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 1-6 of 6 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

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

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).

A322092 Digits of one of the two 13-adic integers sqrt(-3).

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

This square root of -3 in the 13-adic field ends with digit 7. The other, A322091, ends with digit 6.

Examples

			...96AA70B9168BB38376AB76C879074A34388526097.

Links

Crossrefs

Cf. A114525, A286838, A286839, A322087, A322088, A322090, A322091.

Programs

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

Formula

a(n) = (A322090(n+1) - A322090(n))/13^n.
For n > 0, a(n) = 12 - A322091(n).
Equals A286838*A322087 = A286839*A322088.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,7)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

A322089 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-3). Here the 6 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 6, 45, 2073, 15255, 300865, 2899916, 22207152, 273201220, 7614777709, 92450772693, 1333177199334, 4917497987408, 191302178967256, 1705677711928521, 48954194340319989, 202511873382592260, 3529594919298491465, 38131258596823843197, 38131258596823843197, 8809653000849500507259
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 6 modulo 13.
A322090 is the approximation (congruent to 7 mod 13) of another square root of -3 over the 13-adic field.

Examples

			6^2 = 36 = 3*13 - 3.
45^2 = 2025 = 12*13^2 - 3.
2073^2 = 4297329 = 1956*13^3 - 3.

Links

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322090, A322091.

Programs

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

Formula

For n > 0, a(n) = 13^n - A322090(n).
a(n) = Sum_{i=0..n-1} A322091(i)*13^i.
a(n) = A286840(n)*A322086(n) mod 13^n = A286841(n)*A322085(n) mod 13^n.
a(n) == L(13^n,6) (mod 13^n) == (3 + sqrt(10))^(13^n) + (3 - sqrt(10))^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022
Showing 1-6 of 6 results.

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

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