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.

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

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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