This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A269591 #19 Jul 31 2017 12:56:52 %S A269591 1,2,0,2,3,0,4,2,3,3,4,4,3,1,1,3,4,0,3,1,2,0,3,1,1,0,1,1,1,1,2,3,1,1, %T A269591 1,1,4,3,2,2,3,2,4,4,0,3,1,4,0,3,3,1,0,1,3,3,2,3,3,3,4,4,3,1,3,1 %N A269591 Digits of one of the two 5-adic integers sqrt(-4). %C A269591 This is the scaled first difference sequence of A268922. %C A269591 The digits of the other 5-adic integer sqrt(-4), are given in A269592. See also A268922 for the two 5-adic numbers sqrt(-4), called u and -u. %C A269591 a(n) is the unique solution of the linear congruence 2*A268922(n)*a(n) + A269593(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 1 follows from the formula given below. %H A269591 Seiichi Manyama, <a href="/A269591/b269591.txt">Table of n, a(n) for n = 0..10000</a> %H A269591 BCMATH Congruence Programs, <a href="http://www.numbertheory.org/php/p-adic.html">Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.</a> %F A269591 a(n) = (b(n+1) - b(n))/5^n, n >= 0, with b(n) = A268922(n), n >= 0. %F A269591 a(n) = -A269593(n)*(2*A268922(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78. %F A269591 A268922(n+1) = sum(a(k)*5^k, k=0..n), n >= 0. %e A269591 a(4) = 3 because 2*261*3 + 109 = 1675 == 0 (mod 5). %e A269591 a(4) = - 109*(2*261)^3 (mod 5) = -(-1)*(2*1)^3 (mod 5) = 8 (mod 5) = 3. %e A269591 A268922(5) = 2136 = 1*5^0 + 2*5^1 + 0*5^2 + 2*5^3 + 3*5^4. %o A269591 (PARI) a(n) = truncate(sqrt(-4+O(5^(n+1))))\5^n; \\ _Michel Marcus_, Mar 04 2016 %Y A269591 Cf. A210850, A210851, A268922, A269592 (companion), A269593. %K A269591 nonn,base,easy %O A269591 0,2 %A A269591 _Wolfdieter Lang_, Mar 02 2016