A322092 Digits of one of the two 13-adic integers sqrt(-3).

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

Offset: 0

Author

Jianing Song, Nov 26 2018

Keywords

• nonn
• base
• easy

Comments

This square root of -3 in the 13-adic field ends with digit 7. The other, A322091, ends with digit 6.

Examples

			...96AA70B9168BB38376AB76C879074A34388526097.
		

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, Dec 2022.
• Wikipedia, p-adic number

Crossrefs

Cf. A114525, A286838, A286839, A322087, A322088, A322090, A322091.

Programs

PARI

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

Formula

a(n) = (A322090(n+1) - A322090(n))/13^n.

For n > 0, a(n) = 12 - A322091(n).

Equals A286838*A322087 = A286839*A322088.

This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,7)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022