cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, May 02 2012

Keywords

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

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.

A212155 Digits of one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

5, 2, 0, 3, 6, 4, 0, 4, 2, 3, 2, 2, 1, 4, 5, 4, 5, 2, 0, 5, 5, 3, 1, 6, 4, 3, 2, 5, 3, 2, 3, 1, 0, 0, 4, 4, 4, 6, 4, 2, 6, 0, 0, 5, 1, 2, 5, 4, 3, 2, 5, 3, 2, 6, 3, 3, 4, 2, 2, 2, 1, 5, 6, 2, 6, 4, 6, 3, 5, 6, 4, 0, 5, 1, 4, 1, 1, 0, 6, 0, 4, 2, 2, 4, 5, 0, 3, 2, 1, 1, 5, 6, 2, 4, 2, 2, 1, 1, 5, 3
Offset: 0

Views

Author

Wolfdieter Lang, May 02 2012

Keywords

Comments

See A210853 for comments and an approximation to this 7-adic number, called there v. See also A048898 for references on p-adic numbers.
a(n), n>=1, is the (unique) solution of the linear congruence 3 * b(n)^2 * a(n) + c(n) == 0 (mod 7), with b(n):=A212153(n) and c(n):=A212154(n). a(0) = 5, one of the three solutions of X^3+1 == 0 (mod 7).
Since b(n) == 5 (mod 7), a(n) == 4 * c(n) (mod 7) for n>0. - Álvar Ibeas, Feb 20 2017
With a(0) = 4, this is the digits of one of the three cube root of 1, the one that is congruent to 4 modulo 7. - Jianing Song, Aug 26 2022

Links

Crossrefs

Cf. A212153 (approximations of (-1)^(1/3)), A212152 (digits of another cube root of -1), 6*A000012 (digits of -1).
Cf. A210850, A210851 (digits of the 5-adic integers sqrt(-1)); A319297, A319305, A319555 (digits of the 7-adic integers 6^(1/3)).

Programs

  • Mathematica
    Join[{5}, MapIndexed[#/7^#2[[1]] &, Differences[FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 100]]]]] (* Paolo Xausa, Jan 14 2025 *)

Formula

a(n) = (b(n+1) - b(n))/7^n, n>=1, with b(n):=A212153(n), defined by a recurrence given there. One also finds there a Maple program for b(n). a(0)=5.
a(n) = 6 - A212152(n), for n>0. - Álvar Ibeas, Feb 21 2017

A212156 a(n) = ((6*A023000(n))^3 + 1)/7^n, n >= 0.

Original entry on oeis.org

1, 31, 2257, 116623, 5757601, 282424831, 13840934257, 678220602223, 33232913275201, 1628413476849631, 79792265450186257, 3909821042651007823, 191581231339042552801, 9387480337357087274431, 459986536542705291758257
Offset: 0

Views

Author

Wolfdieter Lang, May 02 2012

Keywords

Comments

a(n) is an integer because 6*A023000(n) is one of three solution of X(n)^3+1 == 0 (mod 7^n), namely the one satisfying also X(n) == 6 (mod 7) == -1 (mod 7).
See the comments on A210852, and the Nagell reference given in A210848.

Examples

			a(0) = 1/1 = 1.
a(3) = ((6*57)^3 + 1)/7^3 = 40001689/343  = 116623,  (b(3) = 48^7 (mod 7^3) = 342 = 6*57).

Crossrefs

Cf. A210848, A210849 (the p=5 case). A210853, A212154.

Formula

a(n) = (b(n)^3+1)/7^n, n>=0, with b(n):=6*A023000(n) given by a recurrence obtained from the one of A023000. There also programs for b(n)/6 are given.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.