A138156 -id:A138156 - cp's OEIS frontend

A277295 G.f. A(x,y) satisfies: A( x - yA(x,y)^2, y) = x + (1-y)A(x,y)^2, where the coefficients T(n,k) of x^n*y^k form a triangle read by rows n>=1, for k=0..n-1.

1, 1, 0, 2, 2, 0, 5, 14, 5, 0, 14, 74, 76, 14, 0, 42, 352, 698, 378, 42, 0, 132, 1588, 5088, 5404, 1808, 132, 0, 429, 6946, 32461, 56410, 37546, 8484, 429, 0, 1430, 29786, 189940, 486550, 535410, 244220, 39446, 1430, 0, 4862, 126008, 1046190, 3690410, 6036632, 4597402, 1522466, 182732, 4862, 0, 16796, 527900, 5511440, 25518020, 57890956, 66031704, 36873036, 9227504, 846248, 16796, 0, 58786, 2195580, 28061890, 164565240, 493085566, 784844330, 661152388, 281873618, 54885974, 3926338, 58786, 0

Offset: 1

Paul D. Hanna, Oct 11 2016

More generally, we have the following related identity.
Given functions F and G with F(0)=0, F'(0)=1, G(0)=0, G'(0)=0,
if F(x - y*G(x)) = x + (1-y)*G(x), then
(1) F(x) = x + G( y*F(x) + (1-y)*x ),
(2) y*F(x) + (1-y)*x = Series_Reversion(x - y*G(x)),
(3) F(x) = x + G(x + y*G(x + y*G(x + y*G(x +...)))),
(4) F(x) = x + Sum_{n>=1} y^(n-1) * d^(n-1)/dx^(n-1) G(x)^n / n!.
The g.f. of this sequence A(x,y) equals F(x) in the above when G(x) = F(x)^2.

			G.f.: A(x,y)  = x + x^2 + (2*y + 2)*x^3 + (5*y^2 + 14*y + 5)*x^4 + (14*y^3 + 76*y^2 + 74*y + 14)*x^5 + (42*y^4 + 378*y^3 + 698*y^2 + 352*y + 42)*x^6 + (132*y^5 + 1808*y^4 + 5404*y^3 + 5088*y^2 + 1588*y + 132)*x^7 + (429*y^6 + 8484*y^5 + 37546*y^4 + 56410*y^3 + 32461*y^2 + 6946*y + 429)*x^8 + (1430*y^7 + 39446*y^6 + 244220*y^5 + 535410*y^4 + 486550*y^3 + 189940*y^2 + 29786*y + 1430)*x^9 + (4862*y^8 + 182732*y^7 + 1522466*y^6 + 4597402*y^5 + 6036632*y^4 + 3690410*y^3 + 1046190*y^2 + 126008*y + 4862)*x^10 +...
such that
A( x - y*A(x,y)^2, y)  =  x + (1-y)*A(x,y)^2.
Also,
A(x,y) = x + A( y*A(x,y) + (1-y)*x, y)^2.
...
This triangle of coefficients T(n,k) of x^n*y^k in g.f. A(x,y) begins:
1;
1, 0;
2, 2, 0;
5, 14, 5, 0;
14, 74, 76, 14, 0;
42, 352, 698, 378, 42, 0;
132, 1588, 5088, 5404, 1808, 132, 0;
429, 6946, 32461, 56410, 37546, 8484, 429, 0;
1430, 29786, 189940, 486550, 535410, 244220, 39446, 1430, 0;
4862, 126008, 1046190, 3690410, 6036632, 4597402, 1522466, 182732, 4862, 0;
16796, 527900, 5511440, 25518020, 57890956, 66031704, 36873036, 9227504, 846248, 16796, 0;
58786, 2195580, 28061890, 164565240, 493085566, 784844330, 661152388, 281873618, 54885974, 3926338, 58786, 0; ...
RELATED SEQUENCES.
Given T(n,k) is the coefficient of x^n*y^k in g.f. A(x,y),
if b(n) = Sum_{k=0..n-1} T(n,k) * p^k * q^(n-k-1)
then B(x) = Sum_{n>=1} b(n)*x^n satisfies
(1) B(x - p*B(x)^2) = x + (q-p)*B(x)^2
(2) B(x)  =  x + B( p*B(x) + (q-p)*x )^2.
Examples:
A213591(n) = sum(k=0,n-1, T(n,k) )
A275765(n) = sum(k=0,n-1, T(n,k) * 2^(n-k) )
A276360(n) = sum(k=0,n-1, T(n,k) * 3^(n-k-1) )
A276361(n) = sum(k=0,n-1, T(n,k) * 2^k * 3^(n-k-1) )
A276362(n) = sum(k=0,n-1, T(n,k) * 4^(n-k-1) )
A276363(n) = sum(k=0,n-1, T(n,k) * 3^k * 4^(n-k-1) )
A276365(n) = sum(k=0,n-1, T(n,k) * 2^k )
A277300(n) = sum(k=0,n-1, T(n,k) * 5^(n-k-1) )
A277301(n) = sum(k=0,n-1, T(n,k) * 2^k * 5^(n-k-1) )
A277302(n) = sum(k=0,n-1, T(n,k) * 3^k * 5^(n-k-1) )
A277303(n) = sum(k=0,n-1, T(n,k) * 4^k * 5^(n-k-1) )
A277304(n) = sum(k=0,n-1, T(n,k) * 6^(n-k-1) )
A277305(n) = sum(k=0,n-1, T(n,k) * 5^k * 6^(n-k-1) )
A277306(n) = sum(k=0,n-1, T(n,k) * (-1)^k )
A277307(n) = sum(k=0,n-1, T(n,k) * 3^k )
A277308(n) = sum(k=0,n-1, T(n,k) * 3^k * 2^(n-k-1) )
A277309(n) = sum(k=0,n-1, T(n,k) * 5^k * 2^(n-k-1) )
A277310(n) = sum(k=0,n-1, T(n,k) * 4^k )
A277311(n) = sum(k=0,n-1, T(n,k) * 5^k )
...

Paul D. Hanna, Table of n, a(n) for n = 1..1275, of rows n=1..50 of this triangle in flattened form.

Cf. A000108 (column 0), A138156 (column 1), A277296 (column 2), A277297 (diagonal), A277298 (central terms T(2*n-1,n-1)), A277299 (central terms T(2*n,n-1)).
Cf. A213591 (row sums), A275765, A276360, A276361, A276362, A276363, A276365.
Cf. A277300, A277302, A277303, A277304, A277305, A277306, A277307, A277308, A277309, A277310, A277311.

c[n_] := c[n] = Module[{A}, A[x_] = x; Do[A[x_] = x + A[y A[x] + (1-y) x + x O[x]^j]^2, {j, n}] // Normal; SeriesCoefficient[A[x], {x, 0, n}] // Expand];
T[n_, k_] := SeriesCoefficient[c[n], {y, 0, k}];
Table[T[n, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Sep 30 2019 *)

{T(n,k) = my(A=x); for(i=1, n, A = x + subst(A^2, x, y*A + (1-y)*x +x*O(x^n)) ); polcoeff(polcoeff(A,n,x),k,y)}
for(n=1, 12, for(k=0,n-1, print1(T(n,k), ", "));print(""))

G.f. A(x,y) also satisfies:
(1) A(x,y)  =  x + A( y*A(x,y) + (1-y)*x, y)^2.
(2) y*A(x,y) + (1-y)*x  =  Series_Reversion( x - y*A(x,y)^2 ).
(3) y*x + (1-y)*B(x,y)  =  Series_Reversion( x + (1-y)*A(x,y)^2 ), where B( A(x,y), y) = x.
(4) A(x,y) = x + Sum_{n>=1} y^(n-1) * d^(n-1)/dx^(n-1) A(x,y)^(2*n) / n!.
In formulas 2 and 3, the series reversion is taken with respect to variable x.
T(n+1,0) = T(n+1,n-1) =  binomial(2*n,n)/(n+1) = A000108(n) for n>=1.
T(n+1,1) = 4^n - (3*n+1)*binomial(2*n,n)/(n+1) = A138156(n-1) for n>=1.

A071721 Expansion of (1+x^2C^2)C^2, where C = (1-(1-4x)^(1/2))/(2x) is g.f. for Catalan numbers, A000108.

Original entry on oeis.org

1, 2, 6, 18, 56, 180, 594, 2002, 6864, 23868, 83980, 298452, 1069776, 3863080, 14040810, 51325650, 188574240, 695987820, 2579248980, 9593714460, 35804293200, 134032593240, 503154100020, 1893689067348, 7144084508256

Offset: 0

N. J. A. Sloane, Jun 06 2002

nonn

a(n) = A138156(n) - 4*A138156(n-1). - Alzhekeyev Ascar M, Jul 19 2011
Apparently, for n>=1, the sum of the heights of the first and last peaks in all Dyck n-paths (in paths with one peak the height counts as both first and last). - David Scambler, Oct 05 2012
For n>=1, a(n) is the total number of nonempty subtrees over all binary trees having n+1 internal nodes.  Here, a binary tree is a full (each node has two or zero children), rooted, plane (ordered), unlabeled tree.  An empty subtree is a tree attached to the root that consists only of an external node.  a(n) = 2*A002057(n-2) + A068875(n). - Geoffrey Critzer, Sep 16 2013
From Colin Defant, Sep 15 2018: (Start)
a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, 312, and 321, where s denotes West's stack-sorting map.
a(n) is the number of permutations on [n+1] that avoid the patterns 1342, 2341, 3142, 3241, 3412, and 3421. (End)

			G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 56*x^4 + 180*x^5 + 594*x^6 + 2002*x^7 + ... - _Michael Somos_, Apr 22 2022

Michael De Vlieger, Table of n, a(n) for n = 0..1000
Colin Defant, Stack-sorting preimages of permutation classes, arXiv:1809.03123 [math.CO], 2018.
Stoyan Dimitrov, On permutation patterns with constrained gap sizes, arXiv:2002.12322 [math.CO], 2020.
Ling Gao, Graph assembly for spider and tadpole graphs, Master's Thesis, Cal. State Poly. Univ. (2023). See p. 32.
Boram Park and Seonjeong Park, Shellable posets arising from even subgraphs of a graph, arXiv preprint arXiv:1705.06423 [math.CO], 2017.
Seonjeong Park, Real toric manifolds and shellable posets arising from graphs, 2018.

Cf. A000108, A000245, A001622, A002421, A068875.
Row sums of triangles A319251, A319252.
gf=(1+x^2*C^2)*C^m: A000782 (m=1), this sequence (m=2), A071722 (m=3), A071723 (m=4).

a := n -> `if`(n=0, 1, 6*binomial(2*n, n-1)/(n+2));
seq(a(n), n=0..24); # Peter Luschny, Jun 28 2018

Join[{1},Table[6n CatalanNumber[n]/(n+2),{n,30}]] (* Harvey P. Dale, Jun 05 2012 *)
nn=20;t=(1-(1-4x)^(1/2))/(2x);CoefficientList[Series[D[1+x (y t -y+1)^2,y]/.y->1,{x,0,nn}],x] (* Geoffrey Critzer, Sep 16 2013 *)

{a(n) = if(n<1, n==0, 6*n*(2*n)!/(n!*(n + 1)!*(n + 2)))}; /* Michael Somos, Apr 22 2022 */

a = lambda n: n*(n+1)*hypergeometric([1-n, 2-n], [4], 1) if n>0 else 1
[simplify(a(n)) for n in range(25)] # Peter Luschny, Nov 19 2014

a(n) = 6n * (2n)! / [(n+2)n!(n+1)! ], n>0. In terms of Catalan numbers (A000108), a(n) = 6n*Cat(n)/(n+2), n>0. - Ralf Stephan, Mar 11 2004
a(n) = n*(n+1)*hypergeom([1-n, 2-n], [4], 1) for n>=1. - Peter Luschny, Nov 19 2014
D-finite with recurrence -(n+2)*(n-1)*a(n) +2*n*(2*n-1)*a(n-1)=0. - R. J. Mathar, Jul 18 2017
a(n) = 2*Cat(n+1) - 2*Cat(n) = 2*A000245(n) for n>=1. - Colin Defant, Jun 27 2018
From Amiram Eldar, Mar 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 23/18 + 7*Pi/(27*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 43/50 - 82*sqrt(5)*log(phi)/375, where phi is the golden ratio (A001622). (End)
From Michael Somos, Apr 22 2022: (Start)
G.f.: (1 - 3*x + x^2 - (1 - x) * sqrt(1 - 4*x))/x^2.
G.f.: (2 - 2*x + x^2)/(1 - 3*x + x^2 + (1 - x)*sqrt(1 - 4*x)).
G.f.: 1 + 1/((1 - x)/(1 - sqrt(1 - 4*x)) - 1/2).
a(n) = b(n+1) - b(n) for all n in Z if b(0) = 2, b(-1) = -1, a(0) = 0, a(-1) = 3, a(-2) = -1 where b = A068875.
0 = a(n)*(+16*a(n+1) -58*a(n+2) +18*a(n+3)) +a(n+1)*(+18*a(n+1) +15*a(n+2) -13*a(n+3)) +a(n+2)*(+3*a(n+2) +a(n+3)) for all n in Z if a(0) = 0, a(-1) = 3, a(-2) = -1. (End)

A138157 Triangle read by rows: T(n,k) is the number of binary trees with n edges and path length k; 0<=k<=n*(n+1)/2.

Original entry on oeis.org

1, 0, 2, 0, 0, 1, 4, 0, 0, 0, 0, 4, 2, 8, 0, 0, 0, 0, 0, 0, 6, 8, 8, 4, 16, 0, 0, 0, 0, 0, 0, 0, 0, 4, 24, 4, 28, 16, 16, 8, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 24, 36, 48, 24, 56, 40, 56, 32, 32, 16, 64, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 60, 72, 144, 26, 168, 104, 128, 64, 176, 80, 112

Offset: 0

Emeric Deutsch, Mar 20 2008

Row n has 1+n*(n+1)/2 terms.
Row sums are the Catalan numbers (A000108).
Column sums yield A095830.
Sum_{k>=0} k*T(n,k) = A138156(n).
The g.f. B(w,z) in the Knuth reference is related to the above G(t,z) through B(t,z)=1+zG(t,z).

			T(2,3) = 4 because we have the path trees LL, LR, RL and RR, where L (R) denotes a left (right) edge; each of these four trees has path length 3.
Triangle starts:
  1;
  0,2;
  0,0,1,4;
  0,0,0,0,4,2,8;
  0,0,0,0,0,0,6,8,8,4,16;
  0,0,0,0,0,0,0,0,4,24,4,28,16,16,8,32;

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1997, Vol. 1, p. 405 (exercise 5) and p. 595 (solution).

Alois P. Heinz, Rows n = 0..50, flattened
Marilena Barnabei, Flavio Bonetti, and Niccolò Castronuovo, Motzkin and Catalan Tunnel Polynomials, J. Int. Seq., Vol. 21 (2018), Article 18.8.8.

Cf. A000108, A095830, A138156.

P[0]:=1: for n to 7 do P[n]:=sort(expand(t^n*(2*P[n-1]+add(P[i]*P[n-2-i],i= 0..n-2)))) end do: for n from 0 to 7 do seq(coeff(P[n],t,j),j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form

p[n_] := p[n] = t^n*(2p[n-1] + Sum[p[i]*p[n-2-i], {i, 0, n-2}]); p[0] = 1; Flatten[ Table[ CoefficientList[ p[n], t], {n, 0, 7}]] (* Jean-François Alcover, Jul 22 2011, after Maple prog. *)

G.f.: G(t,z) satisfies G(t,z) = 1 + 2*t*z*G(t,t*z) + (t*z*G(t,t*z))^2. The row generating polynomials P[n] = P[n](t) (n=1,2,...) are given by P[n] = t^n*(2*P[n-1] + Sum_{i=0..n-2} P[i]*P[n-2-i]); P[0] = 1.

A336110 Irregular triangle of Catalan-based numbers, read by rows.

Original entry on oeis.org

1, 1, 2, -2, 5, -14, 5, 14, -74, 74, -14, 42, -352, 668, -352, 42, 132, -1588, 4808, -4808, 1588, -132, 429, -6946, 30371, -48540, 30371, -6946, 429, 1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430

Offset: 1

Sergii Voloshyn, Jul 08 2020

Calculation of the sum over the partitions of r of products of dimensions of two different representations of a symmetric group S_r gives
Sum_{L |- S_r} f(L)*f(l+q^N) = (r+q^N)! * G[N+1] * G[q+1]/(G[N+q+1]) * B_r(1!c_1, ..., r!c_r) where f(L) is the dimension of the symmetric group S_r, G[x] is Barnes function, and B_r() is the complete exponential Bell polynomial.
In the limit N -> infinity the coefficients [are?]
  c_1 = 1/(1+x), c_i = 1/(i*N^(2*(i-1)))*P(i-1), for i >= 2.
Coefficient of x^n in the numerator of P(i) is T(s, i).
This triangle of coefficients was discovered by Borisenko et al. In mathematical physics these coefficients appear as an important ingredient of series that define the free energy of the SU(N) standard lattice model in the large N limit.
They are easily obtained from the g.f.
Some special cases are given by A000108 (first column of the triangle), A138156 (second column of the triangle).
The sum of the numbers in row 2*k+1 is (-1)^k * A000260(k) * 2^(2*k).

			     1;
     1;
     2,     -2;
     5,    -14,      5;
    14,    -74,     74,     -14;
    42,   -352,    668,    -352,     42;
   132,  -1588,   4808,   -4808,   1588,    -132;
   429,  -6946,  30371,  -48540,  30371,   -6946,   429;
  1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430;
  ...

Sergii Voloshyn, Table of n, a(n) for n = 1..120
O. Borisenko, V. Chelnokov, and Sergii Voloshyn, The large N limit of SU(N) integrals in lattice, arXiv:2008.00773 [hep-lat], 2020. See formula (48).
F. Green and S. Samuel, Calculating the large-N phase transition in gauge and matrix models, Nuclear Physics B 194, Issue 1, 11 January 1982, Pages 107-156, See Appendix A.

Cf. A000108, A000309, A138156, A000260.

T[L_, n_] := CatalanNumber[L] Sum[u[L, a]/u[0, a] M[n - 1 - 2*a, L], {a, 0, (n - 1)/2}]-4^L Sum[v[L, a]/v[0, a] M[n - 2 - 2*a, L], {a,0,(n-2)/2}];
M[n_, l_] := Sum[k!/n! BellY[n,k,Table[(-1)^(j-1) j!Binomial[3l+j,j], {j,n}]], {k, 0, n}];
u[k_, n_] := Product[Binomial[2 k + 2 l + 1, 3], {l, 1, n}];
v[k_, n_] := Product[ Binomial[2 k + 2 l + 2, 3], {l, 1, n}];
(* alternate program using coefficients in numerator *)
P[s_] := x CatalanNumber[s] HypergeometricPFQ[{s + 1/2, s + 1, s + 3/2}, {1/2, 3/2}, x^2] - x^2*4^s HypergeometricPFQ[{s + 1, s + 3/2, s + 2}, {3/2, 2}, x^2];
Table[CoefficientList[P[s] // Together // Numerator, x] // Rest, {s, 0, 10}] // Flatten (* amended by Jean-François Alcover, Sep 25 2020 *)
(* another program using coefficients in numerator *)
Needs["Combinatorica`"];
OA[p_,x_]:= (2^p(-(1/(x+1)))^(2p+1))/((2p+1)!(p+1)!) Sum[(-x/(1+x))^(p-r+1)Product[Pochhammer[1+Plus@@Table[3*k[[i]]-1,{i,1,j-1}],3*k[[j]]],{j,1,r}],{r, 1, p},{k,Compositions[p-r,r]+1}]; Table[CoefficientList[OA[s, x] // Together // Numerator, x] //
   Rest, {s, 0, 10}] // Flatten (* Sergii Voloshyn, Sep 03 2021 *)

Coefficients of x^n in the numerator of P(s) = (x * C[s]* 3F2[ s+ 1/2, s+1, s+3/2; 1/2,3/2; x^2] - x^2 * 4^s * 3F2[ s+1,s+3/2, s+2; 3/2, 2; x^2]), where C(s) are Catalan numbers.
or in a more explicit way (only for k >= 1)
T(s, n) = C(s) * U(s, n) - 4^s * V(s, n), where
U(s, n) = Sum_{a=0..(n-1)/2} (u(s, a)/u(0,a)) * M(s, n-1- 2a),
V(s, n) = Sum_{a=0..(n-2)/2} (v(s, a)/v(0,a)) * M(s, n-2- 2a), and
u(s, n) = Product_{L=1..n} binomial(2s+2L+1, 3),
v(s, n) = Product_{L=1..n} binomial(2s+2L+2, 3), and
M(m, n) = Sum_{L=1..n} L!/n! B_{n,l} ( x_1, ..., x_{n-L+1}), and
x_i = (-1)^{1+i} * (3s+i)_i = (-1)^{1+i} * i! * binomial(3s + i, i), where
B_{n,l} (x_1, ..., x_{n-L+1}) is the n-th partial or incomplete exponential Bell polynomial with monomials sorted into graded lexicographic order.
Sum of numbers in the particular row:
Sum_{n=1..2*k+1} T(2*k+1, n) = 2*(4*k+1)!/((k+1)!*(3*k+2)!) *2^(2*k) (odd s);
Sum_{n=1..2*k} T(2*k, n) = 0 (even s).
From Sergii Voloshyn, Oct 22 2020: (Start)
Formulae for particular columns:
T(s, 1) = C(s);
T(s, 2) = C(s)*(3*s+1) - 4^s;
T(s, 3) = C(s)*(binomial(2*s+3,3) + (3*s+1)^2 - binomial(3*s +2,2)) - 4^s*(3*s+1);
T(s, 4) = C(s)*((2*s+1)binomial(2*s+3,3) +(3*s+1)^3 - 2(3*s+1)* binomial(3*s+2,2)+ binomial(3*s+3, 3)) - 4^s*(binomial(3*s+4, 3)/4 + (3*s+1)^2 - binomial(3*s+2,2));
...
T(s, s) = (-1)^(s+1)*C(s). (End)
From Sergii Voloshyn, Mar 17 2021: (Start)
Recursion ( P[0] = x/(1+x) ):
x*(d^3/dx^3)*P[s] = (1/4)*(2*k+2)*(2*k+3)*(2*k+4)*P[s+1]
for P[s_] := x CatalanNumber[s] HypergeometricPFQ[{s + 1/2, s + 1, s + 3/2}, {1/2, 3/2}, x^2] - x^2*4^s HypergeometricPFQ[{s + 1, s + 3/2, s + 2}, {3/2, 2}, x^2].
(End)
Recursion for array ( T(1,1) = 1 ): T(k, p) = (1/(k*(k+1)*(2k+1)))*[p*(p+1)*(p+2)*T(k-1,p+2) - 3*p*(p+1)*(3*k-p-1)*T(k-1,p+1) + 3*p*(3*k-p)*(3*k-p-1)*T(k-1,p) - (3*k-p+1)*(3*k-p)*(3*k-p-1)*T(k-1,p-1)]; p =[1,...,k]. - Sergii Voloshyn, Apr 14 2021
From Sergii Voloshyn, Apr 17 2021: (Start)
G.f.: Sum_{m>=1} x^m Om(k, m) = (1/(1+x)^(3*k+1))*Sum_{n=1..k} x^k * T(k,n);
Om(k, m) = (12^k/((1+k)!*(2k+1)!)) * Product_{L=1..k} binomial(m+2L, 3).
(End)
From Sergii Voloshyn, Apr 25 2021: (Start)
Differential equation for P[k_]:
x*(x^2-1)*(d^3/dx^3)P[s] + (2*k+2)*3*x^2*(d^2/dx^2)P[s] +(2*k+2)*(2*k+1)*3*x (d/dx)P[s] + (2*k+2)(2*k+1)*2*k*P[s] = 0.
Discrete set equation for T(k,n) (n=-1..k-2) at fixed k:
(k-n-1)*(k-n)*(k-n+1)*T(k,n) - (k-n-1)*(k-n)*(8*k+n+5)*T(k,n+1) - (n+1)*(n+2)*(9-n+3)*T(k,n+2) + (n+1)*(n+2)*(n+3)*T(k,n+3) = 0
and
Sum_{m=1..k} (k*(5+7*k) + 12*n*(n-1-k))*T(k,n) = 0. (End)
P[k_]:= (2^k)/((2*k+1)!*(k+1)!)*(-1/(1+x))^{2k+1}[Sum_{r(1)+...+ r(L)=k} (-x/(1+x))^{k-L+1}* 1^{(3*r(1))}*(1+3*r(1)-1)^{(3*r(2))}*... *(1+Sum_{i=1..L-1} 3 r(i) -L+1)^{(3*r(L))}] where Sum_ r_i = k runs over all integer compositions of k, L is a number of parts of this composition and 1^{(3 r(1))} is a rising factorial. - Sergii Voloshyn, Sep 03 2021
Sum_{k} abs(T(n,k)) = A000309(n-1). - Sergii Voloshyn, Nov 20 2024

cp's OEIS Frontend

A277295 G.f. A(x,y) satisfies: A( x - y*A(x,y)^2, y) = x + (1-y)*A(x,y)^2, where the coefficients T(n,k) of x^n*y^k form a triangle read by rows n>=1, for k=0..n-1.

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A071721 Expansion of (1+x^2*C^2)*C^2, where C = (1-(1-4*x)^(1/2))/(2*x) is g.f. for Catalan numbers, A000108.

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A138157 Triangle read by rows: T(n,k) is the number of binary trees with n edges and path length k; 0<=k<=n*(n+1)/2.

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A336110 Irregular triangle of Catalan-based numbers, read by rows.

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A277295 G.f. A(x,y) satisfies: A( x - yA(x,y)^2, y) = x + (1-y)A(x,y)^2, where the coefficients T(n,k) of x^n*y^k form a triangle read by rows n>=1, for k=0..n-1.

A071721 Expansion of (1+x^2C^2)C^2, where C = (1-(1-4x)^(1/2))/(2x) is g.f. for Catalan numbers, A000108.