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#19 by Bruno Berselli at Wed Aug 28 08:15:05 EDT 2019
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#18 by Joerg Arndt at Wed Aug 28 06:46:08 EDT 2019
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#17 by Jean-François Alcover at Wed Aug 28 06:39:44 EDT 2019
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#16 by Jean-François Alcover at Wed Aug 28 06:39:40 EDT 2019
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| MATHEMATICA
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a[n_] := If[n == 0, 1, Sum[MoebiusMu[n/d] (2n Binomial[ 2 Quotient[d, 2], Quotient[d, 2]] + Binomial[2d, d]), {d, Divisors[n]}]/(4n)];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Aug 28 2019, from PARI *)
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| STATUS
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approved
editing
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#15 by Alois P. Heinz at Fri Sep 29 17:57:12 EDT 2017
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#14 by Andrew Howroyd at Fri Sep 29 17:15:23 EDT 2017
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#13 by Andrew Howroyd at Fri Sep 29 17:13:56 EDT 2017
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| FORMULA
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a(n) = 1/(4*n) * Sum_{d|n} mu(n/d) * (2*n*binomial(2*(*floor(d\/2), floor(d\/2) + )) + binomial(2*d, d)) for n > 0. - Andrew Howroyd, Sep 28 2017
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Discussion
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| Fri Sep 29
| 17:14
| Andrew Howroyd: Thanks for spotting that!
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#12 by Alois P. Heinz at Fri Sep 29 17:07:10 EDT 2017
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| FORMULA
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Moebius transform of A005648. - . - _Christian G. Bower_
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| STATUS
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proposed
editing
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Discussion
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| Fri Sep 29
| 17:08
| Alois P. Heinz: could you rewrite "d\2" in formula section please?
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#11 by Andrew Howroyd at Fri Sep 29 16:17:56 EDT 2017
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#10 by Andrew Howroyd at Thu Sep 28 22:01:05 EDT 2017
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| LINKS
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Andrew Howroyd, <a href="/A045628/b045628.txt">Table of n, a(n) for n = 0..200</a>
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| FORMULA
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Moebius transform of A005648 (. - Christian Bower).
a(n) = 1/(4*n) * Sum_{d|n} mu(n/d) * (2*n*binomial(2*(d\2), d\2) + binomial(2*d, d)) for n > 0. - Andrew Howroyd, Sep 28 2017
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| PROG
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(PARI) a(n) = if(n<1, n==0, sumdiv(n, d, moebius(n/d)*(2*n*binomial(2*(d\2), d\2) + binomial(2*d, d)))/(4*n)); \\ Andrew Howroyd, Sep 28 2017
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| CROSSREFS
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Cf. A005648.
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| STATUS
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approved
editing
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