A336810 -id:A336810 - cp's OEIS frontend

A157196 a(n) = (1/2)*(sum of elements of n-th run) using 1 and 2 starting with 1,1.

1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1

Offset: 1

Benoit Cloitre, Feb 24 2009

nonn

We conjecture that the density of 1's in the sequence approaches 2/3 as n -> infinity. This conjecture is proved in the paper of Shallit.

			Write the sums of elements in each run, you obtain: 2,2,4,2,2,2,2,4,2,2,4,2,2,4,4,... dividing by 2 you get: 1,1,2,1,1,1,1,2,1,1,2,1,1,2,2,... the sequence itself.

Daniel Hoyt, Visual representation of the sequence in the style of the Kolakoski sequence.
Daniel Hoyt, Representing the sequence visually.
Jeffrey Shallit, Cloitre's Self-Generating Sequence, arXiv:2501.00784 [math.CO], 2025.

Cf. A000002, A157129, A336810.

mx:= 1000: l:= [1$2]: a:= n-> l[n]:
for h from 2 while nops(l)Alois P. Heinz, May 31 2012

More terms from Alvin Hoover Belt, May 31 2012

A336686 Decimal expansion of Sum_{k>=0} 1/(k!)!.

Original entry on oeis.org

2, 5, 0, 1, 3, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 0, 5, 0, 0, 6, 2, 6, 4, 5, 9, 9, 8, 5, 0, 0, 7, 2, 3, 7, 9, 3, 7, 6, 0, 2, 1, 9, 3, 6, 9, 0, 0, 6, 0, 6, 9, 0, 2, 1, 3, 6, 1, 4, 9, 9, 1, 4, 9, 6, 1, 1, 6, 8, 6, 2, 4, 1, 8, 1, 7, 8, 9, 1, 9, 8, 9, 9, 3, 3, 9, 4, 3, 7, 4, 1, 5, 7, 4, 4, 1

Offset: 1

Daniel Hoyt, Nov 20 2020

The sum has 18 8's in a row within the first 23 decimal places.

1801/720 approximates this constant to 23 significant digits.

			2.5013888888888888888888905006...

Cf. A000197, A336810.

evalf(sum(1/(k!)!, k=0..infinity), 112);  # Alois P. Heinz, Nov 20 2020

RealDigits[Sum[1/(k!)!, {k, 0, 4}], 10, 100][[1]] (* Amiram Eldar, Nov 21 2020 *)

PARI
```
suminf(k=0, 1/(k!)!)
```

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

Original entry on oeis.org

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

Offset: 0

Daniel Hoyt, Jun 23 2023

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

Daniel Hoyt, Table of n, a(n) for n = 0..22
Daniel Hoyt, Python program that generates the continued fraction from formula.

Cf. A363842 (decimal expansion).

Cf. A001511, A336810.

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

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).

A386384 Continued fraction expansion of Sum_{k>=0} (-1)^k/(k!)!.

Original entry on oeis.org

0, 2, 179, 1, 1, 1196852626800230399, 2, 179, 1, 1, 17377308326435956818596067989554034737368967210468674554156131654360754429984573360106123813424835044026977477398690421454067571097599999999999999999999, 2, 179, 2, 1196852626800230399, 1, 1, 179, 2

Offset: 0

Daniel Hoyt, Aug 17 2025

The peak terms have the form P(k) = ((k+1)!)! / ((k!)!)^2 - 1. The sequence is an interleaving between the n-th runs of '2' and '1,1' in A386385, and P(A001511(n)+1).

Daniel Hoyt, Table of n, a(n) for n = 0..39
Daniel Hoyt, Justification for the explicit formula and program provided, Zenodo, 2025.

Cf. A001511, A157196, A336810, A386385.

Cf. A387268 (decimal expansion).

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

import sys #for printing huge factorials
sys.set_int_max_str_digits(0)  # otherwise sys not needed.
def a386384(n):
    import math
    if n==0: return 0
    t=n-1; M=0x18199818
    s=[1,1]; h=2; p=0; r=0
    def g(u):
        nonlocal s,h
        while len(s)>(r&31))&1) else ('A' if xb=='B' else 'B')
        L = 2 if yb=='A' else 1
        if t < L: return 1 if yb=='A' else 2
        t -= L
        if t==0:
            rr=r+1
            K=((rr & -rr).bit_length()-1)+2
            A=math.factorial(math.factorial(K+1))
            B=math.factorial(math.factorial(K))
            return A//(B*B)-1
        t-=1; r+=1

A387268 Decimal expansion of Sum_{k>=0} (-1)^k/(k!)!.

Original entry on oeis.org

4, 9, 8, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 7, 2, 2, 8, 4, 8, 6, 8, 2, 2, 0, 7, 2, 2, 9, 4, 6, 0, 1, 5, 9, 8, 2, 4, 4, 1, 5, 9, 1, 2, 2, 8, 2, 9, 1, 2, 4, 3, 5, 8, 3, 7, 2, 1, 3, 7, 1, 8, 3, 3, 9, 0, 8, 4, 6, 4, 0, 4, 0, 1, 1, 4, 2, 1, 2, 1, 5, 6, 1, 6, 5, 9, 6, 3, 7, 9

Offset: 0

Daniel Hoyt, Aug 24 2025

This constant has an interesting simple continued fraction representation.

359/720 approximates this constant to 19 significant digits.

			0.498611111111111111127228486822072294...

Daniel Hoyt, Table of n, a(n) for n = 0..999

Cf. A386384 (continued fraction expansion).

Cf. A363842, A336810.

RealDigits[Sum[(-1)^k/(k!)!, {k, 0, 6}], 10, 100][[1]]

PARI
```
suminf(k=0, (-1)^k/(k!)!)
```

cp's OEIS Frontend

A157196 a(n) = (1/2)*(sum of elements of n-th run) using 1 and 2 starting with 1,1.

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A336686 Decimal expansion of Sum_{k>=0} 1/(k!)!.

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

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Formula

A386384 Continued fraction expansion of Sum_{k>=0} (-1)^k/(k!)!.

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Mathematica

Python

A387268 Decimal expansion of Sum_{k>=0} (-1)^k/(k!)!.

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Author

Keywords

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Examples

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Crossrefs

Programs

Mathematica

PARI