A007863 -id:A007863 - cp's OEIS frontend

A239108 Number of hybrid 5-ary trees with n internal nodes.

1, 2, 19, 253, 3920, 66221, 1183077, 21981764, 420449439, 8223704755, 163727846678, 3307039145618, 67600147666909, 1395822347989531, 29070233296701815, 609950649080323320, 12881240945694949696, 273590092192962485985, 5840400740191969187922

Offset: 0

N. J. A. Sloane, Mar 26 2014

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..300
SeoungJi Hong and SeungKyung Park, Hybrid d-ary trees and their generalization, Bull. Korean Math. Soc. 51 (2014), No. 1, pp. 229-235. See p. 233.
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin)

Cf. A000045, A007863, A215654, A239107, A239108, A239109.

Column k=5 of A245049.

(1/x InverseSeries[x(1 - x - x^2)^4/(1 + x)^4 + O[x]^20])^(1/4) // CoefficientList[#, x]& (* Jean-François Alcover, Oct 02 2019 *)

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^4)*(1 + x*A^5)); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=polcoeff( ((1/x)*serreverse( x*(1-x-x^2)^4/(1+x +x*O(x^n))^4))^(1/4), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^j)*x^m*A^(3*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^j)*x^m*A^(4*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=polcoeff(((1+x)/(1-x-x^2 +x*O(x^n)))^(4*n+1)/(4*n+1), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

From Paul D. Hanna, Mar 30 2014: (Start)
G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x)^4) * (1 + x*A(x)^5).
(2) A(x) = ( (1/x)*Series_Reversion( x*(1-x-x^2)^4/(1+x)^4 ) )^(1/4).
(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(3*n)/n * Sum_{k=0..n} C(n,k)^2 * A(x)^k ).
(4) A(x) = exp( Sum_{n>=1} x^n*A(x)^(4*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^k ).
(5) A(x) = Sum_{n>=0} Fibonacci(n+2) * x^n * A(x)^(4*n).
(6) A(x) = G(x*A(x)^3) where G(x) = A(x/G(x)^3) is the g.f. of A007863 (number of hybrid binary trees with n internal nodes).
The formal inverse of g.f. A(x) is (sqrt(1-2*x+5*x^2) - (1+x))/(2*x^5).
a(n) = [x^n] ( (1+x)/(1-x-x^2) )^(4*n+1) / (4*n+1).
(End)

A239109 Number of hybrid 6-ary trees with n internal nodes.

Original entry on oeis.org

1, 2, 23, 375, 7138, 148348, 3262975, 74673216, 1759690865, 42412172598, 1040644972314, 25907046248766, 652763779424538, 16614703783094140, 426563932954831827, 11033640140115676862, 287265076610919864178, 7522060666571155198520, 197969862318742854908470

Offset: 0

N. J. A. Sloane, Mar 26 2014

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..300
SeoungJi Hong and SeungKyung Park, Hybrid d-ary trees and their generalization, Bull. Korean Math. Soc. 51 (2014), No. 1, pp. 229-235. See p. 233.

Cf. A000045, A007863, A215654, A239107, A239108, A239109.

Column k=6 of A245049.

(1/x InverseSeries[x*(1 - x - x^2)^5/(1 + x)^5 + O[x]^20])^(1/5) // CoefficientList[#, x]& (* Jean-François Alcover, Oct 02 2019 *)

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^5)*(1 + x*A^6)); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=polcoeff( ((1/x)*serreverse( x*(1-x-x^2)^5/(1+x +x*O(x^n))^5))^(1/5), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^j)*x^m*A^(4*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^j)*x^m*A^(5*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=polcoeff(((1+x)/(1-x-x^2 +x*O(x^n)))^(5*n+1)/(5*n+1), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

From Paul D. Hanna, Mar 30 2014: (Start)
G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x)^5) * (1 + x*A(x)^6).
(2) A(x) = ( (1/x)*Series_Reversion( x*(1-x-x^2)^5/(1+x)^5 ) )^(1/5).
(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(4*n)/n * Sum_{k=0..n} C(n,k)^2 * A(x)^k ).
(4) A(x) = exp( Sum_{n>=1} x^n*A(x)^(5*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^k ).
(5) A(x) = Sum_{n>=0} Fibonacci(n+2) * x^n * A(x)^(5*n).
(6) A(x) = G(x*A(x)^4) where G(x) = A(x/G(x)^4) is the g.f. of A007863 (number of hybrid binary trees with n internal nodes).
The formal inverse of g.f. A(x) is (sqrt(1-2*x+5*x^2) - (1+x))/(2*x^6).
a(n) = [x^n] ( (1+x)/(1-x-x^2) )^(5*n+1) / (5*n+1).
(End)

A215715 G.f. satisfies A(x) = (1 + xA(x)^2) (1 + x*A(x)^4).

Original entry on oeis.org

1, 2, 13, 118, 1242, 14227, 172177, 2165732, 28032668, 370944717, 4995412647, 68239105203, 943278064473, 13169938895473, 185453340189492, 2630813161415976, 37561512615867450, 539336703889993006, 7783290731579783544, 112828761898680983141, 1642222504807143423470

Offset: 0

Paul D. Hanna, Aug 21 2012

nonn

More generally, for fixed parameters p, q, r, and s, if F(x) satisfies:
F(x) = (1 + x^r*F(x)^(p+1)) * (1 + x^(r+s)*F(x)^(p+q+1)), then
F(x) = exp( Sum_{n>=1} x^(n*r)*F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^(k*s)*F(x)^(k*q)] ).
The radius of convergence of g.f. A(x) is r = 0.06368546004073732405169450... with A(r) = 1.3960637117611795281240000742797488619448782873... where y=A(r) satisfies 6*y^5 + 17*y^4 - 46*y^3 + 16*y^2 + 4*y - 8 = 0.

			G.f.: A(x) = 1 + 2*x + 13*x^2 + 118*x^3 + 1242*x^4 + 14227*x^5 + ...
Related expansions.
A(x)^2 = 1 + 4*x + 30*x^2 + 288*x^3 + 3125*x^4 + 36490*x^5 + ...
A(x)^4 = 1 + 8*x + 76*x^2 + 816*x^3 + 9454*x^4 + 115260*x^5 + ...
A(x)^6 = 1 + 12*x + 138*x^2 + 1648*x^3 + 20427*x^4 + 260934*x^5 + ...
where A(x) = 1 + x*(A(x)^2 + A(x)^4) + x^2*A(x)^6.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + A(x)^2)*x*A(x) + (1 + 2^2*A(x)^2 + A(x)^4)*x^2*A(x)^2/2 +
(1 + 3^2*A(x)^2 + 3^2*A(x)^4 + A(x)^6)*x^3*A(x)^3/3 +
(1 + 4^2*A(x)^2 + 6^2*A(x)^4 + 4^2*A(x)^6 + A(x)^8)*x^4*A(x)^4/4 +
(1 + 5^2*A(x)^2 + 10^2*A(x)^4 + 10^2*A(x)^6 + 5^2*A(x)^8 + A(x)^10)*x^5*A(x)^5/5 + ...
Explicitly,
log(A(x)) = 2*x + 22*x^2/2 + 284*x^3/3 + 3878*x^4/4 + 54607*x^5/5 + 784144*x^6/6 + 11414265*x^7/7 + 167819014*x^8/8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..825
Vaclav Kotesovec, Recurrence

Cf. A198953.

Cf. A007863, A069271, A073157, A215654.

CoefficientList[Sqrt[1/x*InverseSeries[Series[x*(1+Sqrt[1-4*x*(1+x)^2])^2/(4*(1+x)^2),{x,0,20}],x]],x] (* Vaclav Kotesovec, Sep 17 2013 *)

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^2)*(1 + x*A^4)); polcoeff(A, n)}
for(n=0,31,print1(a(n),", "))

{a(n)=polcoeff( sqrt((1/x)*serreverse( x*(1 + sqrt(1 - 4*x*(1+x)^2 +x*O(x^n)))^2/(4*(1+x)^2))), n)}

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^(2*j))*x^m*A^m/m))); polcoeff(A, n)}

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^(2*j))*x^m*A^(3*m)/m))); polcoeff(A, n)}

G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x*(1 + sqrt(1 - 4*x*(1+x)^2))^2/(4*(1+x)^2) ) ).
(2) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * Sum_{k=0..n} C(n,k)^2 * A(x)^(2*k) ).
(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(3*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^(2*k) ).
The formal inverse of the g.f. A(x) is (sqrt(x^4 + 4*x^3 - 2*x^2 + 1) - (1+x^2))/(2*x^4).
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 15.70217125479403872... is the root of the equation -1024 - 3840*d + 26368*d^2 - 58644*d^3 + 1933*d^4 + 108*d^5 = 0 and c = 0.320114409... - Vaclav Kotesovec, Sep 17 2013
a(n) = Sum_{k=0..n} binomial(2*n+2*k+1,k) * binomial(2*n+2*k+1,n-k) / (2*n+2*k+1). - Seiichi Manyama, Jul 18 2023

A216314 G.f. satisfies A(x) = (1 + xA(x)) (1 + 2xA(x)^2).

Original entry on oeis.org

1, 3, 17, 121, 965, 8247, 73841, 683713, 6493145, 62898859, 619079889, 6173490857, 62239144525, 633323532783, 6496052173665, 67093423506049, 697181754821297, 7283521984427283, 76455801614169809, 806004056649062937, 8529783421905380629, 90584730265930813607

Offset: 0

Paul D. Hanna, Sep 03 2012

nonn

The radius of convergence of g.f. A(x) is r = 0.08774268876242660659654020... with A(r) = 2.04748732367111203761312028274219344812311691... where y=A(r) satisfies 6*y^3 - 14*y^2 + 4*y - 1 = 0.

r = 1/(((40465 + 387*sqrt(129))^(2/3) + 1174 + 34*(40465 + 387*sqrt(129))^(1/3)) / (40465+387*sqrt(129))^(1/3)/9). - Vaclav Kotesovec, Sep 17 2013

			G.f.: A(x) = 1 + 3*x + 17*x^2 + 121*x^3 + 965*x^4 + 8247*x^5 + 73841*x^6 +...
Related expansions.
A(x)^2 = 1 + 6*x + 43*x^2 + 344*x^3 + 2945*x^4 + 26398*x^5 + 244615*x^6 +...
A(x)^3 = 1 + 9*x + 78*x^2 + 696*x^3 + 6399*x^4 + 60321*x^5 + 580316*x^6 +...
where A(x) = 1 + A(x)*(1+2*A(x))*x + 2*A(x)^3*x^2.
The g.f. also satisfies the series:
A(x) = 1 + 3*x*A(x) + 8*x^2*A(x)^2 + 22*x^3*A(x)^3 + 60*x^4*A(x)^4 + 164*x^5*A(x)^5 + 448*x^6*A(x)^6 +...+ A028859(n)*x^n*A(x)^n +...
The logarithm of the g.f. equals the series:
log(A(x)) = (1*2 + 1/A(x))*x*A(x) + (1*2^2 + 2^2*2/A(x) + 1/A(x)^2)*x^2*A(x)^2/2 +
(1*2^3 + 3^2*2^2/A(x) + 3^2*2/A(x)^2 + 1/A(x)^3)*x^3*A(x)^3/3 +
(1*2^4 + 4^2*2^3/A(x) + 6^2*2^2/A(x)^2 + 4^2*2/A(x)^3 + 1/A(x)^4)*x^4*A(x)^4/4 +
(1*2^5 + 5^2*2^4/A(x) + 10^2*2^3/A(x)^2 + 10^2*2^2/A(x)^3 + 5^2*2/A(x)^4 + 1/A(x)^5)*x^5*A(x)^5/5 +...
Explicitly,
log(A(x)) = 3*x + 25*x^2/2 + 237*x^3/3 + 2361*x^4/4 + 24203*x^5/5 + 252757*x^6/6 + 2674185*x^7/7 + 28567105*x^8/8 +...+ L(n)*x^n/n +...
where L(n) = [x^n] (1+x)^n/(1-2*x-2*x^2)^n.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
R. Bacher, On generating series of complementary plane trees arXiv:math/0409050 [math.CO]

Cf. A007863, A215661, A215654, A028859, A364374.

CoefficientList[1/x * InverseSeries[Series[x*(1 - 2*x - 2*x^2)/(1+x),{x,0,20}],x],x] (* Vaclav Kotesovec, Sep 17 2013 *)

{a(n)=local(A=1+x); for(i=1, n, A=(1 + x*A)*(1 + 2*x*(A+x*O(x^n))^2)); polcoeff(A, n)}

{a(n)=polcoeff( (1/x)*serreverse( x*(1-2*x-2*x^2)/(1+x +x*O(x^n))), n)}

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*2^(m-j)/A^j)*x^m*A^m/m))); polcoeff(A, n)}
for(n=0, 31, print1(a(n), ", "))

G.f. A(x) satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * Sum_{k=0..n} C(n,k)^2 * 2^(n-k)/A(x)^k ).
(2) A(x) = (1/x) * Series_Reversion( x*(1 - 2*x - 2*x^2)/(1+x) ).
(3) A(x) = Sum_{n>=0} A028859(n) * x^n * A(x)^n, where g.f. of A028859 = (1+x)/(1-2*x-2*x^2).
The formal inverse of the g.f. A(x) is (sqrt(1-4*x+12*x^2) - (1+2*x))/(4*x^2).
a(n) = [x^n] ( (1+x)/(1-2*x-2*x^2) )^(n+1) / (n+1).
Recurrence: 3*n*(n+1)*(43*n-76)*a(n) = n*(1462*n^2 - 3315*n + 1274)*a(n-1) + (86*n^3 - 324*n^2 + 523*n - 330)*a(n-2) + (n-2)*(2*n-5)*(43*n-33)*a(n-3)
a(n) ~ 1/516*sqrt(86)*sqrt((1448486261 + 1803807*sqrt(129))^(1/3)*((1448486261 + 1803807*sqrt(129))^(2/3) + 1280110 + 1118*(1448486261 + 1803807*sqrt(129))^(1/3)))/(1448486261 + 1803807*sqrt(129))^(1/3) * (((40465 + 387*sqrt(129))^(2/3) + 1174 + 34*(40465 + 387*sqrt(129) )^(1/3)) / (40465+387*sqrt(129))^(1/3)/9)^n / (n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Sep 17 2013
a(n) = Sum_{k=0..n} 2^k * binomial(n+k+1,k) * binomial(n+k+1,n-k) / (n+k+1). - Seiichi Manyama, Sep 08 2024

A239107 Number of hybrid 4-ary trees with n internal nodes.

Original entry on oeis.org

1, 2, 15, 155, 1854, 24124, 331575, 4736345, 69616637, 1046054129, 15995716263, 248111418112, 3894303176880, 61737213540306, 987116931080661, 15899835212249761, 257758369219909534, 4202381519278498915, 68859442092723799788, 1133401910867109123200

Offset: 0

N. J. A. Sloane, Mar 26 2014

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..800
SeoungJi Hong and SeungKyung Park, Hybrid d-ary trees and their generalization, Bull. Korean Math. Soc. 51 (2014), No. 1, pp. 229-235. See p. 233.
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin)

Cf. A000045, A007863, A215654, A239107, A239108, A239109.

Column k=4 of A245049.

(1/x InverseSeries[x(1 - x - x^2)^3/(1 + x)^3 + O[x]^21])^(1/3) // CoefficientList[#, x]& (* Jean-François Alcover, Oct 02 2019 *)

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^3)*(1 + x*A^4)); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=polcoeff( ((1/x)*serreverse( x*(1-x-x^2)^3/(1+x +x*O(x^n))^3))^(1/3), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^j)*x^m*A^(2*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^j)*x^m*A^(3*m)/m))); polcoeff(A, n) \\ Paul D. Hanna, Mar 30 2014
for(n=0, 20, print1(a(n), ", "))

a(n)=polcoeff(((1+x)/(1-x-x^2 +x*O(x^n)))^(3*n+1)/(3*n+1), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

From Paul D. Hanna, Mar 30 2014: (Start)
G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x)^3) * (1 + x*A(x)^4).
(2) A(x) = ( (1/x)*Series_Reversion( x*(1-x-x^2)^3/(1+x)^3 ) )^(1/3).
(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(2*n)/n * Sum_{k=0..n} C(n,k)^2 * A(x)^k ).
(4) A(x) = exp( Sum_{n>=1} x^n*A(x)^(3*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^k ).
(5) A(x) = Sum_{n>=0} Fibonacci(n+2) * x^n * A(x)^(3*n).
(6) A(x) = G(x*A(x)^2) where G(x) = A(x/G(x)^2) is the g.f. of A007863 (number of hybrid binary trees with n internal nodes).
The formal inverse of g.f. A(x) is (sqrt(1-2*x+5*x^2) - (1+x))/(2*x^4).
a(n) = [x^n] ( (1+x)/(1-x-x^2) )^(3*n+1) / (3*n+1).
(End)
a(n) = 1/(3*n+1) * Sum_{i=0..n} C(3*n+i,i)*C(3*n+i+1,n-i). - Alois P. Heinz, Jul 10 2014

A274735 G.f. satisfies A(x) = (1 + xA(x))^3 (1 + x*A(x)^2).

Original entry on oeis.org

1, 4, 26, 210, 1901, 18445, 187524, 1971672, 21263360, 233907762, 2614446624, 29607343948, 338977591904, 3917185497535, 45629006313280, 535199773167207, 6315789123860388, 74932400322972992, 893276792585933870, 10694510040508714014, 128531711285410216883, 1550159476645634696615, 18755239991772817629972, 227577929298568261967650, 2768820313297861609739979

Offset: 0

Paul D. Hanna, Aug 02 2016

nonn

More generally, if G(x) satisfies
G(x) = (1 + a*x*G(x))^m * (1 + b*x*G(x)^2), then
G(x) = (1/x) * Series_Reversion( x * (1 - b*x*(1 + a*x)^m) / (1 + a*x)^m ).

			G.f.: A(x) = 1 + 4*x + 26*x^2 + 210*x^3 + 1901*x^4 + 18445*x^5 + 187524*x^6 + 1971672*x^7 + 21263360*x^8 +...

Seiichi Manyama, Table of n, a(n) for n = 0..903

Cf. A007863, A274734.

Cf. A274379, A369600.

{a(n) = my(A=1); for(i=1,n, A = (1 + x*A)^3 * (1 + x*A^2) + x*O(x^n) ); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

{a(n) = my(A=1); A = (1/x)*serreverse(x*(1-x*(1+x)^3)/(1+x +x^2*O(x^n) )^3 ); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

G.f.: (1/x) * Series_Reversion( x * (1 - x*(1+x)^3) / (1+x)^3 ).

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(3*n+3*k+3,n-k). - Seiichi Manyama, Jan 27 2024

A364331 G.f. satisfies A(x) = (1 + xA(x)^2) (1 + x*A(x)^5).

Original entry on oeis.org

1, 2, 15, 163, 2070, 28698, 421015, 6425644, 100977137, 1622885389, 26551709946, 440744175801, 7404449354076, 125657625548824, 2150963575012295, 37094953102567208, 643904274979347286, 11241232087809137759, 197247501440314516840, 3476787208220672891388, 61533794803235280779261

Offset: 0

Seiichi Manyama, Jul 18 2023

Cf. A007863, A069271, A073157, A215654, A215715, A364333.
Cf. A215623, A215624, A239108, A364335, A364338.
Cf. A200719.

A364331 := proc(n)
    add( binomial(2*n+3*k+1,k) * binomial(2*n+3*k+1,n-k)/(2*n+3*k+1),k=0..n) ;
end proc:
seq(A364331(n),n=0..70); # R. J. Mathar, Jul 25 2023

a(n) = sum(k=0, n, binomial(2*n+3*k+1, k)*binomial(2*n+3*k+1, n-k)/(2*n+3*k+1));

a(n) = Sum_{k=0..n} binomial(2*n+3*k+1,k) * binomial(2*n+3*k+1,n-k) / (2*n+3*k+1).

x/series_reversion(x*A(x)) = 1 + 2*x + 11*x^2 + 89*x^3 + 836*x^4 + ..., the g.f. of A215623. - Peter Bala, Sep 08 2024

A262910 a(n) = Sum_{k=0..n} binomial(k+n-1,k)binomial(k+n,2k).

Original entry on oeis.org

1, 2, 10, 59, 366, 2337, 15205, 100235, 667222, 4474733, 30188335, 204646532, 1392850785, 9511878729, 65144238981, 447263887479, 3077459618886, 21215286546705, 146500755609415, 1013180180867125, 7016536189029551, 48650933146617728, 337709155342663620

Offset: 0

Vladimir Kruchinin, Oct 04 2015

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..1176

Cf. A007863, A155112.

a := n -> hypergeom([-n, n, n+1], [1/2, 1], -1/4):
seq(round(evalf(a(n), 32)), n=0..21); # Peter Luschny, Oct 08 2015

Table[Sum[Binomial[k+n-1,k]*Binomial[k+n,2*k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Oct 04 2015 *)

B(x):=sum(sum(binomial(i+n-1,i)*binomial(i+n,2*i+1),i,0,n-1)/n*x^n,n,1,30);
taylor(x*diff(B(x),x)/B(x),x,0,20);

a(n) = sum(k=0, n, binomial(k+n-1,k)*binomial(k+n,2*k));
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 04 2015

G.f.: A(x) = x*B'(x)/B(x), where B(x)/x is g.f. of A007863.
Recurrence: 5*(n-1)*n*(35*n^2 - 143*n + 138)*a(n) = 2*(n-1)*(630*n^3 - 2889*n^2 + 3746*n - 1200)*a(n-1) - 2*(70*n^4 - 426*n^3 + 811*n^2 - 589*n + 150)*a(n-2) + 2*(n-3)*(2*n - 3)*(35*n^2 - 73*n + 30)*a(n-3). - Vaclav Kotesovec, Oct 04 2015
a(n) = hypergeom([-n, n, n+1], [1/2, 1], -1/4). - Peter Luschny, Oct 08 2015
a(n) = A155112(2n,n). - Alois P. Heinz, Sep 29 2022

A192131 G.f. satisfies: A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^3A(x)^k) x^n/n ).

Original entry on oeis.org

1, 2, 9, 53, 357, 2611, 20180, 162276, 1344455, 11400944, 98498545, 864068233, 7677040177, 68947431898, 624960856374, 5710352911097, 52542826413590, 486458467209032, 4528570067254485, 42365044032385154, 398081015128641213

Offset: 0

Paul D. Hanna, Jun 24 2011

nonn

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 53*x^3 + 357*x^4 + 2611*x^5 + 20180*x^6 +...
where the g.f. satisfies:
log(A(x)) = (1 + A(x))*x + (1 + 8*A(x) + A(x)^2)*x^2/2 + (1 + 27*A(x) + 27*A(x)^2 + A(x)^3)*x^3/3 + (1 + 64*A(x) + 216*A(x)^2 + 64*A(x)^3 + A(x)^4)*x^4/4 +...

Cf. A007863 (variant).

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^3*(A+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}

A193589 Augmentation of the Fibonacci triangle A193588. See Comments.

Original entry on oeis.org

1, 1, 2, 1, 4, 7, 1, 6, 18, 31, 1, 8, 33, 90, 154, 1, 10, 52, 185, 481, 820, 1, 12, 75, 324, 1065, 2690, 4575, 1, 14, 102, 515, 2006, 6276, 15547, 26398, 1, 16, 133, 766, 3420, 12468, 37711, 92124, 156233, 1, 18, 168, 1085, 5439, 22412, 78030, 230277

Offset: 0

Clark Kimberling, Jul 31 2011

For an introduction to the unary operation augmentation as applied to triangular arrays or sequences of polynomials, see A193091.
Regarding A193589, if the triangle is written as (w(n,k)), then w(n,n)=A007863(n); w(n,n-1)=A011270; and
(col 3)=A033537.

			First 5 rows of A193588:
1
1....2
1....2....3
1....2....3....5
1....2....3....5....8
First 5 rows of A193589:
1
1....2
1....4....7
1....6....18...31
1....8....33...90...154

Cf. A193091, A193588.

p[n_, k_] := Fibonacci[k + 2]
Table[p[n, k], {n, 0, 5}, {k, 0, n}]  (* A193588 *)
m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
TableForm[m[4]]
w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
v[n_] := v[n - 1].m[n]
TableForm[Table[v[n], {n, 0, 6}]]  (* A193589 *)
Flatten[Table[v[n], {n, 0, 8}]]

cp's OEIS Frontend

A239108 Number of hybrid 5-ary trees with n internal nodes.

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A239109 Number of hybrid 6-ary trees with n internal nodes.

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A215715 G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x*A(x)^4).

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A216314 G.f. satisfies A(x) = (1 + x*A(x)) * (1 + 2*x*A(x)^2).

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A239107 Number of hybrid 4-ary trees with n internal nodes.

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A274735 G.f. satisfies A(x) = (1 + x*A(x))^3 * (1 + x*A(x)^2).

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A215715 G.f. satisfies A(x) = (1 + xA(x)^2) (1 + x*A(x)^4).

A216314 G.f. satisfies A(x) = (1 + xA(x)) (1 + 2xA(x)^2).

A274735 G.f. satisfies A(x) = (1 + xA(x))^3 (1 + x*A(x)^2).

A364331 G.f. satisfies A(x) = (1 + xA(x)^2) (1 + x*A(x)^5).

A262910 a(n) = Sum_{k=0..n} binomial(k+n-1,k)binomial(k+n,2k).

A192131 G.f. satisfies: A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^3A(x)^k) x^n/n ).