

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).


8



1, 1, 1, 2, 2, 36, 36, 144, 144, 1440, 1440, 17280, 17280, 241920, 3628800, 29030400, 29030400, 1567641600, 1567641600, 783820800000, 9876142080000, 651825377280000, 217275125760000, 8691005030400000
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OFFSET

1,4


COMMENTS

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 (20072009) 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 padic valuation of rational r. (Here p indicates a prime and H_n is the nth harmonic number.)  Petros Hadjicostas, May 24 2020


LINKS



FORMULA

A formula, conditional on a widely believed conjecture, can be found in Theorem 3 with k = 1 in the article by Krattenthaler and Rivoal (20072009) cited in the references; see the remarks before Theorem 4 in that article.


EXAMPLE

To illustrate the KrattenhalerRivoal 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^{04) * 3^(01) * 5^(10) * 7^(01) * 11^(01) * 13^(01) * 17^(01) * 19^(01) * 23^(01)) since no exponent of a prime is > 2. This product equals 8691005030400000 = a(24). (End)


CROSSREFS



KEYWORD

nonn


AUTHOR

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


STATUS

approved



