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.

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A278880 Triangle where g.f. S = S(x,m) satisfies: S = x/(G(-S^2)*G(-m*S^2)) such that G(x) = 1 + x*G(x)^2 is the g.f. of the Catalan numbers (A000108), as read by rows of coefficients T(n,k) of x^(2*n-1)*m^k in S(x,m) for n>=1, k=0..n-1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 14, 14, 1, 1, 30, 81, 30, 1, 1, 55, 308, 308, 55, 1, 1, 91, 910, 1872, 910, 91, 1, 1, 140, 2268, 8250, 8250, 2268, 140, 1, 1, 204, 4998, 29172, 51425, 29172, 4998, 204, 1, 1, 285, 10032, 87780, 247247, 247247, 87780, 10032, 285, 1, 1, 385, 18711, 233376, 980980, 1565109, 980980, 233376, 18711, 385, 1, 1, 506, 32890, 562419, 3354780, 7970144, 7970144, 3354780, 562419, 32890, 506, 1
Offset: 1

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Author

Paul D. Hanna, Nov 29 2016

Keywords

Comments

T(n,k) = the number of fighting fish with (n-k) left lower free and (k+1) right lower free edges with a marked tail. [See Theorem 3 in the Duchi reference on Fighting Fish: enumerative properties.] - Paul D. Hanna, Dec 08 2016

Examples

			This triangle of coefficients of x^(2*n-1)*m^k in S(x,m) for n>=1, k=0..n-1, begins:
  1;
  1, 1;
  1, 5, 1;
  1, 14, 14, 1;
  1, 30, 81, 30, 1;
  1, 55, 308, 308, 55, 1;
  1, 91, 910, 1872, 910, 91, 1;
  1, 140, 2268, 8250, 8250, 2268, 140, 1;
  1, 204, 4998, 29172, 51425, 29172, 4998, 204, 1;
  1, 285, 10032, 87780, 247247, 247247, 87780, 10032, 285, 1;
  1, 385, 18711, 233376, 980980, 1565109, 980980, 233376, 18711, 385, 1;
  1, 506, 32890, 562419, 3354780, 7970144, 7970144, 3354780, 562419, 32890, 506, 1; ...
Generating function:
S(x,m) = x + (m + 1)*x^3 + (m^2 + 5*m + 1)*x^5 +
 (m^3 + 14*m^2 + 14*m + 1)*x^7 +
 (m^4 + 30*m^3 + 81*m^2 + 30*m + 1)*x^9 +
 (m^5 + 55*m^4 + 308*m^3 + 308*m^2 + 55*m + 1)*x^11 +
 (m^6 + 91*m^5 + 910*m^4 + 1872*m^3 + 910*m^2 + 91*m + 1)*x^13 +
 (m^7 + 140*m^6 + 2268*m^5 + 8250*m^4 + 8250*m^3 + 2268*m^2 + 140*m + 1)*x^15 +...
where S = S(x,m) satisfies:
S = x / ( G(-S^2) * G(-m*S^2) ) such that G(x) = 1 + x*G(x)^2.
Also,
S = x * (1 + x*S) * (1 + m*x*S) / (1 - m*x^2*S^2)^2,
where related series C = C(x,m) and D = D(x,m) satisfy
S = x*C*D, C = 1 + x*S*D, and D = 1 + m*x*S*C,
such that
C = C^2 - S^2,
D = D^2 - m*S^2.
...
The square of the g.f. begins:
S(x,m)^2 = x^2 + (2*m + 2)*x^4 + (3*m^2 + 12*m + 3)*x^6 +
 (4*m^3 + 40*m^2 + 40*m + 4)*x^8 + (5*m^4 + 100*m^3 + 245*m^2 + 100*m + 5)*x^10 +
 (6*m^5 + 210*m^4 + 1008*m^3 + 1008*m^2 + 210*m + 6)*x^12 +
 (7*m^6 + 392*m^5 + 3234*m^4 + 6300*m^3 + 3234*m^2 + 392*m + 7)*x^14 +
 (8*m^7 + 672*m^6 + 8736*m^5 + 29040*m^4 + 29040*m^3 + 8736*m^2 + 672*m + 8)*x^16 +...+ x^(2*n)*Sum_{k=0,n-1} n*A082680(n,k+1)*m^k +...
where A082680(n,k+1) = (n+k)!*(2*n-k-1)!/((k+1)!*(n-k)!*(2*k+1)!*(2*n-2*k-1)!).
		

Crossrefs

Cf. A278881 (C(x,m)), A278882 (D(x,m)), A278883 (central terms).
Cf. A000108, A006013 (row sums), A258313, A278745, A082680.

Programs

  • Mathematica
    T[n_, k_] := (2n-1)/((2n-2k-1)(2k+1)) Binomial[2n-k-2, k] Binomial[n+k-1, n-k-1];
    Table[T[n, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jul 26 2018 *)
  • PARI
    {T(n,k) = my(S=x,C=1,D=1); for(i=0,2*n, S = x*C*D + O(x^(2*n+2)); C = 1 + x*S*D; D = 1 + m*x*S*C;); polcoeff(polcoeff(S,2*n-1,x),k,m)}
    for(n=1,15, for(k=0,n-1, print1(T(n,k),", "));print(""))
    
  • PARI
    /* Explicit formula for T(n,k) */
    {T(n,k) = (2*n-1)/((2*n-2*k-1)*(2*k+1)) * binomial(2*n-k-2,k) * binomial(n+k-1,n-k-1)}
    for(n=1,15, for(k=0,n-1, print1(T(n,k),", "));print(""))

Formula

G.f. S = S(x,m), and related functions C = C(x,m) and D = D(x,m) satisfy:
(1.a) S = x*C*D.
(1.b) C = 1 + x*S*D.
(1.c) D = 1 + m*x*S*C.
...
(2.a) C = C^2 - S^2.
(2.b) D = D^2 - m*S^2.
(2.c) C = (1 + sqrt(1 + 4*S^2))/2.
(2.d) D = (1 + sqrt(1 + 4*m*S^2))/2.
...
(3.a) S = x*(1 + x*S)*(1 + m*x*S) / (1 - m*x^2*S^2)^2.
(3.b) C = (1 + x*S) / (1 - m*x^2*S^2).
(3.c) D = (1 + m*x*S) / (1 - m*x^2*S^2).
(3.d) S = x/((1 - x^2*D^2)*(1 - m*x^2*C^2)).
(3.e) C = 1/(1 - x^2*D^2).
(3.f) D = 1/(1 - m*x^2*C^2).
...
(4.a) x = m^2*x^4*S^5 - 2*m*x^2*S^3 - m*x^3*S^2 + (1 - (m+1)*x^2)*S.
(4.b) 0 = 1 - (1-x^2)*C - 2*m*x^2*C^2 + 2*m*x^2*C^3 + m^2*x^4*C^4 - m^2*x^4*C^5.
(4.c) 0 = 1 - (1-m*x^2)*D - 2*x^2*D^2 + 2*x^2*D^3 + x^4*D^4 - x^4*D^5.
...
(5.a) S(x,m) = Series_Reversion( x*G(-x^2)*G(-m*x^2) ), where G(x) = 1 + x*G(x)^2 is the g.f. of the Catalan numbers (A000108).
Logarithmic derivatives.
(6.a) C'/C = 2*S*S' / (C^2 + S^2).
(6.b) D'/D = 2*m*S*S' / (D^2 + m*S^2).
...
(7.a) S(x,m)^2 = Sum_{n>=1} x^(2*n) * Sum_{k=0,n-1} n*A082680(n,k+1)*m^k, where A082680(n,k+1) = (n+k)!*(2*n-k-1)!/((k+1)!*(n-k)!*(2*k+1)!*(2*n-2*k-1)!).
...
T(n,k) = (2*n-1)/((2*n-2*k-1)*(2*k+1)) * binomial(2*n-k-2,k) * binomial(n+k-1,n-k-1). [From Theorem 3 in the Duchi reference] - Paul D. Hanna, Dec 08 2016
Row sums yield A006013(n-1) = binomial(3*n-2,n-1)/n for n>=1.
Central terms: T(2*n+1, n) = (4*n-3) * ( binomial(3*n-3,n-1)/(2*n-1) )^2 for n>=1.
Sum_{k=0..n-1} 2^k * T(n,k) = A258313(n-1) for n>=1.
Sum_{k=0..2*n-2} (-1)^k * T(2*n-1,k) = A278745(n) for n>=1.
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