cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A185100 Dihedral unlabeled Motzkin numbers: number of ways of drawing any number of nonintersecting chords joining n unlabeled points equally spaced on a circle, up to rotations and reflections of the circle.

Original entry on oeis.org

1, 1, 2, 2, 4, 5, 11, 16, 36, 65, 150, 312, 756, 1743, 4353, 10732, 27489, 70379, 183866, 481952, 1277784, 3402661, 9126689, 24584870, 66567924, 180939737, 493801694, 1352203202, 3715137460, 10237545525, 28291018283, 78384998904, 217715672036, 606103034821, 1691020991782, 4727601528674, 13242641322252, 37162431389051, 104469244613429
Offset: 0

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Author

Max Alekseyev, Feb 07 2011

Keywords

Comments

Unlabeled version of A001006. Another version is given by A175954.
The number of ways of drawing exactly n chords joining 2n unlabeled points up to rotations and reflections is A006082(n+1). - Andrey Zabolotskiy, May 24 2018

Crossrefs

Cf. A001006 (labeled points), A175954 (up to rotations only), A175955, A005773, A006082.

Programs

  • Mathematica
    a1006[0] = 1; a1006[n_Integer] := a1006[n] = a1006[n - 1] + Sum[a1006[k]* a1006[n - 2 - k], {k, 0, n - 2}];
    a142150[n_] := n*(1 + (-1)^n)/4;
    a2426[n_] := Coefficient[(1 + x + x^2)^n, x, n];
    a175954[0] = 1; a175954[n_] := (1/n)*(a1006[n] + a142150[n]*a1006[n/2 - 1] + Sum[EulerPhi[n/d]*a2426[d], {d, Most @Divisors[n]}]);
    a5773[0] = 1; a5773[n_] := Sum[k/n*Sum[Binomial[n, j]*Binomial[j, 2*j - n - k], {j, 0, n}], {k, 1, n}];
    a[0] = 1;
    a[n_?OddQ] := With[{m = (n-1)/2}, (1/2)*(a175954[2*m + 1] + a5773[m + 1])];
    a[n_?EvenQ] := With[{m = n/2}, (1/4)*(2*a175954[2*m] + a5773[m] + a5773[m + 1] + a1006[m - 1])];
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Jul 02 2018, after Andrew Howroyd *)

Formula

a(2n+1) = (1/2) * (A175954(2n+1) + A005773(n+1)). - Andrew Howroyd, Apr 01 2017
a(2n) = (1/4) * (2 * A175954(2n) + A005773(n) + A005773(n+1) + A001006(n-1)) for n > 0. - Andrew Howroyd, Apr 01 2017