A212153 - OEIS

%I #28 Aug 26 2022 23:40:45

%S 0,5,19,19,1048,15454,82682,82682,3376854,14906456,135967277,

%T 700917775,4655571261,18496858462,406052900090,3797168264335,

%U 22787414304107,188952067152112,654213095126526,654213095126526,57648689021992241

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

%C See A210852 for comments and the approximation of one of the other three 7-adic integers (-1)^(1/3), called there u.

%C The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 5^(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 also. The proof can be done by showing that each term in the binomial expansion of (5^(7^(n-1)))^3 + 1 = (2*3^2*7 - 1)^(7^(n-1)) + 1  has a factor 7^n.

%C a(n) == 5 (mod 7), n>=1. This follows from the formula given above, and 5^(7^(n-1)) == 5 (mod 7), n>=1 (proof by induction).

%C The digit t(n), n>=0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to the present 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 = A212154(n). t(0):=5. For these digits see A212155. The 7-adic number is, read from right to left,

%C ...3452150062464440013235234613550254541223240463025 =: v.

%C   a(n) is obtained from reading v in base 7, and adding the first n terms.

%C One can show directly that a(n) = 7^n + 1 - x(n), n>=1, with x(n) = A210852(n), and z(n) = 7^n - 1 = 6*A023000(n), n>=0.

%C Iff a(n+1) = a(n) then t(n) = A212155(n) = 0.

%C 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 x(n) == 3 (mod 7), a(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The x- and z- sequences are given in A210852 and 6*A023000, respectively.

%C 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 From _Jianing Song_, Aug 26 2022: (Start)

%C a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 5 modulo 7 (if n>0).

%C A210852(n) is the multiplicative inverse of a(n) modulo 7^n. (End)

%H Kenny Lau, <a href="/A212153/b212153.txt">Table of n, a(n) for n = 0..1499</a>

%F Recurrence: a(n) = a(n-1)^7 (mod 7^n), n>=2, a(0):=0, a(1)=5.

%F a(n) == 5^(7^(n-1)) (mod 7^n) == 5 (mod 7), n>= 1.

%F a(n+1) = a(n) + A212155(n)*7^n, n>=1.

%F a(n+1) = Sum_{k=0..n} A212155(k)*7^k, n>=1.

%F a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n>=1 (from 3*a(n)^2*A212155(n) + A212154(n) == 0 (mod 7) and the formula two lines above).

%F a(n) = 7^n + 1 - A210852(n), n>=1.

%e a(4) == 19^7 (mod 7^4) = 893871739 (mod 2401) = 1048.

%e a(4) == 5^343 (mod 7^4) = 1048.

%e a(4) = 19 + 3*7^3 = 1048.

%e a(4) = 5*7^0 + 2*7^1 + 0*7^2 + 3*7^3 = 1048.

%e a(4) = 7^4 + 1 - 1354 = 1048.

%e a(3) = a(2) = 19 because A212155(2) = 0.

%p a:=proc(n) option remember: if n=0 then 0 elif n=1 then 5

%p else modp(a(n-1)^7, 7^n) fi end proc:

%o (PARI) a(n) = lift((1+sqrt(-3+O(7^n)))/2) \\ _Jianing Song_, Aug 26 2022

%Y Cf. A212155 (digits of (-1)^(1/3)), A210852 (approximations of another cube root of -1), 6*A023000 (approximations of -1).

%Y Cf. A048898, A048899 (approximations of the 5-adic integers sqrt(-1)); A319097, A319098, A319199 (approximations of the 7-adic integers 6^(1/3)).

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, May 02 2012