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.

A067311 Triangle read by rows: T(n,k) gives number of ways of arranging n chords on a circle with k simple intersections (i.e., no intersections with 3 or more chords) - positive values only.

Original entry on oeis.org

1, 1, 2, 1, 5, 6, 3, 1, 14, 28, 28, 20, 10, 4, 1, 42, 120, 180, 195, 165, 117, 70, 35, 15, 5, 1, 132, 495, 990, 1430, 1650, 1617, 1386, 1056, 726, 451, 252, 126, 56, 21, 6, 1, 429, 2002, 5005, 9009, 13013, 16016, 17381, 16991, 15197, 12558, 9646, 6916, 4641, 2912, 1703, 924, 462, 210, 84, 28, 7, 1
Offset: 0

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Author

Henry Bottomley, Jan 14 2002

Keywords

Comments

Row n contains 1 + n(n-1)/2 entries. - Emeric Deutsch, Jun 03 2009
Row sums are A001147 (double factorials).
Columns include A000108 (Catalan) for k=0 and A002694 for k=1.
Coefficients of Touchard-Riordan polynomials defined on page 3 of the Chakravarty and Kodama paper, related to the array A039599 through the polynomial numerators of Eqn. 2.1. - Tom Copeland, May 26 2016

Examples

			Rows start:
   1;
   1;
   2,   1;
   5,   6,   3,   1;
  14,  28,  28,  20,  10,   4,   1;
  42, 120, 180, 195, 165, 117,  70,  35,  15,   5,   1;
etc.,
i.e., there are 5 ways of arranging 3 chords with no intersections, 6 with one, 3 with two and 1 with three.
		

References

  • P. Flajolet and M. Noy, Analytic combinatorics of chord diagrams; in Formal Power Series and Algebraic Combinatorics, pp. 191-201, Springer, 2000.

Crossrefs

A067310 has a different view of the same table.
Cf. A039599.

Programs

  • Maple
    p := proc (n) options operator, arrow: sort(simplify((sum((-1)^j*q^((1/2)*j*(j-1))*binomial(2*n, n+j), j = -n .. n))/(1-q)^n)) end proc; for n from 0 to 7 do seq(coeff(p(n), q, i), i = 0 .. (1/2)*n*(n-1)) end do; # yields sequence in triangular form; Emeric Deutsch, Jun 03 2009
  • Mathematica
    nmax = 15; se[n_] := se[n] = Series[ Sum[(-1)^j*q^(j(j-1)/2)*Binomial[2 n, n+j], {j, -n, n}]/(1-q)^n , {q, 0, nmax}];
    t[n_, k_] := Coefficient[se[n], q^k]; t[n_, 0] = Binomial[2 n, n]/(n + 1);
    Select[Flatten[Table[t[n, k], {n, 0, nmax}, {k, 0, 2nmax}] ], Positive] [[1 ;; 55]]
    (* Jean-François Alcover, Jun 22 2011, after Emeric Deutsch *)
  • PARI
    M(n)=1/(1-q)^n*sum(k=0, n, (-1)^k * ( binomial(2*n,n-k)-binomial(2*n,n-k-1)) * q^(k*(k+1)/2) );
    for (n=0,10, print( Vec(polrecip(M(n))) ) ); /* print rows */
    /* Joerg Arndt, Oct 01 2012 */

Formula

T(n,k) = Sum_{j=0..n-1} (-1)^j * C((n-j)*(n-j+1)/2-1-k, n-1) * (C(2n, j) - C(2n, j-1)).
Generating polynomial of row n is (1-q)^(-n)*Sum_{j=-n..n} (-1)^j*q^(j*(j-1)/2)*binomial(2*n,n+j). - Emeric Deutsch, Jun 03 2009
O.g.f. as a continued fraction: 1/(1 - t/(1 - (1 + x)*t/(1 - (1 + x + x^2)*t/(1 - (1 + x + x^2 + x^3)*t/(1 - ...))))) = 1 + t + (2 + x)*t^2 + (5 + 6*x +3*x^2 + x^4)*t^3 + .... See Chakravarty and Kodama, equation 3.8. - Peter Bala, Jun 13 2019

Extensions

a(55) onwards from Andrew Howroyd, Nov 22 2024