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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.
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%I #61 Dec 09 2021 01:00:03

%S 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,

%T 910,91,1,1,140,2268,8250,8250,2268,140,1,1,204,4998,29172,51425,

%U 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

%N 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.

%C 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

%H Paul D. Hanna, <a href="/A278880/b278880.txt">Table of n, a(n) for n = 1..1035 for rows 1..45 of the flattened form of this triangle.</a>

%H Enrica Duchi, Veronica Guerrini, Simone Rinaldi, and Gilles Schaeffer, <a href="https://arxiv.org/abs/1611.04625">Fighting Fish: enumerative properties</a>, arXiv:1611.04625 [math.CO], 2016.

%H Thomas Einolf, Robert Muth, and Jeffrey Wilkinson, <a href="https://arxiv.org/abs/2107.13417">Injectively k-colored rooted forests</a>, arXiv:2107.13417 [math.CO], 2021.

%F G.f. S = S(x,m), and related functions C = C(x,m) and D = D(x,m) satisfy:

%F (1.a) S = x*C*D.

%F (1.b) C = 1 + x*S*D.

%F (1.c) D = 1 + m*x*S*C.

%F ...

%F (2.a) C = C^2 - S^2.

%F (2.b) D = D^2 - m*S^2.

%F (2.c) C = (1 + sqrt(1 + 4*S^2))/2.

%F (2.d) D = (1 + sqrt(1 + 4*m*S^2))/2.

%F ...

%F (3.a) S = x*(1 + x*S)*(1 + m*x*S) / (1 - m*x^2*S^2)^2.

%F (3.b) C = (1 + x*S) / (1 - m*x^2*S^2).

%F (3.c) D = (1 + m*x*S) / (1 - m*x^2*S^2).

%F (3.d) S = x/((1 - x^2*D^2)*(1 - m*x^2*C^2)).

%F (3.e) C = 1/(1 - x^2*D^2).

%F (3.f) D = 1/(1 - m*x^2*C^2).

%F ...

%F (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.

%F (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.

%F (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.

%F ...

%F (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).

%F Logarithmic derivatives.

%F (6.a) C'/C = 2*S*S' / (C^2 + S^2).

%F (6.b) D'/D = 2*m*S*S' / (D^2 + m*S^2).

%F ...

%F (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)!).

%F ...

%F 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

%F Row sums yield A006013(n-1) = binomial(3*n-2,n-1)/n for n>=1.

%F Central terms: T(2*n+1, n) = (4*n-3) * ( binomial(3*n-3,n-1)/(2*n-1) )^2 for n>=1.

%F Sum_{k=0..n-1} 2^k * T(n,k) = A258313(n-1) for n>=1.

%F Sum_{k=0..2*n-2} (-1)^k * T(2*n-1,k) = A278745(n) for n>=1.

%e This triangle of coefficients of x^(2*n-1)*m^k in S(x,m) for n>=1, k=0..n-1, begins:

%e 1;

%e 1, 1;

%e 1, 5, 1;

%e 1, 14, 14, 1;

%e 1, 30, 81, 30, 1;

%e 1, 55, 308, 308, 55, 1;

%e 1, 91, 910, 1872, 910, 91, 1;

%e 1, 140, 2268, 8250, 8250, 2268, 140, 1;

%e 1, 204, 4998, 29172, 51425, 29172, 4998, 204, 1;

%e 1, 285, 10032, 87780, 247247, 247247, 87780, 10032, 285, 1;

%e 1, 385, 18711, 233376, 980980, 1565109, 980980, 233376, 18711, 385, 1;

%e 1, 506, 32890, 562419, 3354780, 7970144, 7970144, 3354780, 562419, 32890, 506, 1; ...

%e Generating function:

%e S(x,m) = x + (m + 1)*x^3 + (m^2 + 5*m + 1)*x^5 +

%e (m^3 + 14*m^2 + 14*m + 1)*x^7 +

%e (m^4 + 30*m^3 + 81*m^2 + 30*m + 1)*x^9 +

%e (m^5 + 55*m^4 + 308*m^3 + 308*m^2 + 55*m + 1)*x^11 +

%e (m^6 + 91*m^5 + 910*m^4 + 1872*m^3 + 910*m^2 + 91*m + 1)*x^13 +

%e (m^7 + 140*m^6 + 2268*m^5 + 8250*m^4 + 8250*m^3 + 2268*m^2 + 140*m + 1)*x^15 +...

%e where S = S(x,m) satisfies:

%e S = x / ( G(-S^2) * G(-m*S^2) ) such that G(x) = 1 + x*G(x)^2.

%e Also,

%e S = x * (1 + x*S) * (1 + m*x*S) / (1 - m*x^2*S^2)^2,

%e where related series C = C(x,m) and D = D(x,m) satisfy

%e S = x*C*D, C = 1 + x*S*D, and D = 1 + m*x*S*C,

%e such that

%e C = C^2 - S^2,

%e D = D^2 - m*S^2.

%e ...

%e The square of the g.f. begins:

%e S(x,m)^2 = x^2 + (2*m + 2)*x^4 + (3*m^2 + 12*m + 3)*x^6 +

%e (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 +

%e (6*m^5 + 210*m^4 + 1008*m^3 + 1008*m^2 + 210*m + 6)*x^12 +

%e (7*m^6 + 392*m^5 + 3234*m^4 + 6300*m^3 + 3234*m^2 + 392*m + 7)*x^14 +

%e (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 +...

%e where A082680(n,k+1) = (n+k)!*(2*n-k-1)!/((k+1)!*(n-k)!*(2*k+1)!*(2*n-2*k-1)!).

%t T[n_, k_] := (2n-1)/((2n-2k-1)(2k+1)) Binomial[2n-k-2, k] Binomial[n+k-1, n-k-1];

%t Table[T[n, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* _Jean-François Alcover_, Jul 26 2018 *)

%o (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)}

%o for(n=1,15, for(k=0,n-1, print1(T(n,k),", "));print(""))

%o (PARI) /* Explicit formula for T(n,k) */

%o {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)}

%o for(n=1,15, for(k=0,n-1, print1(T(n,k),", "));print(""))

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

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

%K nonn,tabl

%O 1,5

%A _Paul D. Hanna_, Nov 29 2016