A363841 Continued fraction expansion of Sum_{k>=0} 1/(k!)!^2.

2, 3, 1, 32399, 4, 1432456210278611587930429493084159999, 1, 3, 32399, 1, 3

Offset

0,1

Comments

In general, sums of the form Sum_{k>=0} 1/(k!)!^t, t > 1 in N, have the following continued fraction expansion formulas:

The first term is always 2.

Let P(k) = (((k+1)!)! / ((k!)!)^2)^t - 1.

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).

The next term a(11) has 303 digits. - Stefano Spezia, Jun 24 2023

Links

Daniel Hoyt, Table of n, a(n) for n = 0..22

Daniel Hoyt, Python program that generates the continued fraction from formula.

Formula

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).

Mathematica

ContinuedFraction[Sum[1/(k!)!^2, {k, 0, 6}], 21]

Crossrefs

Cf. A363842 (decimal expansion).

Cf. A001511, A336810.

Sequence in context: A016537 A346381 A106385 * A291293 A259572 A027413

Adjacent sequences: A363838 A363839 A363840 * A363842 A363843 A363844

Keyword

nonn,cofr

Author

Daniel Hoyt, Jun 23 2023

Status

approved