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 16 results. Next

A290563 Coefficients in 5-adic expansion of 3^(1/3).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 06 2017

Keywords

Links

Crossrefs

Cf. A290566, A290568.

Programs

  • PARI
    Vecrev( digits( truncate( (3+O(5^100))^(1/3) ), 5) ) \\ 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 A290563(n)
  A([-3, 0, 0, 1], 2, 5, n)
end
p A290563(100)

A290567 The successive approximations up to 5^n for 5-adic integer 2^(1/3).

Original entry on oeis.org

0, 3, 3, 53, 303, 2178, 5303, 67803, 67803, 849053, 6708428, 6708428, 6708428, 983270928, 983270928, 25397333428, 147467645928, 605231317803, 1368170770928, 5182868036553, 43329840692803, 43329840692803, 43329840692803, 4811701422724053, 52495417243036553
Offset: 0

Views

Author

Seiichi Manyama, Aug 06 2017

Keywords

Comments

x = ...132203,
x^2 = ...344214,
x^3 = ...000002 = 2.

Examples

			a(1) = (   3)_5 = 3,
a(2) = (   3)_5 = 3,
a(3) = ( 203)_5 = 53,
a(4) = (2203)_5 = 303.

Links

Crossrefs

Cf. A290566, A290568.

Programs

  • PARI
    a(n)=truncate((2+O(5^n))^(1/3)); \\ Joerg Arndt, Aug 06 2017

Formula

a(0) = 0 and a(1) = 3, a(n) = a(n-1) + 2 * (a(n-1)^3 - 2) mod 5^n for n > 1.

A321106 Digits of one of the three 13-adic integers 5^(1/3) that is related to A320914.

Original entry on oeis.org

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

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

Examples

			The unique number k in [1, 13^3] and congruent to 7 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 1021 = (607)_13, so the first three terms are 7, 0 and 6.

Links

Crossrefs

Cf. A320914, A320915, A321105, A321107, A321108.
For 5-adic cubic roots, see A290566, A290563, A309443.

Programs

  • PARI
    a(n) = lift(sqrtn(5+O(13^(n+1)), 3) * (-1+sqrt(-3+O(13^(n+1))))/2)\13^n

Formula

a(n) = (A320914(n+1) - A320914(n))/13^n.

A321107 Digits of one of the three 13-adic integers 5^(1/3) that is related to A320915.

Original entry on oeis.org

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

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

Examples

			The unique number k in [1, 13^3] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 177 = (108)_13, so the first three terms are 8, 0 and 1.

Links

Crossrefs

Cf. A320914, A320915, A321105, A321106, A321108.
For 5-adic cubic roots, see A290566, A290563, A309443.

Programs

  • PARI
    a(n) = lift(sqrtn(5+O(13^(n+1)), 3))\13^n

Formula

a(n) = (A320915(n+1) - A320915(n))/13^n.

A321108 Digits of one of the three 13-adic integers 5^(1/3) that is related to A321105.

Original entry on oeis.org

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

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

Examples

			The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 999 = (5BB)_13, so the first three terms are 11, 11 and 5.

Links

Crossrefs

Cf. A320914, A320915, A321105, A321106, A321107.
For 5-adic cubic roots, see A290566, A290563, A309443.

Programs

  • PARI
    a(n) = lift(sqrtn(5+O(13^(n+1)), 3) * (-1-sqrt(-3+O(13^(n+1))))/2)\13^n

Formula

a(n) = (A321105(n+1) - A321105(n))/13^n.

A309443 Coefficients in 5-adic expansion of 4^(1/3).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 03 2019

Keywords

Links

Crossrefs

Cf. A309444.
Digits of p-adic integers:
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3)).

Programs

  • Maple
    op([1,3], padic:-rootp(x^3-4,5,101)); # Robert Israel, Aug 04 2019
  • PARI
    Vecrev(digits(truncate((4+O(5^100))^(1/3)), 5))
  • 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 A309443(n)
  A([-4, 0, 0, 1], 4, 5, n)
end
p A309443(100)

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.

A309445 Coefficients in 7-adic expansion of 2^(1/5).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Aug 03 2019

Keywords

Links

Crossrefs

Cf. A309450.
Digits of p-adic integers:
A290566 (5-adic, 2^(1/3));
A309446 (7-adic, 3^(1/5));
A309447 (7-adic, 4^(1/5));
A309448 (7-adic, 5^(1/5));
A309449 (7-adic, 6^(1/5)).

Programs

  • Maple
    op([1,3], padic:-rootp(x^5-2,7,101)); # Robert Israel, Aug 04 2019
  • PARI
    Vecrev(digits(truncate((2+O(7^100))^(1/5)), 7))
  • 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 A309445(n)
  A([-2, 0, 0, 0, 0, 1], 4, 7, n)
end
p A309445(100)
Showing 1-10 of 16 results. Next

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