A075827 -id:A075827 - cp's OEIS frontend

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.

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

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)).

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

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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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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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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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Programs

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Formula

Extensions

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.

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.

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.