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

0,1

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.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Wikipedia, p-adic number

Formula

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

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

Equals A309989*A322565 = A309990*A322566.

Example

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.

Prog

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

Crossrefs

Cf. A322559, A322560.

Digits of 17-adic square roots:

A309989, A309990 (sqrt(-1));

A322561, this sequence (sqrt(2));

A322565, A322566 (sqrt(-2)).

Sequence in context: A323454 A261353 A087774 * A040118 A099268 A070277

Adjacent sequences: A322559 A322560 A322561 * A322563 A322564 A322565

Keyword

nonn,base

Author

Jianing Song, Aug 29 2019

Status

approved