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 A363841 #63 Aug 05 2023 22:17:01
%S A363841 2,3,1,32399,4,1432456210278611587930429493084159999,1,3,32399,1,3
%N A363841 Continued fraction expansion of Sum_{k>=0} 1/(k!)!^2.
%C A363841 In general, sums of the form Sum_{k>=0} 1/(k!)!^t, t > 1 in N, have the following continued fraction expansion formulas:
%C A363841 The first term is always 2.
%C A363841 Let P(k) = (((k+1)!)! / ((k!)!)^2)^t - 1.
%C A363841 Take the sequence A157196 and replace the runs of '1,1' with 2^t - 1, the odd occurring runs of '2' with 2^t, and the even occurring runs of '2' with 2^t - 2. Finally interleave the modified sequence with a string of 1's and let it be called f(n). To get the continued fraction expansion, interleave the n-th runs of '2^t', '2^t - 1, 1', '1, 2^t - 1' and '1, 2^t - 2, 1' in f(n), and P(A001511(n)+1).
%C A363841 The next term a(11) has 303 digits. - _Stefano Spezia_, Jun 24 2023
%H A363841 Daniel Hoyt, <a href="/A363841/b363841.txt">Table of n, a(n) for n = 0..22</a>
%H A363841 Daniel Hoyt, <a href="/A336810/a336810_2.txt">Python program that generates the continued fraction from formula</a>.
%F A363841 Take the sequence A157196 and replace the runs of '1,1' with '3'. Then replace the odd occurring runs of '2' with '4'. Finally interleave the modified sequence with a string of 1's and let it be called f(n). To get the continued fraction expansion, interleave between the n-th runs of '4', '3, 1', '1, 3' and '1, 2, 1' in f(n), and P(A001511(n)+1).
%t A363841 ContinuedFraction[Sum[1/(k!)!^2, {k, 0, 6}], 21]
%Y A363841 Cf. A363842 (decimal expansion).
%Y A363841 Cf. A001511, A336810.
%K A363841 nonn,cofr
%O A363841 0,1
%A A363841 _Daniel Hoyt_, Jun 23 2023