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 1-10 of 11 results. Next

A322087 Digits of one of the two 13-adic integers sqrt(3).

Original entry on oeis.org

4, 8, 6, 8, 12, 2, 1, 9, 9, 10, 1, 6, 4, 10, 6, 11, 11, 9, 5, 5, 0, 5, 2, 5, 8, 0, 8, 7, 3, 5, 3, 12, 0, 3, 10, 3, 5, 8, 1, 12, 11, 8, 7, 0, 3, 1, 4, 9, 9, 9, 1, 10, 6, 12, 2, 7, 3, 5, 1, 6, 12, 1, 1, 12, 10, 5, 6, 11, 7, 8, 12, 10, 1, 3, 5, 5, 5, 7, 11, 1, 5
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

This square root of 3 in the 13-adic field ends with digit 4. The other, A322088, ends with digit 9.

Examples

			...BC1853A30C35378085250559BB6A461A9912C8684.

Links

Crossrefs

Cf. A322085.
Digits of p-adic integers:
A321074, A321075 (11-adic, sqrt(3));
this sequence, A322088 (13-adic, sqrt(3));
A286838, A286839 (13-adic, sqrt(-1));
A322091, A322092 (13-adic, sqrt(-3)).

Programs

  • PARI
    a(n) = truncate(sqrt(3+O(13^(n+1))))\13^n

Formula

a(n) = (A322085(n+1) - A322085(n))/13^n.
For n > 0, a(n) = 12 - A322088(n).
Equals A286838*A322091 = A286839*A322092.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - Peter Bala, Dec 04 2022

A290557 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(2). These are the numbers congruent to 3 mod 7 (except for the initial 0).

Original entry on oeis.org

0, 3, 10, 108, 2166, 4567, 38181, 155830, 1802916, 24862120, 266983762, 1961835256, 5916488742, 19757775943, 116646786350, 116646786350, 9611769806236, 42844700375837, 275475214363044, 6789129606004840, 75182500718243698, 154974767015855699
Offset: 0

Views

Author

Seiichi Manyama, Aug 05 2017

Keywords

Comments

x = ...216213,
x^2 = ...000002 = 2.

Examples

			a(1) = (    3)_7 = 3,
a(2) = (   13)_7 = 10,
a(3) = (  213)_7 = 108,
a(4) = ( 6213)_7 = 2166,
a(5) = (16213)_7 = 4567.

Links

Crossrefs

Cf. A034945, A051277, A290559.

Programs

Formula

a(0) = 0 and a(1) = 3, a(n) = a(n-1) + (a(n-1)^2 - 2) mod 7^n for n > 1.
a(n) == 2*T(7^n, 3/2) (mod 7^n) == ((3 + sqrt(5))/2)^(7^n) + ((3 - sqrt(5))/2)^(7^n) (mod 7^n), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 03 2022

A290559 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(2). These are the numbers congruent to 4 mod 7 (except for the initial 0).

Original entry on oeis.org

0, 4, 39, 235, 235, 12240, 79468, 667713, 3961885, 15491487, 15491487, 15491487, 7924798459, 77131234464, 561576286499, 4630914723593, 23621160763365, 189785813611370, 1352938383547405, 4609765579368303, 4609765579368303, 403571097067428308
Offset: 0

Views

Author

Seiichi Manyama, Aug 05 2017

Keywords

Comments

x = ...450454,
x^2 = ...000002 = 2.

Examples

			a(1) = (    4)_7 = 4,
a(2) = (   54)_7 = 39,
a(3) = (  454)_7 = 235,
a(4) = (  454)_7 = 235,
a(5) = (50454)_7 = 12240.

Links

Crossrefs

Cf. A051277, A290557, A290558.

Programs

  • PARI
    a(n) = if (n==0, 0, 7^n - truncate(sqrt(2+O(7^n)))); \\ Michel Marcus, Aug 06 2017

Formula

If n > 0, a(n) = 7^n - A290557(n).
a(0) = 0 and a(1) = 4, a(n) = a(n-1) + 6 * (a(n-1)^2 - 2) mod 7^n for n > 1.
a(n) == 2*T(7^n, 2) (mod 7^n) == (2 + sqrt(3))^(7^n) + (2 - sqrt(3))^(7^n) (mod 7^n), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 03 2022

A322088 Digits of one of the two 13-adic integers sqrt(3).

Original entry on oeis.org

9, 4, 6, 4, 0, 10, 11, 3, 3, 2, 11, 6, 8, 2, 6, 1, 1, 3, 7, 7, 12, 7, 10, 7, 4, 12, 4, 5, 9, 7, 9, 0, 12, 9, 2, 9, 7, 4, 11, 0, 1, 4, 5, 12, 9, 11, 8, 3, 3, 3, 11, 2, 6, 0, 10, 5, 9, 7, 11, 6, 0, 11, 11, 0, 2, 7, 6, 1, 5, 4, 0, 2, 11, 9, 7, 7, 7, 5, 1, 11, 7
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

This square root of 3 in the 13-adic field ends with digit 9. The other, A322087, ends with digit 4.

Examples

			...10B47929C097954C47A7C773116286B233BA04649.

Links

Crossrefs

Cf. A286838, A286839, A322086, A322087, A322091, A322092.

Programs

  • PARI
    a(n) = truncate(-sqrt(3+O(13^(n+1))))\13^n

Formula

a(n) = (A322086(n+1) - A322086(n))/13^n.
For n > 0, a(n) = 12 - A322087(n).
Equals A286838*A322092 = A286839*A322091.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,9/2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - Peter Bala, Dec 04 2022

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

A309989 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

4, 2, 10, 5, 12, 16, 12, 8, 13, 3, 14, 0, 6, 1, 0, 15, 1, 8, 14, 5, 7, 16, 14, 1, 5, 13, 9, 6, 5, 12, 16, 15, 9, 16, 14, 12, 16, 1, 3, 6, 4, 10, 15, 5, 16, 12, 2, 1, 5, 4, 0, 15, 2, 11, 14, 9, 5, 1, 11, 16, 15, 7, 5, 6, 14, 3, 12, 0, 0, 11, 12, 13, 9, 5, 4, 16, 13
Offset: 0

Views

Author

Jianing Song, Aug 26 2019

Keywords

Comments

This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).

Examples

			The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.

Links

Crossrefs

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
this sequence, A309990 (17-adic, sqrt(-1)).

Programs

  • PARI
    a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n

Formula

a(n) = (A286877(n+1) - A286877(n))/17^n.
For n > 0, a(n) = 16 - A309990(n).

A309990 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

13, 14, 6, 11, 4, 0, 4, 8, 3, 13, 2, 16, 10, 15, 16, 1, 15, 8, 2, 11, 9, 0, 2, 15, 11, 3, 7, 10, 11, 4, 0, 1, 7, 0, 2, 4, 0, 15, 13, 10, 12, 6, 1, 11, 0, 4, 14, 15, 11, 12, 16, 1, 14, 5, 2, 7, 11, 15, 5, 0, 1, 9, 11, 10, 2, 13, 4, 16, 16, 5, 4, 3, 7, 11, 12, 0
Offset: 0

Views

Author

Jianing Song, Aug 26 2019

Keywords

Comments

This square root of -1 in the 17-adic field ends with digit 13 (D when written as a 17-adic number). The other, A309989, ends with digit 4.

Examples

			The solution to x^2 == -1 (mod 17^4) such that x == 13 (mod 17) is x == 56028 (mod 17^4), and 56028 is written as B6ED in heptadecimal, so the first four terms are 13, 14, 6 and 11.

Links

Crossrefs

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
A309989, this sequence (17-adic, sqrt(-1)).

Programs

  • PARI
    a(n) = truncate(-sqrt(-1+O(17^(n+1))))\17^n

Formula

a(n) = (A286878(n+1) - A286878(n))/17^n.
For n > 0, a(n) = 16 - A309989(n).

A290558 Coefficients in 7-adic expansion of sqrt(2).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 05 2017

Keywords

Links

Crossrefs

Cf. A051277, A290559.

Programs

  • Maple
    t := proc(n) option remember; if n = 1 then 4 else irem(t(n-1)^7 - 7*t(n-1)^5 + 14*t(n-1)^3 - 7*t(n-1), 7^n) end if; end:
convert(t(100), base, 7); # Peter Bala, Nov 20 2022
  • PARI
    { my(v=Vecrev( digits( truncate( (2+O(7^100))^(1/2) ), 7) )); vector(#v,k,6-v[k]+(k==1)) } \\ Joerg Arndt, Aug 06 2017
  • Ruby
    require 'OpenSSL'
def f_a(ary, a)
  (0..ary.size - 1).inject(0){|s, i| s + ary[i] * a ** i}
end
def df(ary)
  (1..ary.size - 1).map{|i| i * ary[i]}
end
def A(c_ary, k, m, n)
  x = OpenSSL::BN.new((-f_a(df(c_ary), k)).to_s).mod_inverse(m).to_i % m
  f_ary = c_ary.map{|i| x * i}
  f_ary[1] += 1
  d_ary = []
  ary = [0]
  a, mod = k, m
  (n + 1).times{|i|
    b = a % mod
    d_ary << (b - ary[-1]) / m ** i
    ary << b
    a = f_a(f_ary, b)
    mod *= m
  }
  d_ary
end
def A290558(n)
  A([-2, 0, 1], 4, 7, n)
end
p A290558(100)

Formula

a(n) = 6 - A051277(n) for n > 0.
Equals the 7-adic limit as n -> oo of 2*T(7^n,2) = the 7-adic limit as n -> oo of (2 + sqrt(3))^(7^n) + (2 - sqrt(3))^(7^n), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Nov 20 2022

A321074 Digits of one of the two 11-adic integers sqrt(3).

Original entry on oeis.org

5, 2, 6, 8, 1, 9, 9, 4, 3, 9, 2, 8, 3, 4, 9, 1, 9, 3, 3, 0, 5, 5, 0, 9, 8, 4, 1, 9, 6, 9, 3, 0, 7, 5, 8, 6, 3, 9, 0, 9, 7, 7, 9, 8, 10, 5, 8, 6, 9, 3, 5, 9, 4, 7, 2, 1, 1, 0, 1, 0, 8, 1, 6, 5, 7, 10, 8, 2, 4, 7, 8, 7, 2, 3, 3, 1, 10, 6, 0, 10, 0, 6, 2, 5, 1, 10, 3
Offset: 0

Views

Author

Jianing Song, Oct 27 2018

Keywords

Comments

This square root of 3 in the 11-adic field ends with digit 5. The other, A321075, ends with digit 6.

Examples

			...9093685703969148905503391943829349918625.

Links

Crossrefs

Cf. A321072, A321075, A322087, A322088.

Programs

  • PARI
    a(n) = truncate(sqrt(3+O(11^(n+1))))\11^n
  • PARI
    seq(n)={Vecrev(digits(truncate(sqrt(3 + O(11^n))), 11), n)} \\ Andrew Howroyd, Nov 03 2018

Formula

a(n) = (A321072(n+1) - A321072(n))/11^n.
For n > 0, a(n) = 10 - A321075(n).
This 11-adic integer equals the 11-adic limit as n -> oo of 2*T(11^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 05 2022
Showing 1-10 of 11 results. Next

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