A075829 -id:A075829 - cp's OEIS frontend

A024168 a(n) = n! * (1 + Sum_{j=1..n} (-1)^j/j).

1, 0, 1, 1, 10, 26, 276, 1212, 14736, 92304, 1285920, 10516320, 166112640, 1680462720, 29753498880, 359124192000, 7053661440000, 98989454592000, 2137497610752000, 34210080898560000, 805846718380032000, 14489879077804032000, 369868281883398144000

Offset: 0

Clark Kimberling

nonn

a(n) is the number of permutations of n letters all cycles of which have length <= n/2, a quantity which arises in the solution to the One Hundred Prisoners problem. - Jim Ferry (jferry(AT)alum.mit.edu), Mar 29 2007

Alois P. Heinz, Table of n, a(n) for n = 0..450
Wikipedia, One hundred prisoners.
Index entries for sequences related to factorial numbers

A075829(n) = a(n-1)/gcd(a(n-1), a(n)).

Cf. A000142, A024167, A244009.

a := n -> n!*((-1)^n*LerchPhi(-1, 1, n + 1) + 1 - log(2));
seq(simplify(a(n)), n=0..21); # Peter Luschny, Dec 27 2018

f[k_] := (k + 1) (-1)^(k + 1)
t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 22}]    (* A024168 signed *)
(* Clark Kimberling, Dec 30 2011 *)

x='x+O('x^33); concat([0],Vec(serlaplace((x-log(1+x))/(1-x)))) \\ Joerg Arndt, Dec 27 2018

From Michael Somos, Oct 29 2002: (Start)
E.g.f.: (log(x+1)-1)/(x-1).
a(n) = a(n-1)+a(n-2)*(n-1)^2, n>=2. (End)
a(0) = 1, a(n) = a(n-1)*n + (-1)^n*(n-1)!. - Daniel Suteu, Feb 06 2017
a(n) = n!*((-1)^n*LerchPhi(-1, 1, n+1) + 1 - log(2)). - Peter Luschny, Dec 27 2018
Limit_{n->oo} a(n)/n! = 1 - log(2) = A244009. - Alois P. Heinz, Jul 08 2022

More terms from Michael Somos, Oct 29 2002

a(0)=1 prepended and edited by Alois P. Heinz, Sep 24 2023

A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)x + c(n))/(a(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

0, 1, 1, 5, 7, 47, 37, 319, 533, 1879, 1627, 20417, 18107, 263111, 237371, 52279, 95549, 1768477, 1632341, 33464927, 155685007, 166770367, 156188887, 3825136961, 3602044091, 19081066231, 18051406831, 57128792093, 7751493599

Offset: 1

Benoit Cloitre, Oct 14 2002

nonn

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n)-n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]
From Petros Hadjicostas, May 05 2020: (Start)
Given x > 0, u(n) = (A075827(n)*x + A075828(n))/(a(n)*x + A075829(n)) = (b(n)*x + c(n))/(a(n)*x + d(n)) with gcd(gcd(b(n), c(n)), gcd(a(n), d(n))) = 1 for each n >= 1.
Conjecture 1: Define the sequences (A(n): n >= 1) and (B(n): n >= 1) by A(n+1) = n^2/A(n) + 1 for n >= 2 with A(1) = infinity and A(2) = 1, and B(n+1) = n^2/B(n) + 1 for n >= 3 with B(1) = 0, B(2) = infinity, and B(3) = 1. Then a(n) = denominator(A(n)), b(n) = numerator(A(n)), c(n) = numerator(B(n)), and d(n) = denominator(B(n)) (assuming infinity = 1/0). Also, gcd(a(n), d(n)) = 1.
In 2002, Michael Somos claimed that d(n) = A024168(n-1)/gcd(A024168(n-1), A024168(n)) for n >= 2. In 2006, N. J. A. Sloane claimed that a(n) = A058313(n-1) for n >= 2 while Alexander Adamchuk claimed that d(n) = A058312(n-1) - A058313(n-1) for n >= 2.
Conjecture 2: a(n) = A024167(n-1)/gcd(A024167(n-1), A024167(n)).
Conjecture 3: b(p) = a(p+1) for p = 1 or prime. In general, it seems that b(n) = A048671(n)*a(n+1) for all n for which A048671(n) < n.
Conjecture 4: c(n) = n*(a(n) + d(n)) - b(n) for n >= 1. (End)
All conjectures are proved in the link below except for the second part of Conjecture 3. - Petros Hadjicostas, May 21 2020

Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Apart from the leading term, same as A058313.
Cf. A075827 (= b), A075828 (= c), A075829 (= d).
Cf. A024167, A024168, A048671, A058312.

u(n)=if(n<2,x,(n-1)^2/u(n-1)+1);
a(n)=polcoeff(denominator(u(n)),1,x);

Name edited by Petros Hadjicostas, May 04 2020

A334958 GCD of consecutive terms of the factorial times the alternating harmonic series.

Original entry on oeis.org

1, 1, 1, 2, 2, 12, 12, 48, 144, 1440, 1440, 17280, 17280, 241920, 18144000, 145152000, 145152000, 2612736000, 2612736000, 10450944000, 219469824000, 4828336128000, 4828336128000, 115880067072000, 579400335360000, 15064408719360000, 135579678474240000, 26573616980951040000, 26573616980951040000

Offset: 1

Petros Hadjicostas, May 17 2020

nonn

For n = 1..14, we have a(n) = A025527(n), but a(15) = 18144000 <> 3628800 = A025527(15).
It appears that A025527(n) | a(n) for all n >= 1 and A025527(n) = a(n) for infinitely many n. In addition, it seems that a(n)/a(n-1) = A048671(n) for infinitely many n >= 2. However, I have not established these claims.
This sequence appears in formulas for sequences A075827, A075828, A075829, and A075830 (the first one of which was established in 2002 by Michael Somos).
Conjecture: a(n) = n! * Product_{p <= n} p^min(0, v_p(H'(n))), where the product ranges over primes p, H'(n) = Sum_{k=1..n} (-1)^(k+1)/k, and v_p(r) is the p-adic valuation of rational r (checked for n < 1100).

			A024167(4) = 4!*(1 - 1/2 + 1/3 - 1/4) = 14, A024167(5) = 5!*(1 - 1/2 + 1/3 - 1/4 + 1/5) = 94, A024168(4) = 4!*(1/2 - 1/3 + 1/4) = 10, and A024168(5) = 5!*(1/2 - 1/3 + 1/4 - 1/5) = 26. Then a(4) = gcd(14, 94) = gcd(10, 26) = gcd(14, 4!) = gcd(10, 4!) = gcd(14, 10) = 2.

Cf. A000142, A024167, A024168, A025527, A048671, A058312, A058313, A075827, A075828, A075829, A075830.

Cf. A056612 (similar sequence for the harmonic series).

b:= proc(n) b(n):= (-(-1)^n/n +`if`(n=1, 0, b(n-1))) end:
a:= n-> (f-> igcd(b(n)*f, f))(n!):
seq(a(n), n=1..30);  # Alois P. Heinz, May 18 2020

b[n_] := b[n] = -(-1)^n/n + If[n == 1, 0, b[n-1]];
a[n_] := GCD[b[n] #, #]&[n!];
Array[a, 30] (* Jean-François Alcover, Oct 27 2020, after Alois P. Heinz *)

def A():
    a, b, n = 1, 1, 2
    while True:
        yield gcd(a, b)
        b, a = a, a + b * n * n
        n += 1
a = A(); print([next(a) for  in range(29)]) # _Peter Luschny, May 19 2020

a(n) = gcd(A024167(n+1), A024167(n)) = gcd(A024168(n+1), A024168(n)) = gcd(A024167(n), n!) = gcd(A024168(n), n!) = gcd(A024167(n), A024168(n)).

A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)x + b(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

1, 1, 5, 14, 47, 222, 319, 2132, 5637, 16270, 20417, 217284, 263111, 3323194, 3920925, 764392, 1768477, 29382138, 33464927, 622740028, 3502177707, 3436155514, 3825136961, 86449058184, 95405331155, 469336577606, 514159128837, 1519292745404, 236266661971, 6755272778730, 7313175618421

Offset: 1

Benoit Cloitre, Oct 14 2002

nonn

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n) - n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]

Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Cf. A075828 (= b), A075829 (= d), A075830 (= c).

Cf. A024167, A024168, A334958.

u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
a(n) = polcoeff(numerator(u(n)), 1, x);
for(n=1, 30, print1(a(n)", ")) \\ Petros Hadjicostas, May 06 2020

From Petros Hadjicostas, May 18 2020: (Start)
a(n) = A024167(n)/gcd(A024167(n), A024167(n-1)) = A024167(n)/A334958(n-1) for n >= 2. (Cf. Michael Somos's result for d = A075829 using A024168.)
u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)

Name edited by Petros Hadjicostas, May 06 2020

More terms from Michel Marcus, Aug 01 2025

A075828 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)x + a(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

0, 1, 1, 10, 13, 138, 101, 1228, 1923, 8930, 7303, 115356, 97249, 1721846, 1484475, 388760, 681971, 14725926, 13093585, 308430212, 1386466053, 1685280806, 1529091919, 42052434936, 38450390845, 226713176794, 208661769963

Offset: 1

Benoit Cloitre, Oct 14 2002

nonn

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n) - n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]

Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Cf. A075827 (= b), A075829 (= d), A075830 (= c).

Cf. A024167, A024168, A334958.

u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
a(n) = polcoeff(numerator(u(n)), 0 ,x)

From Petros Hadjicostas, May 18 2020: (Start)
a(n) = A024168(n)/gcd(A024168(n), A024168(n-1)) = A024168(n)/A334958(n) for n >= 2. (Cf. Michael Somos's claim for d = A075829 using A024168.)
u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)

Name edited by Petros Hadjicostas, May 06 2020

A119248 a(n) is the difference between denominator and numerator of the n-th alternating harmonic number Sum_{k=1..n} (-1)^(k+1)/k = A058313(n)/A058312(n).

Original entry on oeis.org

0, 1, 1, 5, 13, 23, 101, 307, 641, 893, 7303, 9613, 97249, 122989, 19793, 48595, 681971, 818107, 13093585, 77107553, 66022193, 76603673, 1529091919, 1752184789, 7690078169, 8719737569, 23184641107, 3721854001, 96460418429

Offset: 1

Alexander Adamchuk, Jul 22 2006

a(n)/A058312(n) = 1 - A058313(n)/A058312(n) appears in the locker puzzle (see the links in A364317) for the probability of success with the strategy used there for n lockers and allowed openings of up to floor(n/2) lockers. Note that gcd(a(n), A058312(n)) = 1. - Wolfdieter Lang, Aug 12 2023

Cf. A058312, A058313, A075829, A364317.

Denominator[Table[Sum[(-1)^(k+1)/k,{k,1,n}],{n,1,30}]]-Numerator[Table[Sum[(-1)^(k+1)/k,{k,1,n}],{n,1,30}]]

a(n) = my(x=sum(k=1, n, (-1)^(k+1)/k)); denominator(x) - numerator(x); \\ Michel Marcus, May 07 2020

a(n) = denominator(Sum_{k=1..n} (-1)^(k+1)/k) - numerator(Sum_{k=1..n} (-1)^(k+1)/k).
a(n) = A058312(n) - A058313(n).
a(n) = A075829(n+1).
a(n) = numerator(Sum_{k=2..n} (-1)^k/k). (Cf. A024168.) - Petros Hadjicostas, May 17 2020

cp's OEIS Frontend

A024168 a(n) = n! * (1 + Sum_{j=1..n} (-1)^j/j).

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A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)*x + c(n))/(a(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

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A334958 GCD of consecutive terms of the factorial times the alternating harmonic series.

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A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)*x + b(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

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A075828 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)*x + a(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

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A119248 a(n) is the difference between denominator and numerator of the n-th alternating harmonic number Sum_{k=1..n} (-1)^(k+1)/k = A058313(n)/A058312(n).

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A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)x + c(n))/(a(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)x + b(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

A075828 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)x + a(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.