A322091 - OEIS

A322091

Digits of one of the two 13-adic integers sqrt(-3).

6, 3, 12, 6, 10, 7, 4, 4, 9, 8, 9, 2, 8, 5, 12, 3, 5, 4, 0, 6, 5, 1, 2, 6, 5, 9, 4, 9, 1, 1, 4, 6, 11, 3, 1, 12, 5, 2, 2, 6, 3, 11, 11, 8, 4, 5, 10, 10, 7, 9, 5, 7, 7, 7, 8, 0, 1, 0, 7, 7, 0, 9, 12, 10, 8, 1, 6, 1, 2, 10, 2, 9, 7, 2, 1, 10, 11, 4, 3, 5, 6

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OFFSET

0,1

COMMENTS

This square root of -3 in the 13-adic field ends with digit 6. The other, A322092, ends with digit 7.

LINKS

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

Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3/a>, Dec 2022.

Wikipedia, p-adic number

FORMULA

a(n) = (A322089(n+1) - A322089(n))/13^n.

For n > 0, a(n) = 12 - A322092(n).

Equals A286838*A322088 = A286839*A322087.

This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,6)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

EXAMPLE

...36225C13B64119495621560453C582989447A6C36.

PROG

(PARI) a(n) = truncate(sqrt(-3+O(13^(n+1))))\13^n

CROSSREFS

Cf. A114525, A286838, A286839, A322087, A322088, A322089, A322092.

Sequence in context: A335393 A040033 A165998 * A329583 A050132 A320038

Adjacent sequences: A322088 A322089 A322090 * A322092 A322093 A322094

KEYWORD

nonn,base,easy

AUTHOR

Jianing Song, Nov 26 2018

STATUS

approved