A322566 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322564.

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

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 10 (A when written as a 17-adic number). The other, A322565, ends with digit 7.

Examples

			The solution to x^2 == -2 (mod 17^4) such that x == 10 (mod 17) is x == 80029 (mod 17^4), and 80029 is written as G4FA in heptadecimal, so the first four terms are 10, 15, 4 and 16.
		

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));

A322565, this sequence (sqrt(-2)).

Programs

PARI

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

Formula

a(n) = (A322564(n+1) - A322564(n))/17^n.

For n > 0, a(n) = 16 - A322565(n).

Equals A309989*A322562 = A309990*A322561.