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.

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

1,5

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

Links

Paul D. Hanna, Table of n, a(n) for n = 1..1035 for rows 1..45 of the flattened form of this triangle.

Enrica Duchi, Veronica Guerrini, Simone Rinaldi, and Gilles Schaeffer, Fighting Fish: enumerative properties, arXiv:1611.04625 [math.CO], 2016.

Thomas Einolf, Robert Muth, and Jeffrey Wilkinson, Injectively k-colored rooted forests, arXiv:2107.13417 [math.CO], 2021.

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.

Example

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)!).

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 *)

Prog

(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(""))

Crossrefs

Cf. A278881 (C(x,m)), A278882 (D(x,m)), A278883 (central terms).

Cf. A000108, A006013 (row sums), A258313, A278745, A082680.

Sequence in context: A300342 A119725 A239279 * A111910 A181143 A144438

Adjacent sequences: A278877 A278878 A278879 * A278881 A278882 A278883

Keyword

nonn,tabl

Author

Paul D. Hanna, Nov 29 2016

Status

approved