A322088 - OEIS

%I #17 Dec 04 2022 12:37:21

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

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

%U 11,11,0,2,7,6,1,5,4,0,2,11,9,7,7,7,5,1,11,7

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

%C This square root of 3 in the 13-adic field ends with digit 9. The other, A322087, ends with digit 4.

%H Seiichi Manyama, <a href="/A322088/b322088.txt">Table of n, a(n) for n = 0..10000</a>

%H Peter Bala, <a href="/A051277/a051277.pdf">Using Chebyshev polynomials to find the p-adic square roots of 2 and 3</a>, Dec 2022.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

%F a(n) = (A322086(n+1) - A322086(n))/13^n.

%F For n > 0, a(n) = 12 - A322087(n).

%F Equals A286838*A322092 = A286839*A322091.

%F This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,9/2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - _Peter Bala_, Dec 04 2022

%e ...10B47929C097954C47A7C773116286B233BA04649.

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

%Y Cf. A286838, A286839, A322086, A322087, A322091, A322092.

%K nonn,base,easy

%O 0,1

%A _Jianing Song_, Nov 26 2018