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

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

Offset: 0

Author

Daniel Hoyt, Jun 23 2023

Keywords

• nonn
• cofr

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.

Crossrefs

Cf. A363842 (decimal expansion).

Cf. A001511, A336810.

Programs

Mathematica

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

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).