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

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)

A320914 One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 7 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 7, 7, 1021, 20794, 77916, 4533432, 57628331, 810610535, 8967917745, 40781415864, 592215383260, 22098140111704, 208482821091552, 3842984100198588, 23529866028695033, 586574689183693360, 5244490953465952247, 74447818308516655711, 524269446116346228227, 9295791188369022892289
Offset: 0

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 7 mod 13 such that k^3 - 5 is divisible by 13^n.
For k not divisible by 13, 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, 13^2] and congruent to 7 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 7, so a(2) = 7.
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, so a(3) = 1021.

Links

Crossrefs

Cf. A320915, A321105, A321106, A321107, A321108.
For 5-adic cubic roots, see A290567, A290568, A309444.

Programs

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

A320915 One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 8 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 8, 8, 177, 11162, 211089, 211089, 24345134, 777327338, 7303173106, 113348166836, 1629791577175, 12382753941397, 222065520043726, 1130690839820485, 16880196382617641, 272809661453071426, 5596142534918510154, 14246558454299848087, 576523593214086813732, 4962284464340425145763
Offset: 0

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 8 mod 13 such that k^3 - 5 is divisible by 13^n.
For k not divisible by 13, 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, 13^2] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 8, so a(2) = 8.
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, so a(3) = 177.

Links

Crossrefs

Cf. A320914, A321105, A321106, A321107, A321108.
For 5-adic cubic roots, see A290567, A290568, A309444.

Programs

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

A321105 One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 11 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 11, 154, 999, 25166, 82288, 82288, 43523569, 43523569, 4937907895, 121587400998, 1362313827639, 12115276191861, 175201872049228, 2901077831379505, 10775830602778083, 471448867729594896, 6460198350378213465, 23761030189140889331, 361127251045013068718, 4746888122171351400749
Offset: 0

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 11 mod 13 such that k^3 - 5 is divisible by 13^n.
For k not divisible by 13, 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, 13^2] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 154, so a(2) = 154.
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, so a(3) = 999.

Links

Crossrefs

Cf. A320914, A320915, A321106, A321107, A321108.
For 5-adic cubic roots, see A290567, A290568, A309444.

Programs

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

A319097 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 3 (mod 7) case (except for n = 0).

Original entry on oeis.org

0, 3, 24, 122, 808, 10412, 111254, 817148, 1640691, 24699895, 186114323, 186114323, 6118094552, 6118094552, 490563146587, 2525232365134, 26263039914849, 59495970484450, 1222648540420485, 6107889334151832, 74501260446390690, 234085793041614692, 1351177521208182706, 24810103812706111000, 134285093173029776372
Offset: 0

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 3 mod 7 such that k^3 - 6 is divisible by 7^n.
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.

Examples

			The unique number k in [1, 7^2] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 24, so a(2) = 24.
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, so a(3) = 122.

Links

Crossrefs

Cf. A319297, A319305, A319555.
Approximations of p-adic cubic roots:
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
this sequence, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

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

Formula

a(n) = A319098(n)*(A210852(n)-1) mod 7^n = A319098(n)*A210852(n)^2 mod 7^n.
a(n) = A319199(n)*(A212153(n)-1) mod 7^n = A319199(n)*A212153(n)^2 mod 7^n.

A319098 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 5 (mod 7) case (except for n = 0).

Original entry on oeis.org

0, 5, 40, 138, 824, 3225, 87260, 793154, 793154, 29617159, 191031587, 1320932583, 7252912812, 7252912812, 7252912812, 2041922131359, 16284606661188, 82750467800390, 1013272523749218, 9155340513301463, 31953130884047749, 111745397181659750, 670291261264943757
Offset: 0

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 5 mod 7 such that k^3 - 6 is divisible by 7^n.
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.

Examples

			The unique number k in [1, 7^2] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 40, so a(2) = 40.
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, so a(3) = 138.

Links

Crossrefs

Cf. A319297, A319305, A319555.
Approximations of p-adic cubic roots:
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, this sequence, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

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

Formula

a(n) = A319097(n)*(A212153(n)-1) mod 7^n = A319097(n)*A212153(n)^2 mod 7^n.
a(n) = A319199(n)*(A210852(n)-1) mod 7^n = A319199(n)*A210852(n)^2 mod 7^n.

A319199 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 6 (mod 7) case (except for n = 0).

Original entry on oeis.org

0, 6, 34, 83, 769, 3170, 36784, 36784, 3330956, 26390160, 187804588, 470279837, 470279837, 83518003043, 180407013450, 180407013450, 23918214563165, 90384075702367, 1020906131651195, 7534560523292991, 53130141264785563, 212714673860009565, 1888352266109861586
Offset: 0

Views

Author

Jianing Song, Aug 27 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 6 mod 7 such that k^3 - 6 is divisible by 7^n.
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.

Examples

			The unique number k in [1, 7^2] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 34, so a(2) = 24.
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, so a(3) = 122.

Links

Crossrefs

Cf. A319297, A319305, A319555.
Approximations of p-adic cubic roots:
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, this sequence (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

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

Formula

a(n) = A319097(n)*(A210852(n)-1) mod 7^n = A319097(n)*A210852(n)^2 mod 7^n.
a(n) = A319098(n)*(A212153(n)-1) mod 7^n = A319098(n)*A212153(n)^2 mod 7^n.

A322701 The successive approximations up to 2^n for 2-adic integer 3^(1/3).

Original entry on oeis.org

0, 1, 3, 3, 11, 27, 59, 123, 123, 379, 379, 379, 379, 4475, 12667, 29051, 61819, 127355, 127355, 127355, 127355, 127355, 2224507, 2224507, 2224507, 19001723, 52556155, 119665019, 253882747, 253882747, 253882747, 1327624571, 3475108219, 7770075515
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 3 (mod 2^n) in the range [0, 2^n - 1].

Examples

			11^3 = 1331 = 83*2^4 + 3;
27^3 = 19683 = 615*2^5 + 3;
59^3 = 205379 = 3209*2^6 + 3.

Links

Crossrefs

For the digits of 3^(1/3), see A323000.
Approximations of p-adic cubic roots:
this sequence (2-adic, 3^(1/3));
A322926 (2-adic, 5^(1/3));
A322934 (2-adic, 7^(1/3));
A322999 (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(3+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 3 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).

A322926 The successive approximations up to 2^n for 2-adic integer 5^(1/3).

Original entry on oeis.org

0, 1, 1, 5, 13, 29, 29, 93, 93, 93, 605, 1629, 3677, 3677, 3677, 20061, 20061, 20061, 151133, 151133, 151133, 151133, 151133, 4345437, 4345437, 21122653, 54677085, 54677085, 188894813, 457330269, 457330269, 457330269, 2604813917, 6899781213, 6899781213
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 5 (mod 2^n) in the range [0, 2^n - 1].

Examples

			13^3 = 2197 = 137*2^4 + 5;
29^3 = 24389 = 762*2^5 + 5 = 381*2^6 + 5;
93^3 = 804357 = 6284*2^7 + 5 = 3142*2^8 + 5 = 1571*2^9 + 5.

Links

Crossrefs

For the digits of 5^(1/3), see A323045.
Approximations of p-adic cubic roots:
A322701 (2-adic, 3^(1/3));
this sequence (2-adic, 5^(1/3));
A322934 (2-adic, 7^(1/3));
A322999 (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(5+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 5 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).

A322934 The successive approximations up to 2^n for 2-adic integer 7^(1/3).

Original entry on oeis.org

0, 1, 3, 7, 7, 23, 23, 23, 151, 407, 407, 1431, 3479, 3479, 11671, 11671, 44439, 109975, 241047, 503191, 1027479, 2076055, 2076055, 6270359, 6270359, 6270359, 6270359, 6270359, 6270359, 274705815, 811576727, 1885318551, 1885318551, 6180285847
Offset: 0

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

a(n) is the unique solution to x^3 == 7 (mod 2^n) in the range [0, 2^n - 1].

Examples

			7^3 = 343 = 21*2^4 + 7;
23^3 = 12167 = 380*2^5 + 7 = 190*2^6 + 7 = 95*2^7 + 7;
151^3 = 3442951 = 13449*2^8 + 7.

Links

Crossrefs

For the digits of 7^(1/3), see A323095.
Approximations of p-adic cubic roots:
A322701 (2-adic, 3^(1/3));
A322926 (2-adic, 5^(1/3));
this sequence (2-adic, 7^(1/3));
A322999 (2-adic, 9^(1/3));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A309444 (5-adic, 4^(1/3));
A319097, A319098, A319199 (7-adic, 6^(1/3));
A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

  • PARI
    a(n) = lift(sqrtn(7+O(2^n), 3))

Formula

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 7 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
Showing 1-10 of 11 results. Next

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