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.

Showing 1-2 of 2 results.

A227543 Triangle defined by g.f. A(x,q) such that: A(x,q) = 1 + x*A(q*x,q)*A(x,q), as read by terms k=0..n*(n-1)/2 in rows n>=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 2, 1, 1, 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1, 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1, 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1, 1, 7, 21, 41, 65, 86, 102, 115, 118, 118, 113, 106, 96, 85, 73, 63, 53, 42, 34, 26, 20, 15, 11, 7, 5, 3, 2, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Jul 15 2013

Keywords

Comments

See related triangle A138158.
Row sums are the Catalan numbers (A000108), set q=1 in the g.f. to see this.
Antidiagonal sums equal A005169, the number of fountains of n coins.
The maximum in each row of the triangle is A274291. - Torsten Muetze, Nov 28 2018
The area between a Dyck path and the x-axis may be decomposed into unit area triangles of two types - up-triangles with vertices at the integer lattice points (x, y), (x+1, y+1) and (x+2, y) and down-triangles with vertices at the integer lattice points (x, y), (x-1, y+1) and (x+1, y+1). The table entry T(n,k) equals the number of Dyck paths of semilength n containing k down triangles. See the illustration in the Links section. Cf. A239927. - Peter Bala, Jul 11 2019
The row polynomials of this table are a q-analog of the Catalan numbers due to Carlitz and Riordan. For MacMahon's q-analog of the Catalan numbers see A129175. - Peter Bala, Feb 28 2023

Examples

			G.f.: A(x,q) = 1 + x*(1) + x^2*(1 + q) + x^3*(1 + 2*q + q^2 + q^3)
 + x^4*(1 + 3*q + 3*q^2 + 3*q^3 + 2*q^4 + q^5 + q^6)
 + x^5*(1 + 4*q + 6*q^2 + 7*q^3 + 7*q^4 + 5*q^5 + 5*q^6 + 3*q^7 + 2*q^8 + q^9 + q^10)
 + x^6*(1 + 5*q + 10*q^2 + 14*q^3 + 17*q^4 + 16*q^5 + 16*q^6 + 14*q^7 + 11*q^8 + 9*q^9 + 7*q^10 + 5*q^11 + 3*q^12 + 2*q^13 + q^14 + q^15) +...
where g.f.A(x,q) = Sum_{k=0..n*(n-1)/2, n>=0} T(n,k)*x^n*q^k
satisfies A(x,q) = 1 + x*A(q*x,q)*A(x,q).
This triangle of coefficients T(n,k) in A(x,q) begins:
 1;
 1;
 1, 1;
 1, 2, 1, 1;
 1, 3, 3, 3, 2, 1, 1;
 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1;
 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1;
 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1;
 1, 7, 21, 41, 65, 86, 102, 115, 118, 118, 113, 106, 96, 85, 73, 63, 53, 42, 34, 26, 20, 15, 11, 7, 5, 3, 2, 1, 1;
 1, 8, 28, 63, 112, 167, 219, 268, 303, 326, 338, 338, 331, 314, 293, 268, 245, 215, 190, 162, 139, 116, 97, 77, 63, 48, 38, 28, 22, 15, 11, 7, 5, 3, 2, 1, 1; ...
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := Module[{P, Q},
    P = Sum[q^(m^2) (-x)^m/Product[1-q^j, {j, 1, m}] + x O[x]^n, {m, 0, n}];
    Q = Sum[q^(m(m-1)) (-x)^m/Product[1-q^j, {j, 1, m}] + x O[x]^n, {m, 0, n}];
    SeriesCoefficient[P/Q, {x, 0, n}, {q, 0, k}]
    ];
    Table[T[n, k], {n, 0, 10}, {k, 0, n(n-1)/2}] // Flatten (* Jean-François Alcover, Jul 27 2018, from PARI *)
  • PARI
    /* From g.f. A(x,q) = 1 + x*A(q*x,q)*A(x,q): */
    {T(n, k)=local(A=1); for(i=1, n, A=1+x*subst(A, x, q*x)*A +x*O(x^n)); polcoeff(polcoeff(A, n, x), k, q)}
    for(n=0, 10, for(k=0, n*(n-1)/2, print1(T(n, k), ", ")); print(""))
    
  • PARI
    /* By Ramanujan's continued fraction identity: */
    {T(n,k)=local(P=1,Q=1);
    P=sum(m=0,n,q^(m^2)*(-x)^m/prod(k=1,m,1-q^k)+x*O(x^n));
    Q=sum(m=0,n,q^(m*(m-1))*(-x)^m/prod(k=1,m,1-q^k)+x*O(x^n));
    polcoeff(polcoeff(P/Q,n,x),k,q)}
    for(n=0, 10, for(k=0, n*(n-1)/2, print1(T(n, k), ", ")); print(""))
    
  • PARI
    P(x, n) =
    {
        if ( n<=1, return(1) );
        return( sum( i=0, n-1, P(x, i) * P(x, n-1 -i) * x^((i+1)*(n-1 -i)) ) );
    }
    for (n=0, 10, print( Vec( P(x, n) ) ) ); \\ Joerg Arndt, Jan 23 2024
    
  • PARI
    \\ faster with memoization:
    N=11;
    VP=vector(N+1);  VP[1] =VP[2] = 1;  \\ one-based; memoization
    P(n) = VP[n+1];
    for (n=2, N, VP[n+1] = sum( i=0, n-1, P(i) * P(n-1 -i) * x^((i+1)*(n-1-i)) ) );
    for (n=0, N, print( Vec( P(n) ) ) ); \\ Joerg Arndt, Jan 23 2024

Formula

G.f.: A(x,q) = 1/(1 - x/(1 - q*x/(1 - q^2*x/(1 - q^3*x/(1 - q^4*x/(1 -...)))))), a continued fraction.
G.f. satisfies: A(x,q) = P(x,q)/Q(x,q), where
P(x,q) = Sum_{n>=0} q^(n^2) * (-x)^n / Product_{k=1..n} (1-q^k),
Q(x,q) = Sum_{n>=0} q^(n*(n-1)) * (-x)^n / Product_{k=1..n} (1-q^k),
due to Ramanujan's continued fraction identity.
...
Sum_{k=0..n*(n-1)/2} T(n,k)*k = 2^(2*n-1) - C(2*n+1,n) + C(2*n-1,n-1) = A006419(n-1) for n>=1.
Logarithmic derivative of the g.f. A(x,q), wrt x, yields triangle A227532.
From Peter Bala, Jul 11 2019: (Start)
(n+1)th row polynomial R(n+1,q) = Sum_{k = 0..n} q^k*R(k,x)*R(n-k,q), with R(0,q) = 1.
1/A(q*x,q) is the generating function for the triangle A047998. (End)
Conjecture: b(n) = P(n, n) where b(n) is an integer sequence with g.f. B(x) = 1/(1 - f(0)*x/(1 - f(1)*x/(1 - f(2)*x/(1 - f(3)*x/(1 - f(4)*x/(1 -...)))))), P(n, k) = P(n-1, k) + f(n-k)*P(n, k-1) for 0 < k <= n with P(n, k) = 0 for k > n, P(n, 0) = 1 for n >= 0 and where f(n) is an arbitrary function. In fact for this sequence we have f(n) = q^n. - Mikhail Kurkov, Sep 26 2024

A138158 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and path length k; 0 <= k <= n(n+1)/2.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 2, 1, 1, 0, 0, 0, 0, 1, 3, 3, 3, 2, 1, 1, 0, 0, 0, 0, 0, 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1, 0, 0, 0, 0, 0, 0, 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1
Offset: 0

Views

Author

Emeric Deutsch, Mar 21 2008

Keywords

Comments

T(n,k) is the number of Dyck paths of semilength n for which the sum of the heights of the vertices that terminate an upstep (i.e. peaks and doublerises) is k. Example: T(4,7)=3 because we have UUDUDUDD, UDUUUDDD and UUUDDDUD.
See related triangle A227543.
Row n contains 1+n(n+1)/2 terms.
The maximum in each row of the triangle is A274291. - Torsten Muetze, Nov 28 2018
It appears that for j = 0,1,...,n-1 the first j terms of the rows in reversed order are given by A000041(j), the partition numbers. - Geoffrey Critzer, Jul 14 2020

Examples

			T(2,2)=1 because /\ is the only ordered tree with 2 edges and path length 2.
Triangle starts
 1,
 0, 1,
 0, 0, 1, 1,
 0, 0, 0, 1, 2, 1, 1,
 0, 0, 0, 0, 1, 3, 3, 3, 2, 1, 1,
 0, 0, 0, 0, 0, 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1,
 0, 0, 0, 0, 0, 0, 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1,
 0, 0, 0, 0, 0, 0, 0, 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1,
... [_Joerg Arndt_, Feb 21 2014]
		

Crossrefs

Programs

  • Maple
    P[0]:=1: for n to 7 do P[n]:=sort(expand(t*(sum(P[j]*P[n-j-1]*t^(n-j-1),j= 0.. n-1)))) end do: for n from 0 to 7 do seq(coeff(P[n], t, j),j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
  • Mathematica
    nmax = 7;
    P[0] = 1; P[n_] := P[n] = t*Sum[P[j]*P[n-j-1]*t^(n-j-1), {j, 0, n-1}];
    row[n_] := row[n] = CoefficientList[P[n] + O[t]^(n(n+1)/2 + 1), t];
    T[n_, k_] := row[n][[k+1]];
    Table[T[n, k], {n, 0, nmax}, {k, 0, n(n+1)/2}] // Flatten (* Jean-François Alcover, Jul 11 2018, from Maple *)
    nn = 10; f[z_, u_] := Sum[Sum[a[n, k] u^k z^n, {k, 0, Binomial[n, 2]}], {n, 1, nn}]; sol = SolveAlways[Series[0 == f[z, u] - z/(1 - f[u z, u]) , {z, 0, nn}], {z, u}];Level[Table[Table[a[n, k], {k, 0, Binomial[n, 2]}], {n, 1, nn}] /.
    sol, {2}] // Grid (* Geoffrey Critzer, Jul 14 2020 *)

Formula

G.f. G(t,z) satisfies G(t,z) = 1+t*z*G(t,z)*G(t,t*z).
Row generating polynomials P[n]=P[n](t) are given by P[0]=1, P[n] = t * Sum( P[j]*P[n-j-1]*t^(n-1-j), j=0..n-1 ) (n>=1).
Row sums are the Catalan numbers (A000108).
Sum of entries in column n = A005169(n).
Sum_{k=0..n(n+1)/2} k*T(n,k) = A000346(n-1).
T(n,k) = A047998(k,n).
G.f.: 1/(1 - x*y/(1 - x*y^2/(1 - x*y^3/(1 - x*y^4/(1 - x*y^5)/(1 - ... ))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
Showing 1-2 of 2 results.