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%I #63 Aug 05 2023 22:17:01
%S 2,3,1,32399,4,1432456210278611587930429493084159999,1,3,32399,1,3
%N Continued fraction expansion of Sum_{k>=0} 1/(k!)!^2.
%C In general, sums of the form Sum_{k>=0} 1/(k!)!^t, t > 1 in N, have the following continued fraction expansion formulas:
%C The first term is always 2.
%C Let P(k) = (((k+1)!)! / ((k!)!)^2)^t - 1.
%C 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 The next term a(11) has 303 digits. - _Stefano Spezia_, Jun 24 2023
%H Daniel Hoyt, <a href="/A363841/b363841.txt">Table of n, a(n) for n = 0..22</a>
%H Daniel Hoyt, <a href="/A336810/a336810_2.txt">Python program that generates the continued fraction from formula</a>.
%F 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 ContinuedFraction[Sum[1/(k!)!^2, {k, 0, 6}], 21]
%Y Cf. A363842 (decimal expansion).
%Y Cf. A001511, A336810.
%K nonn,cofr
%O 0,1
%A _Daniel Hoyt_, Jun 23 2023
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