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A322087 Digits of one of the two 13-adic integers sqrt(3).

%I A322087 #25 Dec 04 2022 12:32:24
%S A322087 4,8,6,8,12,2,1,9,9,10,1,6,4,10,6,11,11,9,5,5,0,5,2,5,8,0,8,7,3,5,3,
%T A322087 12,0,3,10,3,5,8,1,12,11,8,7,0,3,1,4,9,9,9,1,10,6,12,2,7,3,5,1,6,12,1,
%U A322087 1,12,10,5,6,11,7,8,12,10,1,3,5,5,5,7,11,1,5
%N A322087 Digits of one of the two 13-adic integers sqrt(3).
%C A322087 This square root of 3 in the 13-adic field ends with digit 4. The other, A322088, ends with digit 9.
%H A322087 Seiichi Manyama, <a href="/A322087/b322087.txt">Table of n, a(n) for n = 0..10000</a>
%H A322087 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 A322087 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322087 a(n) = (A322085(n+1) - A322085(n))/13^n.
%F A322087 For n > 0, a(n) = 12 - A322088(n).
%F A322087 Equals A286838*A322091 = A286839*A322092.
%F A322087 This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - _Peter Bala_, Dec 04 2022
%e A322087 ...BC1853A30C35378085250559BB6A461A9912C8684.
%o A322087 (PARI) a(n) = truncate(sqrt(3+O(13^(n+1))))\13^n
%Y A322087 Cf. A322085.
%Y A322087 Digits of p-adic integers:
%Y A322087 A321074, A321075 (11-adic, sqrt(3));
%Y A322087 this sequence, A322088 (13-adic, sqrt(3));
%Y A322087 A286838, A286839 (13-adic, sqrt(-1));
%Y A322087 A322091, A322092 (13-adic, sqrt(-3)).
%K A322087 nonn,base
%O A322087 0,1
%A A322087 _Jianing Song_, Nov 26 2018