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.

User: Kevin Liu

Kevin Liu's wiki page.

Kevin Liu has authored 3 sequences.

A371659 Triangle read by rows: T(n,k) is the number of planar tanglegrams of size n with irreducible component of size k.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 3, 3, 5, 0, 13, 9, 20, 34, 0, 90, 46, 70, 170, 273, 0, 747, 312, 360, 680, 1638, 2436, 0, 7040, 2580, 2435, 3570, 7371, 17052, 23391, 0, 71736, 24056, 19800, 23970, 39858, 85260, 187128, 237090, 0, 774738, 243483, 182850, 193664, 267813, 477456, 1029204, 2133810, 2505228
Offset: 1

Author

Kevin Liu, Apr 01 2024

Keywords

Comments

A proper subtanglegram of a planar tanglegram is a pair of subtrees whose leaves are matched in the tanglegram, and the irreducible component of a planar tanglegram is formed by contracting each maximal proper subtanglegram into a pair of matched leaves.

Examples

			Triangle begins
  1;
  0,    1;
  0,    1,    1;
  0,    3,    3,    5;
  0,   13,    9,   20,   34;
  0,   90,   46,   70,  170,  273;
  0,  747,  312,  360,  680, 1638,  2436;
  0, 7040, 2580, 2435, 3570, 7371, 17052, 23391;
  ...
		

Crossrefs

Cf. A349408 (diagonal), A257887 (row sums).

Formula

G.f.: F(x,y) = H(F(x),y) + x*y + y^2*(F(x)^2 + F(x^2))/2 where the coefficient of x^n*y^k is the number of planar tanglegrams of size n with irreducible component of size k, F(x) is the g.f. for A349408, and H(x)/x^2 is the g.f. for A257887.

A349409 Triangle read by rows: T(n,k) is the number of planar tanglegrams of size n with 0 <= k < n leaf-matched pairs. A leaf matched pair is a pair of non-leaf vertices (u,v) in the tanglegram such that the induced subtrees rooted and u and v also form a tanglegram (equivalently, the leaves in these two subtrees are matched by the matching that forms the original tanglegram).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 5, 4, 2, 0, 34, 28, 11, 3, 0, 273, 239, 102, 29, 6, 0, 2436, 2283, 1045, 325, 73, 11, 0, 23391, 23475, 11539, 3852, 968, 181, 23, 0, 237090, 254309, 133690, 47640, 12923, 2756, 444, 46, 0, 2505228, 2864283, 1605280, 607743, 175976, 40903, 7650, 1085, 98
Offset: 1

Author

Kevin Liu, Nov 16 2021

Keywords

Comments

The generating function can be proven using a generalization of the proof for A349408.

Examples

			Triangle begins
      1;
      0,      1;
      0,      1,     1;
      0,      5,     4,     2;
      0,     34,    28,    11,    3;
      0,    273,   239,   102,   29,   6;
      0,   2436,  2283,  1045,  325,  73,  11;
      0,  23391, 23475, 11539, 3852, 968, 181, 23;
      ...
		

Crossrefs

Cf. A257887 (2nd column), A349408 (row sums), A001190 (diagonal).

Programs

  • PARI
    \\ here H(n)/x^2 is g.f. of A257887.
    H(n)={(x - x^2 - serreverse(sum(k=0, n+1, (binomial(2*k, k)/(k+1))^2*x^(k+1)) + O(x^(n+3))))/2}
    F(n)={my(h=H(n-2), p=O(x)); for(n=1, n, p = x + y*subst(h + O(x*x^n), x, p) + y*(p^2 + subst(subst(p,x,x^2),y,y^2))/2); p}
    T(n)={[Vecrev(p) | p<-Vec(F(n))]}
    {my(v=T(10)); for(n=1, #v, print(v[n]))} \\ Andrew Howroyd, Nov 18 2021

Formula

G.f.: F(x,q) = q*H(F(x,q)) + x + q*(F(x,q)^2 + F(x^2,q^2))/2 where coefficient of x^n*q^k is the number of planar tanglegrams with size n and k leaf-matched pairs, and H(x)/x^2 is the g.f. for A257887.

A349408 Number of planar tanglegrams of size n.

Original entry on oeis.org

1, 1, 2, 11, 76, 649, 6173, 63429, 688898, 7808246, 91537482, 1102931565, 13594564857, 170804438005, 2181426973452, 28257128116954, 370581034530685, 4913238656392058, 65773613137623085, 888155942037325535, 12086555915234897267, 165641209243876120135
Offset: 1

Author

Kevin Liu, Nov 16 2021

Keywords

Examples

			For n=4, there are 11 planar tanglegrams of size 4.
		

Crossrefs

Row sums of A349409.

Programs

  • PARI
    \\ here H(n)/x^2 is g.f. of A257887.
    H(n)={(x - x^2 - serreverse(sum(k=0, n+1, (binomial(2*k, k)/(k+1))^2*x^(k+1)) + O(x^(n+3))))/2}
    seq(n)={my(h=H(n-2), p=O(x)); for(n=1, n, p = subst(h + O(x*x^n), x, p) + x + (p^2 + subst(p,x,x^2))/2); Vec(p)} \\ Andrew Howroyd, Nov 18 2021

Formula

G.f.: F(x) satisfies F(x) = H(F(x)) + x + (F(x)^2 + F(x^2))/2 where H(x)/x^2 is the g.f. of A257887.

Extensions

Terms a(11) and beyond from Andrew Howroyd, Nov 18 2021