A212153 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

0, 5, 19, 19, 1048, 15454, 82682, 82682, 3376854, 14906456, 135967277, 700917775, 4655571261, 18496858462, 406052900090, 3797168264335, 22787414304107, 188952067152112, 654213095126526, 654213095126526, 57648689021992241, 456610020510052246, 2132247612759904267

Offset: 0

Author

Wolfdieter Lang, May 02 2012

Keywords

• nonn
• easy

Comments

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

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.

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).

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,

...3452150062464440013235234613550254541223240463025 =: v.

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

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.

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

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.

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

From Jianing Song, Aug 26 2022: (Start)

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

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

Examples

			a(4) == 19^7 (mod 7^4) = 893871739 (mod 2401) = 1048.
a(4) == 5^343 (mod 7^4) = 1048.
a(4) = 19 + 3*7^3 = 1048.
a(4) = 5*7^0 + 2*7^1 + 0*7^2 + 3*7^3 = 1048.
a(4) = 7^4 + 1 - 1354 = 1048.
a(3) = a(2) = 19 because A212155(2) = 0.
		

Links

• Kenny Lau, Table of n, a(n) for n = 0..1499

Crossrefs

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

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

Programs

Maple

a:=proc(n) option remember: if n=0 then 0 elif n=1 then 5
else modp(a(n-1)^7, 7^n) fi end proc:

Mathematica

Join[{0}, FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 25]]] (* Paolo Xausa, Jan 14 2025 *)

PARI

a(n) = lift((1+sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022

Formula

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

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

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

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

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).

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