A113788 Number of irreducible multiple zeta values at weight n.
0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 7, 8, 11, 13, 17, 21, 28, 34, 45, 56, 73, 92, 120, 151, 197, 250, 324, 414, 537, 687, 892, 1145, 1484, 1911, 2479, 3196, 4148, 5359, 6954, 9000, 11687, 15140, 19672, 25516, 33166, 43065, 56010, 72784, 94716, 123185
Offset: 1
Links
- Danny Rorabaugh, Table of n, a(n) for n = 1..8000
- Kam Cheong Au, Evaluation of one-dimensional polylogarithmic integral, with applications to infinite series, arXiv:2007.03957 [math.NT], 2020. See 1st line of Table 1 (p. 6).
- R. Bisdorff and J.-L. Marichal, Counting non-isomorphic maximal independent sets of the n-cycle graph, arXiv:0701647 [math.CO], 2007-2008.
- R. Bisdorff and J.-L. Marichal, Counting non-isomorphic maximal independent sets of the n-cycle graph, JIS 11 (2008), #08.5.7.
- D. J. Broadhurst and D. Kreimer, Association of multiple zeta values with positive knots via Feynman diagrams up to 9 loops, UTAS-PHYS-96-44; arXiv:hep-th/9609128, 1996.
- D. J. Broadhurst and D. Kreimer, Associated multiple zeta values with positive knots via Feynman diagrams up to 9 knots, Phys. Lett B, 393 (1997), 403-412.
- M. Waldschmidt, Lectures on Multiple Zeta Values, IMSC 2011.
Programs
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Maple
A113788 := proc(n::integer) local resul,d; resul :=0; for d from 1 to n do if n mod d = 0 then resul := resul +numtheory[mobius](n/d)*A001608(d); fi; od: RETURN(resul/n); end: # R. J. Mathar, Apr 25 2006
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Mathematica
(* p = A001608 *) p[n_] := p[n] = p[n-2] + p[n-3]; p[0] = 3; p[1] = 0; p[2] = 2; a[n_] := (1/n)*Sum[MoebiusMu[n/d]*p[d], {d, Divisors[n]}]; Table[a[n], {n, 1, 56}] (* Jean-François Alcover, Jul 16 2012, from first formula *)
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Sage
z = PowerSeriesRing(ZZ, 'z').gen().O(30) r = (1 - (z**2 + z**3)) F = -z*r.derivative()/r [sum(moebius(n//d)*F[d] for d in divisors(n))//n for n in range(1, 24)] # F. Chapoton, Apr 24 2020
Formula
a(n) = (1/n) * Sum_{d|n} mu(n/d)*Perrin(d), where Perrin(d) = A001608 starting with 0, 2, 3, ... .
a(n) = Sum_{d|n} mu(n/d)*A127687(d) = (1/n) * Sum_{d|n} mu(n/d)*A001608(d). - Jean-Luc Marichal (jean-luc.marichal(AT)uni.lu), Jan 24 2007
For p an odd prime, a(p) = Sum_{i=0..floor((p-3)/6)} (A(i)+B(i)-1)!/(A(i)!*B(i)!), where A(i) = (p-3)/2 - 3*i, and B(i) = 1 + 2*i. - Richard Turk, Sep 08 2015
a(n) ~ A060006^n / n. - Vaclav Kotesovec, Oct 09 2019
Comments