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 A349834 #21 Jul 27 2022 08:56:12 %S A349834 1,6,22,92,358,1460,5756,23288,92294,372036,1478420,5947272,23671516, %T A349834 95102088,378922552,1521039088,6064766662,24329781988,97059838372, %U A349834 389194630888,1553243997172,6226104229528,24855484384072,99604902663568,397733491426972 %N A349834 Expansion of sqrt(1 + 4*x)/(1 - 4*x). %C A349834 Let b(n) = a(n)/4^n, {b(n)} = {1, 3/2, 11/8, 23/16, 179/128, 365/256, 1439/1024, ...}. Since a(n) >= 4^n, Sum_{n>=0} b(n) is divergent. Let c(n) = A349835(n)/(-4)^n, {c(n)} = {1, -3/2, 7/8, -11/16, 75/128, -133/256, 483/1024, ...}. Since |c(n)| ~ 2/sqrt(Pi*n) and |c(n+1)|/|c(n)| = ((4*n+3)*(2*n-1)) / ((4*n-1)*(2*n+2)) < 1, Sum_{n>=0} c(n) is conditionally convergent by Leibniz's criterion. Note that Sum_{n>=0} b(n)*x^n = sqrt(1 + x)/(1 - x), Sum_{n>=0} c(n)*x^n = (1 - x)/sqrt(1 + x), hence the Cauchy product of Sum_{n>=0} b(n) and Sum_{n>=0} c(n) is 1 + 0 + 0 + .... {b(n)} and {c(n)} serve as an example such that the Cauchy product of a divergent series and a conditionally convergent series can be absolutely convergent. %H A349834 Wikipedia, <a href="https://en.m.wikipedia.org/wiki/Cauchy_product">Cauchy product</a> %F A349834 a(n) = 2^(2*n) + (Sum_{k=0..n-1} (-1)^k * 2^(2*n-1-2*k) * CatalanNumber(k)). %F A349834 a(n) = 2^(2*n + 1/2) - ((-1)^n * CatalanNumber(n) * hypergeom([1, n + 1/2], [n + 2], -1)) / 2. - _Peter Luschny_, Dec 03 2021 %F A349834 D-finite with recurrence n*a(n) -6*a(n-1) +8*(-2*n+3)*a(n-2)=0. - _R. J. Mathar_, Jul 27 2022 %e A349834 Let C(n) denote the Catalan numbers. %e A349834 a(0) = 2^0 = 1; %e A349834 a(1) = 2^2 + 2^1 * C(0) = 6; %e A349834 a(2) = 2^4 + 2^3 * C(0) - 2^1 * C(1) = 22; %e A349834 a(3) = 2^6 + 2^5 * C(0) - 2^3 * C(1) + 2^1 * C(2) = 92; %e A349834 a(4) = 2^8 + 2^7 * C(0) - 2^5 * C(1) + 2^3 * C(2) - 2^1 * C(3) = 358. %o A349834 (PARI) C(n) = binomial(2*n, n)/(n+1) %o A349834 a(n) = 2^(2*n) + sum(k=0, n-1, (-1)^k * 2^(2*n-1-2*k) * C(k)) %Y A349834 Cf. A000108, A349835. %K A349834 nonn %O A349834 0,2 %A A349834 _Jianing Song_, Dec 01 2021