A278881 -id:A278881 - cp's OEIS frontend

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

Paul D. Hanna, Nov 29 2016

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

			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)!).

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.
Zai-Ting Huang, Congruence Properties From Fish Numbers, Master's thesis, Nat'l Taiwan Normal Univ. (2025) 31944792.

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

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

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 *)

{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(""))

/* 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(""))

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.

A278882 Triangle where g.f. D = D(x,m) and related series S = S(x,m) and C = C(x,m) satisfy S = xCD, C = 1 + xSD, and D = 1 + mxS*C, as read by rows of coefficients T(n,k) of x^(2n)m^k in C(x,m) for n>=1, k=0..n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 3, 8, 1, 0, 4, 30, 20, 1, 0, 5, 80, 147, 40, 1, 0, 6, 175, 672, 504, 70, 1, 0, 7, 336, 2310, 3600, 1386, 112, 1, 0, 8, 588, 6552, 18150, 14520, 3276, 168, 1, 0, 9, 960, 16170, 72072, 102245, 48048, 6930, 240, 1, 0, 10, 1485, 35904, 240240, 546546, 455455, 137280, 13464, 330, 1, 0, 11, 2200, 73359, 700128, 2382380, 3179904, 1701700, 350064, 24453, 440, 1, 0, 12, 3146, 140140, 1833975, 8868288, 17672928, 15148224, 5542680, 815100, 42042, 572, 1

Offset: 0

Paul D. Hanna, Nov 29 2016

			This triangle of coefficients of x^(2*n)*m^k in D(x,m) for n>=0, k=0..n, begins:
1;
0, 1;
0, 2, 1;
0, 3, 8, 1;
0, 4, 30, 20, 1;
0, 5, 80, 147, 40, 1;
0, 6, 175, 672, 504, 70, 1;
0, 7, 336, 2310, 3600, 1386, 112, 1;
0, 8, 588, 6552, 18150, 14520, 3276, 168, 1;
0, 9, 960, 16170, 72072, 102245, 48048, 6930, 240, 1;
0, 10, 1485, 35904, 240240, 546546, 455455, 137280, 13464, 330, 1;
0, 11, 2200, 73359, 700128, 2382380, 3179904, 1701700, 350064, 24453, 440, 1;
0, 12, 3146, 140140, 1833975, 8868288, 17672928, 15148224, 5542680, 815100, 42042, 572, 1; ...
Generating function:
D(x,m) = 1 + m*x^2 + (2*m + m^2)*x^4 + (3*m + 8*m^2 + m^3)*x^6 +
(4*m + 30*m^2 + 20*m^3 + m^4)*x^8 +
(5*m + 80*m^2 + 147*m^3 + 40*m^4 + m^5)*x^10 +
(6*m + 175*m^2 + 672*m^3 + 504*m^4 + 70*m^5 + m^6)*x^12 +
(7*m + 336*m^2 + 2310*m^3 + 3600*m^4 + 1386*m^5 + 112*m^6 + m^7)*x^14 +
(8*m + 588*m^2 + 6552*m^3 + 18150*m^4 + 14520*m^5 + 3276*m^6 + 168*m^7 + m^8)*x^16 +...
where g.f. D = D(x,m) and related series S = S(x,m) and C = C(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:
D(x,m)^2 = 1 + 2*m*x^2 + (3*m^2 + 4*m)*x^4 +
(4*m^3 + 20*m^2 + 6*m)*x^6 +
(5*m^4 + 60*m^3 + 70*m^2 + 8*m)*x^8 +
(6*m^5 + 140*m^4 + 392*m^3 + 180*m^2 + 10*m)*x^10 +
(7*m^6 + 280*m^5 + 1512*m^4 + 1680*m^3 + 385*m^2 + 12*m)*x^12 +
(8*m^7 + 504*m^6 + 4620*m^5 + 9900*m^4 + 5544*m^3 + 728*m^2 + 14*m)*x^14 +
(9*m^8 + 840*m^7 + 12012*m^6 + 43560*m^5 + 47190*m^4 + 15288*m^3 + 1260*m^2 + 16*m)*x^16 +
(10*m^9 + 1320*m^8 + 27720*m^7 + 156156*m^6 + 286286*m^5 + 180180*m^4 + 36960*m^3 + 2040*m^2 + 18*m)*x^18 +...

Paul D. Hanna, Table of n, a(n) for n = 0..1080 for rows 0..45 of the flattened form of this triangle.

Cf. A278880 (S(x,m)), A278881 (C(x,m)), A278884 (central terms).

Cf. A001764 (row sums), A000108, A258315, A243863.

{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(D,2*n,x),k,m)}
for(n=0,15, for(k=0,n, print1(T(n,k),", "));print(""))

/* Explicit formula for T(n, k) */
{T(n, k) = if(n==k, 1, if(k==0, 0, (2*n-k)!*(n+k-1)!/(k!*(n-k)!*(2*k-1)!*(2*n-2*k+1)!) ))}
for(n=0, 15, for(k=0, n, print1(T(n, k), ", ")); print("")) \\ Paul D. Hanna, Dec 11 2016

G.f. D = D(x,m), and related functions S = S(x,m) and C = C(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).
...
T(n,k) = (n-k+1) * A082680(n+1,n-k+1) for n>=0 with T(0,0) = 1 and T(n,0) = 0 for n>0. - Paul D. Hanna, Dec 11 2016
T(n,k) = (2*n-k)!*(n+k-1)!/(k!*(n-k)!*(2*k-1)!*(2*n-2*k+1)!) for n>k>0 with T(n,0) = 1 and T(n,n) = 0 for n>0. - Paul D. Hanna, Dec 11 2016
Row sums yield A001764(n) = binomial(3*n,n)/(2*n+1).
Central terms: T(2*n,n) = binomial(3*n-1,n) * binomial(3*n,n)/(2*n+1).
Sum_{k=0..n} 2^k * T(n,k) = A258315(n-1) for n>=0.
Sum_{k=0..n} (-1)^k * T(n,k) = (-1)^n * A243863(n) for n>=0.

A243863 G.f. satisfies: A(x) = 1/(1 - x*A(-x)^2).

Original entry on oeis.org

1, 1, -1, -4, 7, 33, -68, -344, 767, 4035, -9425, -50832, 122436, 671804, -1653776, -9189488, 22992655, 129001239, -326863667, -1847900500, 4729547023, 26903463697, -69424933968, -396930961632, 1031309398852, 5921685690388, -15474833826028, -89179284390112, 234201961398776, 1353916407418200

Offset: 0

Paul D. Hanna, Jun 12 2014

sign

Compare to: G(x) = 1/(1 - x*G(x)^2) where G(x) = 1 + x*G(x)^3.

Compare to: G(x) = 1/(1 - x*G(-x)) where G(x) = 1 + x*C(-x^2) and C(x) = 1 + x*C(x)^2.

			G.f.: A(x) = 1 + x - x^2 - 4*x^3 + 7*x^4 + 33*x^5 - 68*x^6 - 344*x^7 + 767*x^8 + 4035*x^9 - 9425*x^10 - 50832*x^11 + 122436*x^12 +...
where
A(x)^2 = 1 + 2*x - x^2 - 10*x^3 + 7*x^4 + 88*x^5 - 68*x^6 - 946*x^7 + 767*x^8 + 11298*x^9 - 9425*x^10 - 144024*x^11 + 122436*x^12 +...
1/A(x) = 1 - x + 2*x^2 + x^3 - 10*x^4 - 7*x^5 + 88*x^6 + 68*x^7 - 946*x^8 - 767*x^9 + 11298*x^10 + 9425*x^11 - 144024*x^12 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..440

Cf. A278745, A278881, A278882.

{a(n)=local(A=1+x);for(i=1,n,A=1/((1-x*subst(A^2,x,-x +x*O(x^n)))));polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

{a(n) = if(n==0,1, sum(k=0,n-1,(-1)^k* (n+k)!*(2*n-k-1)!/(k!*(n-k)!*(2*k+1)!*(2*n-2*k-1)!)))}
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Dec 15 2016

G.f. satisfies: x*(A(x) + A(-x))^3 - x*(A(x) + A(-x))^2 = 2*(A(x) - A(-x)).
From Vaclav Kotesovec, Jun 15 2014: (Start)
G.f. satisfies: (1-x)*y - 2*x*y^2 + 2*x*y^3 - x^2*y^4 + x^2*y^5 = 1, where y=A(x).
Recurrence: 64*(n-1)*n*(2*n-1)*(2*n+1)*(97344*n^7 - 1687296*n^6 + 12307620*n^5 - 48939592*n^4 + 114477342*n^3 - 157378523*n^2 + 117615351*n - 36821682)*a(n) =  -48*(n-1)*(2*n-1)*(259584*n^7 - 4255264*n^6 + 28913612*n^5 - 105040520*n^4 + 218720756*n^3 - 257978161*n^2 + 156355143*n - 36051210)*a(n-1) - 12*(35043840*n^11 - 747601920*n^10 + 7046084448*n^9 - 38652188640*n^8 + 136786604012*n^7 - 326872151500*n^6 + 535977209166*n^5 - 599646116255*n^4 + 445036801288*n^3 - 206251878585*n^2 + 52769481426*n - 5487093360)*a(n-2) + 12*(n-3)*(1460160*n^8 - 21131136*n^7 + 128186214*n^6 - 422759974*n^5 + 821575770*n^4 - 951902899*n^3 + 631809561*n^2 - 214667176*n + 27671400)*a(n-3) + 3*(n-4)*(n-3)*(3*n-11)*(3*n-10)*(97344*n^7 - 1005888*n^6 + 4228068*n^5 - 9303892*n^4 + 11456294*n^3 - 7773065*n^2 + 2627695*n - 329436)*a(n-4).
a(n) ~ sqrt(1+sqrt(3)) * (3*(45+26*sqrt(3)))^(n/2) * (cos(Pi*n/2) + sqrt(1+2/sqrt(3))*sin(Pi*n/2)) / (sqrt(Pi) * n^(3/2) * 2^(2*n+2)).
(End)
a(n) = Sum_{k=0..n} (-1)^k * A278881(n,k) for n>=0. - Paul D. Hanna, Dec 01 2016
a(n) = Sum_{k=0..n-1} (-1)^k * (n+k)!*(2*n-k-1)!/(k!*(n-k)!*(2*k+1)!*(2*n-2*k-1)!) for n>0 with a(0)=1. - Paul D. Hanna, Dec 15 2016
a(0) = 1; a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} (-1)^(n-i-1) * a(i) * a(j) * a(n-i-j-1). - Ilya Gutkovskiy, Jul 28 2021

A278884 a(n) = binomial(3n-1,n) binomial(3n,n)/(2n+1).

Original entry on oeis.org

1, 2, 30, 672, 18150, 546546, 17672928, 600935040, 21212454582, 770748371250, 28657235757150, 1085694550387200, 41778588391394400, 1628982594897249312, 64234570537702934400, 2557710564063135005184, 102714012593435476016982, 4155894894567674772785250, 169274181059121504574121550, 6935873114065443534326340000, 285716428631735196825345889350, 11826871503027977442890882704050, 491714173272153004121882711232000

Offset: 0

Paul D. Hanna, Nov 29 2016

Central terms of triangles A278881 and A278882; a(n) = A278881(2*n,n) for n>=0.

Seiichi Manyama, Table of n, a(n) for n = 0..606

Cf. A278881, A278882.

Table[(Binomial[3n-1,n]Binomial[3n,n])/(2n+1),{n,0,50}] (* Harvey P. Dale, Mar 26 2023 *)

{a(n) = binomial(3*n-1,n) * binomial(3*n,n) / (2*n+1)}
for(n=0,20,print1(a(n),", "))

4*n^2*(2*n-1)*(2*n+1)*a(n) - 9*(3*n-1)^2*(3*n-2)^2*a(n-1) = 0. - R. J. Mathar, Dec 02 2016
From Stefano Spezia, Sep 04 2025: (Start)
G.f.: (1 + 2*hypergeom([1/3, 1/3, 2/3, 2/3], [1/2, 1, 3/2],9^3*x/2^4])/3.
a(n) ~ 4^(-2*n-1)*9^(3*n)/(n^2*Pi). (End)

cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A278882 Triangle where g.f. D = D(x,m) and related series S = S(x,m) and C = C(x,m) satisfy S = x*C*D, C = 1 + x*S*D, and D = 1 + m*x*S*C, as read by rows of coefficients T(n,k) of x^(2*n)*m^k in C(x,m) for n>=1, k=0..n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A243863 G.f. satisfies: A(x) = 1/(1 - x*A(-x)^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A278884 a(n) = binomial(3*n-1,n) * binomial(3*n,n)/(2*n+1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

A278882 Triangle where g.f. D = D(x,m) and related series S = S(x,m) and C = C(x,m) satisfy S = xCD, C = 1 + xSD, and D = 1 + mxS*C, as read by rows of coefficients T(n,k) of x^(2n)m^k in C(x,m) for n>=1, k=0..n.

A278884 a(n) = binomial(3n-1,n) binomial(3n,n)/(2n+1).