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 1-5 of 5 results.

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

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).

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

A321073 One of the two successive approximations up to 11^n for 11-adic integer sqrt(3). Here the 6 (mod 11) case (except for n = 0).

Original entry on oeis.org

0, 6, 94, 578, 3240, 135009, 296060, 2067621, 118990647, 1619502814, 3977450505, 211476847313, 782100188535, 22751098825582, 229887371689168, 609637205272409, 38204870730013268, 84154600593585429, 3622283800088641826, 42541704994534262193, 654132609478679725103
Offset: 0

Views

Author

Jianing Song, Oct 27 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == 3 (mod 11^n) in the range [0, 11^n - 1] and congruent to 6 modulo 11.
A321072 is the approximation (congruent to 5 mod 11) of another square root of 3 over the 11-adic field.

Examples

			6^2 = 36 = 3 + 3*11.
94^2 = 8836 = 3 + 73*11^2.
578^2 = 334084 = 3 + 251*11^3.

Links

Crossrefs

Cf. A321072, A321075.

Programs

  • PARI
    a(n) = truncate(-sqrt(3+O(11^n)))

Formula

For n > 0, a(n) = 11^n - A321072(n).
a(n) = Sum_{i=0..n-1} A321075(i)*11^i.
a(n) == (3 + sqrt(8))^(11^n) + (3 - sqrt(8))^(11^n) (mod 11^n). - Peter Bala, Dec 04 2022
Showing 1-5 of 5 results.

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

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