A286839 -id:A286839 - cp's OEIS frontend

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

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

Jianing Song, Aug 26 2019

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).

			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.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

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)).

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

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

Jianing Song, Aug 26 2019

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.

			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.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

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)).

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

a(n) = (A286878(n+1) - A286878(n))/17^n.

For n > 0, a(n) = 16 - A309989(n).

A322091 Digits of one of the two 13-adic integers sqrt(-3).

Original entry on oeis.org

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

Offset: 0

Jianing Song, Nov 26 2018

This square root of -3 in the 13-adic field ends with digit 6. The other, A322092, ends with digit 7.

			...36225C13B64119495621560453C582989447A6C36.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, p-adic number

Cf. A114525, A286838, A286839, A322087, A322088, A322089, A322092.

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

a(n) = (A322089(n+1) - A322089(n))/13^n.
For n > 0, a(n) = 12 - A322092(n).
Equals A286838*A322088 = A286839*A322087.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,6)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

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

Jianing Song, Nov 26 2018

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

			...96AA70B9168BB38376AB76C879074A34388526097.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, p-adic number

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

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

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

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

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

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PARI

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A322091 Digits of one of the two 13-adic integers sqrt(-3).

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

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