A056612 -id:A056612 - cp's OEIS frontend

A335024 Ratios of consecutive terms of A056612.

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).

A131657 For n >= 1, put A_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j and B_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j * (Sum_{k=1..jn} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)A_n(z))) has integral coefficients. The sequence is b(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, 783820800000, 9876142080000, 651825377280000, 217275125760000, 8691005030400000

Offset: 1

Christian Krattenthaler (Christian.Krattenthaler(AT)univie.ac.at), Sep 12 2007, Sep 30 2007

nonn

Different from A131658 and A056612. The first difference between A056612 and this sequence occurs for n = 20, while the first difference between A056612 and A131658 occurs for n = 21.

Krattenhaler and Rivoal (2007-2009) conjecture that a(n) = n!*Xi(n), where Xi(1) = 1, Xi(7) = 1/140, and Xi(n) = Product_{p <= n} p^min(2, v_p(H_n)) for n <> 1, 7, where v_p(r) is the p-adic valuation of rational r. (Here p indicates a prime and H_n is the n-th harmonic number.) - Petros Hadjicostas, May 24 2020

			From _Petros Hadjicostas_, May 24 2020: (Start)
To illustrate the Krattenhaler-Rivoal conjecture consider the case n = 24. Then H_24 = Sum_{k=1..24} 1/k = 1347822955/356948592 and {p <= 24} = {2, 3, 5, 7, 11, 13, 17, 19, 23} with {v_p(numerator): p <= 24} = {0, 0, 1, 0, 0, 0, 0, 0, 0} and {v_p(denominator): p <= 24} = {4, 1, 0, 1, 1, 1, 1, 1, 1}.
Thus, the conjectured value for a(24) is 24! * (2^(0-4) * 3^(0-1) * 5^(1-0) * 7^(0-1) * 11^(0-1) * 13^(0-1) * 17^(0-1) * 19^(0-1) * 23^(0-1)) since no exponent of a prime is > 2. This product equals 8691005030400000 = a(24). (End)

Christian Krattenthaler, Table of n, a(n) for n = 1..40
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, arXiv:0709.1432 [math.NT], 2007-2009.
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, II, Communications in Number Theory and Physics, 3(3) (2009), 555-591. [Part II appeared before Part I.]
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, Duke Math. J. 151(2) (2010), 175-218.

Cf. A007757, A056612, A131658.

A formula, conditional on a widely believed conjecture, can be found in Theorem 3 with k = 1 in the article by Krattenthaler and Rivoal (2007-2009) cited in the references; see the remarks before Theorem 4 in that article.

A131658 For n >= 1, put A_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j and B_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j * (Sum__{k=j+1..jn} (1/k)), and let u(n) be the largest integer for which exp(B_n(z)/(u(n)A_n(z))) has integral coefficients. The sequence is u(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, 49380710400000, 217275125760000, 1086375628800000, 1738201006080000

Offset: 1

Christian Krattenthaler (Christian.Krattenthaler(AT)univie.ac.at), Sep 12 2007, Sep 30 2007

nonn

Different from A131657 and A056612.

Christian Krattenthaler, Table of n, a(n) for n = 1..40
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, arXiv:0709.1432 [math.NT], 2007-2009.
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, II, Communications in Number Theory and Physics, 3(3) (2009), 555-591. [Part II appeared before Part I.]
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, Duke Math. J. 151(2) (2010), 175-218.

Cf. A007757 (bisection at even integers), A056612, A131657.

A formula, conditional on a widely believed conjecture, can be found in the article by Krattenthaler and Rivoal (2007-2009) cited in the references: see Theorem 4 and the accompanying remarks.

A007757 Dwork-Kontsevich sequence evaluated at 2*n.

Original entry on oeis.org

1, 2, 36, 144, 1440, 17280, 241920, 29030400, 1567641600, 156764160000, 217275125760000, 1738201006080000, 45193226158080000, 3796230997278720000, 113886929918361600000, 1822190878693785600000, 22489479824838701875200000, 28336744579296764362752000000, 1076796294013277045784576000000, 1679802218660712191423938560000000

Offset: 1

Richard E. Borcherds (reb(AT)math.berkeley.edu)

nonn

For n positive, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum_{k=j+1..j*n} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)*A_n(z))) has integral coefficients. The sequence is b(2*n).

A formula, conditional on a widely believed conjecture, can be found in the Krattenthaler-Rivoal (2007-2009) paper; see Theorem 4 with k = 1 and the remarks on the top of page 8. Since R. E. Borcherds defined a sequence b(n), but then only entered b(2*n) in the OEIS, the formula has to be taken with n replaced by 2*n. - Christian Krattenthaler (Christian.Krattenthaler(AT)univie.ac.at), Sep 12 2007

			G.f. = x + 2*x^2 + 36*x^3 + 144*x^4 + 1440*x^5 + 17280*x^6 + 241920*x^7 + ...

Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, arXiv:0709.1432 [math.NT], 2007-2009.
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, II, Communications in Number Theory and Physics, 3(3) (2009), 555-591. [Part II appeared before Part I.]
Christian Krattenthaler and Tanguy Rivoal, On the integrality of the Taylor coefficients of mirror maps, Duke Math. J. 151(2) (2010), 175-218.

Cf. A131657, A131658, A056612.

a[n0_] := Module[{A, MM = 2, n = 2n0, c1, c2}, A = Exp[Sum[x^j (n j)!/ (j!^n) Sum[1/k, {k, j+1, j n}], {j, 0, MM}]/Sum[x^j (n j)!/(j!^n), {j, 0, MM}]]; c1 = SeriesCoefficient[A, {x, 0, 1}]; c2 = SeriesCoefficient[A, {x, 0, 2}]; GCD[c1, (c1 + c1^2)/2 - c2]];
Array[a, 20] (* Jean-François Alcover, Dec 17 2018, from PARI *)

{a(n) = my(A, MM=2, c1, c2); if(n<1, 0, n*=2; A = x * O(x^MM); A = exp( sum(j=0, MM, x^j * (n*j)! / (j!^n) * sum(k=j+1, j*n, 1/k), A) / sum(j=0, MM, x^j * (n*j)! / (j!^n), A)); c1 = polcoeff(A, 1); c2 = polcoeff(A, 2); gcd(c1, (c1 + c1^2)/2 - c2))}; /* Michael Somos, Nov 16 2006 */

Definition in comment line, PARI code and terms of sequence corrected by Christian Krattenthaler (christian.krattenthaler(AT)univie.ac.at), Sep 30 2007

a(8) corrected by Sean A. Irvine, Jan 22 2018

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)).

A335023 Ratios of consecutive terms of A334958.

Original entry on oeis.org

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).

cp's OEIS Frontend

A335024 Ratios of consecutive terms of A056612.

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A131657 For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum_{k=1..j*n} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)*A_n(z))) has integral coefficients. The sequence is b(n).

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A131658 For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum__{k=j+1..j*n} (1/k)), and let u(n) be the largest integer for which exp(B_n(z)/(u(n)*A_n(z))) has integral coefficients. The sequence is u(n).

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A007757 Dwork-Kontsevich sequence evaluated at 2*n.

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PARI

Extensions

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

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SageMath

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A335023 Ratios of consecutive terms of A334958.

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Maple

Mathematica

PARI

Formula

A131657 For n >= 1, put A_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j and B_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j * (Sum_{k=1..jn} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)A_n(z))) has integral coefficients. The sequence is b(n).

A131658 For n >= 1, put A_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j and B_n(z) = Sum_{j>=0} (nj)!/(j!^n) z^j * (Sum__{k=j+1..jn} (1/k)), and let u(n) be the largest integer for which exp(B_n(z)/(u(n)A_n(z))) has integral coefficients. The sequence is u(n).