A322562 Digits of one of the two 17-adic integers sqrt(2) that is related to A322560.

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

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 11 (B when written as a 17-adic number). The other, A322561, ends with digit 6.

Examples

			The solution to x^2 == 2 (mod 17^4) such that x == 11 (mod 17) is x == 39927 (mod 17^4), and 39927 is written as 822B in heptadecimal, so the first four terms are 11, 2, 2 and 8.
		

Links

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

Crossrefs

Cf. A322559, A322560.

Digits of 17-adic square roots:

A309989, A309990 (sqrt(-1));

A322561, this sequence (sqrt(2));

A322565, A322566 (sqrt(-2)).

Programs

PARI

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

Formula

a(n) = (A322560(n+1) - A322560(n))/17^n.

For n > 0, a(n) = 16 - A322561(n).

Equals A309989*A322565 = A309990*A322566.