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

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

Author

Jianing Song, Aug 26 2019

Keywords

• nonn
• base

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

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

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).