A334958 -id:A334958 - cp's OEIS frontend

A335023 Ratios of consecutive terms of A334958.

1, 1, 2, 1, 6, 1, 4, 3, 10, 1, 12, 1, 14, 75, 8, 1, 18, 1, 4, 21, 22, 1, 24, 5, 26, 9, 196, 1, 30, 1, 16, 33, 34, 5, 36, 1, 38, 39, 40, 1, 42, 1, 44, 45, 46, 1, 48, 7, 50, 51, 52, 1, 54, 55, 56, 57, 58, 1, 60, 1, 62, 63, 32, 65, 66, 1, 68, 69, 70, 1, 72, 1, 74, 375, 76, 847

Offset: 1

Petros Hadjicostas, May 19 2020

nonn

Conjecture: a(n) = 1 if and only if n+1 is prime.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A056612, A334958.

b:= proc(n) b(n):= (-(-1)^n/n +`if`(n=1, 0, b(n-1))) end:
g:= proc(n) g(n):= (f-> igcd(b(n)*f, f))(n!) end:
a:= n-> g(n+1)/g(n):
seq(a(n), n=1..80);  # Alois P. Heinz, May 20 2020

b[n_] := b[n] = -(-1)^n/n + If[n==1, 0, b[n-1]];
g[n_] := GCD[b[n] #, #]&[n!];
a[n_] := g[n+1]/g[n];
Array[a, 80] (* Jean-François Alcover, Nov 30 2020, after Alois P. Heinz *)

f(n) = n!*sum(k=2, n, (-1)^k/k); \\ A024168
g(n) = gcd(f(n+1), f(n)); \\ A334958
a(n) = g(n+1)/g(n); \\ Michel Marcus, May 20 2020

a(n) = A334958(n+1)/A334958(n).

A056612 a(n) = gcd(n!, n!*(1 + 1/2 + 1/3 + ... + 1/n)).

Original entry on oeis.org

1, 1, 1, 2, 2, 36, 36, 144, 144, 1440, 1440, 17280, 17280, 241920, 3628800, 29030400, 29030400, 1567641600, 1567641600, 156764160000, 9876142080000, 217275125760000, 217275125760000, 1738201006080000, 1738201006080000

Offset: 1

Leroy Quet, Aug 08 2000

The first difference between this sequence and A131657 occurs for n = 20, while the first difference between this sequence and A131658 occurs for n = 21. - Christian Krattenthaler, Sep 30 2007

			a(4) = gcd(4!, 4!*(1 + 1/2 + 1/3 + 1/4)) = gcd(24, 50) = 2.
a(4) = gcd(A000254(5), A000254(4)) = gcd(5!*(1 + 1/2 + 1/3 + 1/4 + 1/5), 4!*(1 + 1/2 + 1/3 + 1/4)) = gcd(274, 50) = 2. - _Petros Hadjicostas_, May 18 2020

Michael De Vlieger, Table of n, a(n) for n = 1..535

Cf. A000142, A001008, A002805.
Cf. A007757, A131657, A131658.
Cf. A334958 (similar sequence for the alternative harmonic series).

Table[GCD[#, # Total@ Map[1/# &, Range@ n]] &[n!], {n, 25}] (* Michael De Vlieger, Sep 23 2017 *)
a[n_] := n!/Denominator@ HarmonicNumber@ n; Array[a, 25] (* Robert G. Wilson v, Jun 30 2018 *)

a(n) = gcd(n!, n!*sum(k=1, n, 1/k)); \\ Michel Marcus, Jul 14 2018

a(n) = gcd(stirling(n+1, 2, 1), n!); \\ Michel Marcus, May 20 2020

a(n) = A000142(n)/A002805(n) = A000254(n)/A001008(n). - Franz Vrabec, Sep 13 2005
a(n) = gcd(A000254(n+1), A000254(n)). - Petros Hadjicostas, May 18 2020
a(n) = gcd(Stirling1(n+1, 2), n!). - Michel Marcus, May 20 2020

A075829 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))/(d(n)x + a(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

1, 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

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]
Difference between the denominator and the numerator of the (n-1)-th alternating harmonic number Sum_{k=1..n-1} (-1)^(k+1)*1/k = A058313(n-1)/A058312(n-1). - Alexander Adamchuk, Jul 22 2006
From Petros Hadjicostas, May 06 2020: (Start)
Inspired by Michael Somos's result below, we established the following formulas (valid for n >= 2). All the denominators in the first three formulas are equal to A334958(n).
b(n) = A024167(n)/gcd(A024167(n-1), A024167(n)).
c(n) = A024168(n)/gcd(A024168(n-1), A024168(n)).
d(n) = A024167(n-1)/gcd(A024167(n-1), A024167(n)).
b(n) + c(n) = n*(d(n) + a(n)).
u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)). (End)

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

Cf. A075827 (= b), A075828 (= c), A075830 (= d).

Cf. A024167, A024168, A058312, A058313, A334958.

Denominator[Table[Sum[(-1)^(k+1)*1/k,{k,1,n-1}],{n,1,30}]]-Numerator[Table[Sum[(-1)^(k+1)*1/k,{k,1,n-1}],{n,1,30}]] (* Alexander Adamchuk, Jul 22 2006 *)

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

a(n) = A024168(n-1)/gcd(A024168(n-1), A024168(n)). - Michael Somos, Oct 29 2002
From Alexander Adamchuk, Jul 22 2006: (Start)
a(n) = A058312(n-1) - A058313(n-1) for n > 1 with a(1) = 1.
a(n) = denominator(Sum_{k=1..n-1} (-1)^(k+1)*1/k) - numerator(Sum_{k=1..n-1}(-1)^(k+1)*1/k). (End)

Name edited by Petros Hadjicostas, May 06 2020

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

A335024 Ratios of consecutive terms of A056612.

Original entry on oeis.org

1, 1, 2, 1, 18, 1, 4, 1, 10, 1, 12, 1, 14, 15, 8, 1, 54, 1, 100, 63, 22, 1, 8, 1, 26, 3, 28, 1, 30, 1, 16, 363, 34, 35, 36, 1, 38, 39, 40, 1, 294, 1, 4, 45, 46, 1, 48, 1, 50, 51, 52, 1, 162, 55, 56, 57, 58, 1, 60, 1, 62, 189, 32, 65, 198, 1, 68, 23, 70, 1, 24, 1, 74, 75, 76, 847, 78, 1, 80

Offset: 1

Petros Hadjicostas, May 19 2020

nonn

Conjecture 1: a(n) = 1 if and only if n + 1 = p^k for some prime p and some positive integer k < p.

Conjecture 2 (due to Alois P. Heinz): a(n) = 1 <=> n+1 in A136327.

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Cf. A056612, A136327, A334958.

b:= proc(n) b(n):= (1/n +`if`(n=1, 0, b(n-1))) end:
g:= proc(n) g(n):= (f-> igcd(b(n)*f, f))(n!) end:
a:= n-> g(n+1)/g(n):
seq(a(n), n=1..80);  # Alois P. Heinz, May 20 2020

g[n_] := GCD[n!, n! Sum[1/k, {k, 1, n}]];
a[n_] := g[n + 1]/g[n];
Array[a, 80] (* Jean-François Alcover, Dec 01 2020, after PARI *)

g(n) = gcd(n!, n!*sum(k=1, n, 1/k)); \\ A056612
a(n) = g(n+1)/g(n); \\ Michel Marcus, May 20 2020

a(n) = A056612(n+1)/A056612(n).

cp's OEIS Frontend

A335023 Ratios of consecutive terms of A334958.

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A056612 a(n) = gcd(n!, n!*(1 + 1/2 + 1/3 + ... + 1/n)).

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A075829 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))/(d(n)*x + a(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

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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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Author

Keywords

Comments

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Programs

PARI

Formula

Extensions

A335024 Ratios of consecutive terms of A056612.

Views

Author

Keywords

Comments

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Programs

Maple

Mathematica

PARI

Formula

A075829 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))/(d(n)x + a(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.