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 1-7 of 7 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.

A319297 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.

Original entry on oeis.org

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

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots. [Typo corrected by Keyang Li, Nov 04 2024]

Examples

			The unique number k in [1, 7^3] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 122 = (233)_7, so the first three terms are 3, 3 and 2.

Links

Crossrefs

Cf. A319097, A319098, A319199.
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
this sequence, A319305, A319555 (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1+sqrt(-3+O(7^(n+1))))/2)\7^n

Formula

Equals A319305*(A212152-1) = A319305*A212152^2, where each A-number represents a 7-adic number.
Equals A319555*(A212155-1) = A319555*A212155^2.

A319305 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098.

Original entry on oeis.org

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

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots. [Typo corrected by Keyang Li, Nov 04 2024]

Examples

			The unique number k in [1, 7^3] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 138 = (255)_7, so the first three terms are 5, 5 and 2.

Links

Crossrefs

Cf. A319097, A319098, A319199.
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
A319297, this sequence, A319555 (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1-sqrt(-3+O(7^(n+1))))/2)\7^n

Formula

Equals A319297*(A212155-1) = A319297*A212155^2, where each A-number represents a 7-adic number.
Equals A319555*(A212152-1) = A319555*A212152^2.

A319555 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319199.

Original entry on oeis.org

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

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.

Examples

			The unique number k in [1, 7^3] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 83 = (146)_7, so the first three terms are 6, 4 and 1.

Links

Crossrefs

Cf. A319097, A319098, A319199.
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
A319297, A319305, this sequence (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(6+O(7^(n+1)), 3))\7^n

Formula

Equals A319297*(A212152-1) = A319297*A212152^2, where each A-number represents a 7-adic number.
Equals A319305*(A212155-1) = A319305*A212155^2.

A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 0

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the former one. See A318960 for detailed information.

Examples

			...10110001110011100100110001100000010110101.

Links

Crossrefs

Cf. A318960.
Digits of p-adic integers:
this sequence, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

Programs

  • PARI
    a(n) = truncate(-sqrt(-7+O(2^(n+2))))\2^n

Formula

a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if A318960(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318963(n) for n >= 1.
For n >= 2, a(n) = (A318960(n+1) - A318960(n))/2^n.

Extensions

Corrected by Jianing Song, Aug 28 2019

A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0
Offset: 0

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the latter one. See A318961 for detailed information.

Examples

			...01001110001100011011001110011111101001011.

Links

Crossrefs

Cf. A318961.
Digits of p-adic integers:
A318962, this sequence (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

Programs

  • PARI
    a(n) = if(n==1, 1, truncate(sqrt(-7+O(2^(n+2))))\2^n)

Formula

a(0) = a(1) = 1; for n >= 2, a(n) = 0 if A318961(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318962(n) for n >= 1.
For n >= 2, a(n) = (A318961(n+1) - A318961(n))/2^n.

Extensions

Corrected by Jianing Song, Aug 28 2019

A212152 Digits of one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, May 02 2012

Keywords

Comments

See A210852 for comments and an approximation to this 7-adic number, called there u. 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):=A210852(n) and c(n):=A210853(n). a(0) = 3, one of the three solutions of x^3+1 == 0 (mod 7).
Since b(n) == 3 (mod 7), a(n) == c(n) (mod 7) for n>0. - Álvar Ibeas, Feb 20 2017
With a(0) = 2, this is the digits of one of the three cube root of 1, the one that is congruent to 2 modulo 7. - Jianing Song, Aug 26 2022

Links

Crossrefs

Cf. A210852 (approximations of (-1)^(1/3)), A212155 (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

  • Maple
    op([1,1,3],select(t -> padic:-ratvaluep(t,1)=3, [padic:-rootp(x^3+1,7,100)])); # Robert Israel, Mar 27 2018
  • Mathematica
    Join[{3}, MapIndexed[#/7^#2[[1]] &, Differences[FoldList[PowerMod[#, 7, 7^#2] &, 3, Range[2, 100]]]]] (* Paolo Xausa, Jan 14 2025 *)

Formula

a(n) = (b(n+1) - b(n))/7^n, n>=1, with b(n):=A210852(n), defined by a recurrence given there. One also finds a Maple program for b(n) there. a(0)=3.
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