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.

A247255 Triangular array read by rows: T(n,k) is the number of weakly unimodal partitions of n in which the greatest part occurs exactly k times, n>=1, 1<=k<=n.

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 6, 1, 0, 1, 12, 2, 0, 0, 1, 21, 4, 1, 0, 0, 1, 38, 6, 2, 0, 0, 0, 1, 63, 11, 3, 1, 0, 0, 0, 1, 106, 16, 5, 2, 0, 0, 0, 0, 1, 170, 27, 7, 3, 1, 0, 0, 0, 0, 1, 272, 40, 11, 4, 2, 0, 0, 0, 0, 0, 1, 422, 63, 16, 6, 3, 1, 0, 0, 0, 0, 0, 1, 653, 92, 24, 8, 4, 2, 0, 0, 0, 0, 0, 0, 1, 986, 141, 34, 12, 5, 3, 1, 0, 0, 0, 0, 0, 0, 1
Offset: 1

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Author

Geoffrey Critzer, Nov 29 2014

Keywords

Comments

These are called stack polyominoes in the Flajolet and Sedgewick reference.

Examples

			    1;
    1,  1;
    3,  0, 1;
    6,  1, 0, 1;
   12,  2, 0, 0, 1;
   21,  4, 1, 0, 0, 1;
   38,  6, 2, 0, 0, 0, 1;
   63, 11, 3, 1, 0, 0, 0, 1;
  106, 16, 5, 2, 0, 0, 0, 0, 1;
  170, 27, 7, 3, 1, 0, 0, 0, 0, 1;
		

References

  • P. Flajolet and R Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, page 46.

Crossrefs

Row sums give A001523.
Main diagonal gives A000012.

Programs

  • Maple
    b:= proc(n, i) option remember; local r; expand(
          `if`(i>n, 0, `if`(irem(n, i, 'r')=0, x^r, 0)+
          add(b(n-i*j, i+1)*(j+1), j=0..n/i)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, 1)):
    seq(T(n), n=1..14);  # Alois P. Heinz, Nov 29 2014
  • Mathematica
    nn = 14; Table[
      Take[Drop[
         CoefficientList[
          Series[ Sum[
            u z^k/(1 - u z^k) Product[1/(1 - z^i), {i, 1, k - 1}]^2, {k,
             1, nn}], {z, 0, nn}], {z, u}], 1], n, {2, n + 1}][[n]], {n,
       1, nn}] // Grid

Formula

G.f.: Sum_{k>=1} y*x^k/(1 - y*x^k)/(Product_{i=1..k-1} (1 - x^i))^2.
For fixed k>=1, T(n,k) ~ Pi^(k-1) * (k-1)! * exp(2*Pi*sqrt(n/3)) / (2^(k+2) * 3^(k/2 + 1/4) * n^(k/2 + 3/4)). - Vaclav Kotesovec, Oct 24 2018