A322561 Digits of one of the two 17-adic integers sqrt(2) that is related to A322559.

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

Offset

0,1

Comments

This square root of 2 in the 17-adic field ends with digit 6. The other, A322562, ends with digit 11 (B when written as a 17-adic number).

Links

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

Wikipedia, p-adic number

Formula

a(n) = (A322559(n+1) - A322559(n))/17^n.

For n > 0, a(n) = 16 - A322562(n).

Equals A309989*A322566 = A309990*A322565.

Example

The solution to x^2 == 2 (mod 17^4) such that x == 6 (mod 17) is x == 43594 (mod 17^4), and 43594 is written as 8EE6 in heptadecimal, so the first four terms are 6, 14, 14 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));

this sequence, A322562 (sqrt(2));

A322565, A322566 (sqrt(-2)).

Sequence in context: A329065 A338419 A184998 * A079010 A324814 A015822

Adjacent sequences: A322558 A322559 A322560 * A322562 A322563 A322564

Keyword

nonn,base

Author

Jianing Song, Aug 29 2019

Status

approved