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

Author

Jianing Song, Aug 26 2019

Keywords

• nonn
• base

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

• 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));

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).