Daniel Hoyt - cp's OEIS frontend

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

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!)!)
```

A386385 Period-32 block rewriting of A157196 (blocks 11 and 2): for block index i, keep if i mod 32 in {3,4,11,12,15,16,19,20,27,28}, else swap 11<->2.

Original entry on oeis.org

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

Offset: 0

Daniel Hoyt, Aug 17 2025

The runs of '2', and '1,1' are interleaved in the continued fraction expansion of Sum_{k>=0} (-1)^k/(k!)! (A386384).
Parse A157196 into blocks A = 11 and B = 2, and index the blocks i = 0,1,2,.... For each block i, keep it when i mod 32 in {3, 4, 11, 12, 15, 16, 19, 20, 27, 28}; otherwise swap A <-> B (i.e, swap 1,1 <-> 2) at all other i. Finally expand blocks by A -> 1,1 and B -> 2.
We index the blocks starting at i=0. "Keep" residues are {3,4,11,12,15,16,19,20,27,28} (mod 32); at all other residues we swap (11) <-> (2). After that, expand by (11)->1,1 and (2)->2.
Each block is handled independently by its own i mod 32.

			Starting from A157196 parsed as (11)(2)(11)(11)(2)(11)(2)... = ABAABAB...
i=0: (11), residue 0 not in keep set -> swap to (2)  -> output 2.
i=1: (2),  residue 1 not in keep set -> swap to (11) -> output 1, 1.
i=2: (11), residue 2 not in keep set -> swap to (2)  -> output 2.
i=3: (11), residue 3 is in keep set  -> keep (11)    -> output 1, 1.
i=4: (2),  residue 4 is in keep set  -> keep (2)     -> output 2.
i=5: (11), residue 5 not in keep set -> swap to (2)  -> output 2.
i=6: (2),  residue 6 not in keep set -> swap to (11) -> output 1, 1.
Concatenating: 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, ...

Daniel Hoyt, Table of n, a(n) for n = 0..1000
Daniel Hoyt, Justification for the explicit formula and program provided.

Cf. A157196, A386384.

def a386385(n):
    m=n+1
    if m<=0: raise ValueError("n>=0")
    M=0x18199818
    def run(N,mode):
        s=[1,1];h=2;p=0;k=0;c=0;y='B'
        def g(u):
            nonlocal s,h
            while len(s)>(k&31))&1) else ('A' if xb=='B' else 'B')
            c += 2 if y=='A' else 1; k+=1
        return c if mode==0 else y
    lo,hi=0,m
    while lo=m: hi=md
        else: lo=md+1
    return 2 if run(lo,1)=='B' else 1

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).

A363842 Decimal expansion of Sum_{k>=0} 1/(k!)!^2.

Original entry on oeis.org

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

Offset: 1

Daniel Hoyt, Jun 23 2023

1166401/518400 approximates this constant to 48 significant digits.

			2.25000192901234567901234567901234567901234567901...

Cf. A363841 (continued fraction).

Cf. A336686.

RealDigits[Sum[1/(k!)!^2, {k, 0, 4}], 10, 100][[1]]

PARI
```
suminf(k=0, 1/k!!^2)
```

A363704 Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).

Original entry on oeis.org

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

Offset: 0

Daniel Hoyt, Jun 16 2023

That this constant is less than one allows Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1) = floor(x^2), when x is the square root of any natural number greater than 1.

The limit converges slowly.

			0.98854960114226875064475410833997126442199868380...

Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x), 2022, pp. 1-4.
Wikipedia, Stieltjes Constants.

Cf. A098572, A350862, A329117, A351885, A001620, A363716.

digits = 120; d = 1; j = 2; s = StieltjesGamma[1]; While[Abs[d] > 10^(-digits - 5), d = (-1)^j / j! * Derivative[j][Zeta][j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* Vaclav Kotesovec, Jun 17 2023 *)

# Gives 14 correct digits
from mpmath import stieltjes,fac
def limgen(n):
    terms = []
    for y in range(3, n):
        for x in range(y, n):
            terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))
    n,o_sum = 2,0
    while True:
        n_term = 1/((n-1)**(n+1))
        n_sum = o_sum + n_term
        if o_sum == n_sum:
            break
        o_sum = n_sum
        n += 1
    return sum(terms) + 0.5 - stieltjes(0) + n_sum
print(str(limgen(60))[:-1])

Equals 1/2 - A001620 + Sum_{k>=2} (1/(k-1)^(k+1)) + Sum_{k>=3} Sum_{n>=k} (((-1)^k)*Stieltjes(n)*(n-k+1)^(k-2))/((n-k+2)!*(k-2)!).
From Vaclav Kotesovec, Jun 17 2023: (Start)
Equals lim_{n->oo} (Sum_{m=1..n} m^(1/m)) - n - log(n)^2/2.
Equals sg1 + Sum_{k>=2} (-1)^k / k! * k-th derivative of zeta(k), where sg1 is the first Stieltjes constant (see A082633). (End)

A351885 Decimal expansion of lim_{n -> infinity} (Sum_{x=1..n} x^(1/x) - Integral_{k=0..n} x^(1/x) dx).

Original entry on oeis.org

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

Offset: 0

Daniel Hoyt, Feb 23 2022

The limiting difference between the integral and sum of x^(1/x). The limit converges slowly.

			0.5681800123590664525123147265218830744...

Daniel Hoyt, Graphical representation of the limit.
Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x).
Wikipedia, Stieltjes Constants

Cf. A350862, A329117, A175999, A001620.

# Gives 15 correct digits
from mpmath import stieltjes,fac,quad
def limgen(n):
    terms = []
    for y in range(3, n):
        for x in range(y, n):
            terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))
    return terms
f = lambda x: x**(1/x)
int01 = quad(f, [0,1])
limit = sum(limgen(60)) + 1.5 - stieltjes(0) - int01
print(limit)

Equals 3/2 - A001620 - A175999 + Sum_{k>=3} Sum_{n>=k} (((-1)^k)*Stieltjes(n)*(n-k+1)^(k-2))/((n-k+2)!*(k-2)!).

A350862 Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).

Original entry on oeis.org

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

Offset: 7

Daniel Hoyt, Jan 19 2022

If x is a whole number greater than 1, the Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) = x + C, where C is a constant less than 1.
The above relation was tested for all 1 < x < 10^7.
If x = 1, the sum is A329117.
This sum demonstrates this relationship: setting sqrt(x) = 1111 generates the sum 1111^2 + C or 1234321 + C. Another example would be Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(1729))) - 1) = 1729.84841430674....
Evaluating the sum at larger x values converges slower and slower. Monotonically changing extrapolation methods such as Richardson's Extrapolation must be used to compute these values.
Since the output (x + C) will be the square of the input (sqrt(x)) plus a constant less than 1, this implies that Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) diverges as x tends to infinity, or simplified to Sum_{k>=1} (k^(1/k) - 1).

			1234321.98281389093336864244004887748682691258771548...

Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x), 2022, pp. 3-4.

Cf. A329117, A308511.

digits = 120; d = 1; j = 1; s = 0; While[Abs[d] > 10^(-digits - 5), d = (-1)^j/j!*Derivative[j][Zeta][(1 + 1/1111)*j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* Vaclav Kotesovec, Jun 18 2023 *)

PARI
```
sumpos(k=1, k^(1/(k^(1 + 1/1111))) - 1)
```

A344906 Decimal expansion of Sum_{k>=0} arctan(1/2^k).

Original entry on oeis.org

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

Offset: 1

Daniel Hoyt, Jun 01 2021

This number can be interpreted geometrically as the angle in radians of a fan made of stacked right triangles, with the length to height ratio doubling each successive triangle as seen in the illustration.

Since this angle exceeds Pi/2, the set of rotation angles used in the CORDIC algorithm covers an angle range sufficient to compute sine and cosine for any angle between 0 and Pi/2. This means the algorithm can converge to any angle in that range through appropriate combinations of these basic rotations. - Daniel Hoyt, Oct 25 2024

			1.743286620472340003...

Daniel Hoyt, Illustration of this angle's arctan relationship.
Wikipedia, CORDIC.

Cf. A003881, A065445, A073000, A195727, A195782.

Digits:= 140:
evalf(sum(arccot(2^k), k=0..infinity));  # Alois P. Heinz, Jun 02 2021

PARI
```
suminf(k=0, atan(1/2^k))
```

sumalt(k=1, ((-1)^(k+1))*2^(2*k-1)/((2^(2*k-1)-1)*(2*k-1)))

Equals Sum_{k>=1} (-1)^(k+1)*2^(2*k-1)/((2^(2*k-1)-1)*(2*k-1)).

A336810 Continued fraction expansion of Sum_{k>=0} 1/(k!)!.

Original entry on oeis.org

2, 1, 1, 179, 2, 1196852626800230399, 1, 1, 179, 1, 1

Offset: 0

Daniel Hoyt, Nov 20 2020

a(11), a(21), and a(41) have 152, 1349, and 12981 digits, respectively.

Georg Fischer, Table of n, a(n) for n = 0..20
Georg Fischer, Table of n, a(n) for n = 0..139
Daniel Hoyt, Python program that generates the continued fraction from formula.
Alfred J. van der Poorten and Jeffrey Shallit, Folded continued fractions, Journal of Number Theory, Vol. 40, Issue 2, 1992, pp. 237-250 (cf. prop. 2).

Cf. A336686 (decimal expansion).

Cf. A001511, A157196, A363841.

ContinuedFraction[Sum[1/(k!)!, {k, 0, 6}], 21] (* Amiram Eldar, Nov 22 2020 *)

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

The peak terms have the form ((k+1)!)! / ((k!)!)^2 - 1. - Georg Fischer, Oct 19 2022 [pers. comm. with J. Shallit]

Let P(k) = ((k+1)!)! / ((k!)!)^2 - 1. After the first term, the rest of the sequence is an interleaving between the n-th runs of '1, 1' and '2' in A157196, and P(A001511(n)+1). - Daniel Hoyt, Jun 26 2023

cp's OEIS Frontend

User: Daniel Hoyt

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

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

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A386385 Period-32 block rewriting of A157196 (blocks 11 and 2): for block index i, keep if i mod 32 in {3,4,11,12,15,16,19,20,27,28}, else swap 11<->2.

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

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A363842 Decimal expansion of Sum_{k>=0} 1/(k!)!^2.

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A363704 Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).

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A351885 Decimal expansion of lim_{n -> infinity} (Sum_{x=1..n} x^(1/x) - Integral_{k=0..n} x^(1/x) dx).

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A350862 Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).

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