^3+1)/7^n = A210853(n). t(0):=3, one of the three solutions of X^3 + 1 == 0 (mod 7). For these digits see A212152. The 7-adic number is, read from right to left, ...3143214516604202226653431432053116412125443426203643 =: u.
%C A210852   a(n) is obtained from reading u in base 7, and adding the first n terms.
%C A210852 One can show directly that a(n) = 7^n + 1 - y(n), n >= 1, with y(n) = A212153(n) and z(n) = 7^n - 1 = 6*A023000(n), n >= 0.
%C A210852 Iff a(n+1) = a(n) then t(n) = A212152(n) = 0.
%C A210852 See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 + 1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 + 1 == 0 (mod 7^n) exactly three solutions for each n >= 1, which can be chosen as a(n) == 3 (mod 7), y(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The y- and z- sequences are given in A212153 and 6*A023000, respectively.
%C A210852 For n > 0, a(n) - 1 (== a(n)^2 (mod 7^n)) and 7^n - a(n) (== a(n)^4 (mod 7^n)) are the two primitive cubic roots of unity in Z/(7^n Z). - _Álvar Ibeas_, Feb 20 2017
%C A210852 From _Jianing Song_, Aug 26 2022: (Start)
%C A210852 a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 3 modulo 7 (if n>0).
%C A210852 A212153(n) is the multiplicative inverse of a(n) modulo 7^n. (End)
%H A210852 Kenny Lau, <a href="/A210852/b210852.txt">Table of n, a(n) for n = 0..1183</a>
%F A210852 Recurrence: a(n) = a(n-1)^7 (mod 7^n), n >= 2, a(0)=0, a(1)=3.
%F A210852 a(n) == 3^(7^(n-1)) (mod 7^n) == 3 (mod 7), n >= 1.
%F A210852 a(n+1) = a(n) + A212152(n)*7^n, n >= 1.
%F A210852 a(n+1) = Sum_{k=0..n} A212152(k)*7^k, n >= 1.
%F A210852 a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n >= 1 (from 3*a(n)^2* A212152(n) + A210853(n) == 0 (mod 7) and the second-to-last formula from above).
%F A210852 a(n) = 7^n + 1 - A212153(n), n >= 1.
%e A210852 a(3) == 31^7 (mod 7^3) == 27512614111 (mod 343) = 325.
%e A210852 a(3) == 3^49 (mod 7^3) = 325.
%e A210852 a(3) = 31 + 6*7^2 = 325.
%e A210852 a(3) = 3*7^0 + 4*7^1 + 6*7^2 = 325.
%e A210852 a(3) = 7^3 +1 - 19 = 325.
%e A210852 a(5) = a(4) = 1354 because A212152(4) = 0.
%p A210852 a:=proc(n) option remember: if n=0 then 0 elif n=1 then 3
%p A210852 else modp(a(n-1)^7, 7^n) fi end proc: [seq(a(n),n=0..30)];
%t A210852 a[n_] := a[n] = Which[n == 0, 0, n == 1, 3, True, Mod[a[n-1]^7, 7^n]]; Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Mar 05 2014, after Maple *)
%o A210852 (PARI) a(n) = lift((1-sqrt(-3+O(7^n)))/2) \\ _Jianing Song_, Aug 26 2022
%Y A210852 Cf. A212152 (digits of (-1)^(1/3)), A212153 (approximations of another cube root of -1), 6*A023000 (approximations of -1).
%Y A210852 Cf. A048898, A048899 (approximations of the 5-adic integers sqrt(-1)); A319097, A319098, A319199 (approximations of the 7-adic integers 6^(1/3)).
%K A210852 nonn,easy
%O A210852 0,2
%A A210852 _Wolfdieter Lang_, May 02 2012