A321078 Digits of one of the two 11-adic integers sqrt(5).

4, 4, 10, 4, 0, 9, 5, 8, 7, 7, 3, 9, 7, 4, 1, 8, 10, 5, 10, 8, 10, 2, 5, 6, 10, 10, 3, 10, 2, 2, 1, 6, 6, 5, 7, 10, 0, 6, 9, 8, 3, 1, 6, 5, 3, 1, 9, 0, 5, 8, 8, 10, 4, 8, 9, 1, 9, 7, 1, 2, 1, 0, 5, 0, 2, 9, 1, 5, 2, 1, 3, 1, 7, 10, 3, 5, 0, 4, 10, 5, 1, 2, 10, 2, 1, 3, 0

Offset

0,1

Comments

This square root of 5 in the 11-adic field ends with digit 4. The other, A321079, ends with digit 7.

In an unpublished notebook, Gauss (circa 1800) computes this square root of 5 in the 11-adic integers. See Lemmermeyer, Section 2. - Peter Bala, Dec 10 2025

Links

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

Franz Lemmermeyer, Gauss and p-adic numbers, arXiv:2510.11559 [math.NT], 2025. See pp. 4-5.

Wikipedia, p-adic number

Formula

a(n) = (A321076(n+1) - A321076(n))/11^n.

For n > 0, a(n) = 10 - A321079(n).

Example

...8960A7566122A3AA652A8A5A8147937785904A44.

Maple

# computes both 11-adic square roots of 5 up to O(11^86)
padic:-rootp(_Z^2-5, 11, 87); # Peter Bala, Dec 10 2025

Prog

(PARI) a(n) = truncate(sqrt(5+O(11^(n+1))))\11^n

Crossrefs

Cf. A321076, A321079.

Sequence in context: A198025 A205549 A120395 * A160723 A255486 A286779

Adjacent sequences: A321075 A321076 A321077 * A321079 A321080 A321081

Keyword

nonn,base

Author

Jianing Song, Oct 27 2018

Status

approved