A322565 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322563.

7, 1, 12, 0, 9, 0, 16, 8, 5, 16, 14, 0, 1, 15, 11, 16, 15, 8, 13, 15, 11, 5, 11, 3, 9, 16, 16, 15, 3, 3, 0, 15, 7, 15, 16, 3, 14, 9, 12, 5, 2, 2, 4, 12, 12, 11, 11, 0, 9, 15, 12, 2, 9, 14, 2, 10, 6, 0, 8, 5, 15, 6, 6, 14, 9, 2, 10, 1, 7, 2, 13, 12, 3, 13, 6, 16

Offset: 0

Author

Jianing Song, Aug 29 2019

Keywords

• nonn
• base

Comments

This square root of -2 in the 17-adic field ends with digit 7. The other, A322566, ends with digit 10 (A when written as a 17-adic number).

Examples

			The solution to x^2 == -2 (mod 17^4) such that x == 7 (mod 17) is x == 3492 (mod 17^4), and 3492 is written as C17 in heptadecimal, so the first four terms are 7, 1, 12 and 0.
		

Links

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

Crossrefs

Cf. A322563, A322564.

Digits of 17-adic square roots:

A309989, A309990 (sqrt(-1));

A322561, A322562 (sqrt(2));

this sequence, A322566 (sqrt(-2)).

Programs

PARI

a(n) = truncate(sqrt(-2+O(17^(n+1))))\17^n

Formula

a(n) = (A322563(n+1) - A322563(n))/17^n.

For n > 0, a(n) = 16 - A322566(n).

Equals A309989*A322561 = A309990*A322562.