A219535 -id:A219535 - cp's OEIS frontend

A219537 G.f. satisfies A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4).

1, 1, 3, 13, 66, 366, 2148, 13115, 82449, 530095, 3469401, 23037642, 154820262, 1050999343, 7196493255, 49644745965, 344704716018, 2407157839593, 16895247295947, 119121868831235, 843306880720218, 5992060655349521, 42718501097385207, 305476181765843358

Offset: 0

Paul D. Hanna, Nov 21 2012

nonn

a(n) is the number of noncrossing partial matchings on points 1, 2, ... , 3*n where point 1 is unmatched if n>0 and only points congruent modulo 3 can be matched. See Example 57 on p. 47 of the Burstein-Shapiro reference. - Alexander Burstein, Jun 03 2022

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 66*x^4 + 366*x^5 + 2148*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 32*x^3 + 167*x^4 + 942*x^5 + 5593*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 58*x^3 + 312*x^4 + 1794*x^5 + 10794*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 92*x^3 + 511*x^4 + 3000*x^5 + 18316*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 775*x^4 + 4651*x^5 + 28845*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 188*x^3 + 1116*x^4 + 6852*x^5 + 43204*x^6 +...
where A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4),
and A(x)^2 = 1 + x*(A(x)^2 + A(x)^5),
and A(x)^3 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^6),
and A(x)^4 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^5 + A(x)^7),
and A(x)^5 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^5 + A(x)^6 + A(x)^8), etc.
The g.f. satisfies A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2) where
F(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + 51*x^7 +...+ A001006(n-1)*x^n +...
is a g.f. of the Motzkin numbers (A001006, shifted right 1 place).
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 80*x^5 + 322*x^6 +...+ A106228(n)*x^n +...
satisfies G(x) = 1 + x*G(x)/(1 - x*G(x)^2).

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Alexander Burstein and Louis W. Shapiro, Pseudo-Involutions in the Riordan Group, J. Integer Sequences 25 (2022), Article 22.3.6.

Cf. A219534, A219535, A219536, A219538, A363573, A366400.

Cf. A001006, A106228, A271469, A364765.

rec := {(36*n^4+126*n^3+126*n^2+36*n)*a(n)+(-276*n^4-1548*n^3-3198*n^2-2898*n-972)*a(n+1)+(940*n^4+7090*n^3+19916*n^2+24650*n+11316)*a(n+2)+(-845*n^4-9000*n^3-34159*n^2-53004*n-26136)*a(n+3)+(-260*n^4-5200*n^3-37454*n^2-116538*n-133128)*a(n+4)+(459*n^4+9774*n^3+77955*n^2+276012*n+366060)*a(n+5)+(-54*n^4-1242*n^3-10686*n^2-40758*n-58140)*a(n+6), a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 13, a(4) = 66, a(5) = 366}:
f:= gfun:-rectoproc(rec,a(n),remember):
map(f, [$0..50]); # Robert Israel, Feb 25 2018

nmax = 23; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x (A[x]^2 - A[x]^3 + A[x]^4)) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)

/* Formula A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4): */
{a(n)=local(A=1);for(i=1,n,A=1+x*(A^2-A^3+A^4) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion involving Motzkin numbers: */
{a(n)=local(A=1);A=(1+x-sqrt(1-2*x-3*x^2+x^3*O(x^n)))/(2*x); polcoeff(sqrt(1/x*serreverse(x/A^2)), n)}
for(n=0,25,print1(a(n),", "))

G.f. A(x) satisfies [from Paul D. Hanna, Mar 21 2016]: (Start)
(1) A(x)^2 = 1 + x*(A(x)^2 + A(x)^5).
(2) A(x)^3 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^6).
Let F(x) = (1+x - sqrt(1 - 2*x - 3*x^2)) / (2*x), then g.f. A(x) satisfies:
(3) A(x) = sqrt( (1/x)*Series_Reversion(x/F(x)^2) ),
(4) A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2),
where F(x) = 1 + x*M(x) such that M(x) = 1 + x*M(x) + x^2*M(x)^2 is the g.f. of the Motzkin numbers (A001006).
Let G(x) = 1 + x*G(x)/(1 - x*G(x)^2), then g.f. A(x) satisfies:
(5) A(x) = (1/x)*Series_Reversion(x/G(x)),
(6) A(x) = G(x*A(x)) and G(x) = A(x/G(x)).
where G(x) is the g.f. of A106228. (End)
Recurrence: 3*n*(3*n-1)*(3*n+1)*(5*n-11)*(5*n-8)*(5*n-6)*a(n) = 6*(5*n-11)*(900*n^5 - 3870*n^4 + 6033*n^3 - 4165*n^2 + 1238*n - 120)*a(n-1) - 2*(n-2)*(5*n-1)*(950*n^4 - 5510*n^3 + 11199*n^2 - 9207*n + 2430)*a(n-2) + 6*(n-3)*(n-2)*(2*n-5)*(5*n-6)*(5*n-3)*(5*n-1)*a(n-3). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ sqrt(300+75*10^(2/3)+30*10^(1/3))/90 * (5/9*10^(2/3)+10/9*10^(1/3)+8/3)^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 19 2013
Recurrence: 18*n*(2*n+1)*(n+2)*(n+1)*a(n)-(6*(n+1))*(n+2)*(46*n^2+120*n+81)*a(n+1)+(2*(n+2))*(470*n^3+2605*n^2+4748*n+2829)*a(n+2)-(n+3)*(845*n^3+6465*n^2+14764*n+8712)*a(n+3)-(2*(n+4))*(130*n^3+2080*n^2+10407*n+16641)*a(n+4)+(3*(n+5))*(153*n^3+2493*n^2+13520*n+24404)*a(n+5)-(6*(n+5))*(3*n+17)*(3*n+19)*(n+6)*a(n+6) = 0. - Robert Israel, Feb 25 2018
G.f. A(x) satisfies: A(-x*A(x)^5) = 1/A(x). - Alexander Burstein, Jun 03 2022
a(n) = (1/n) * Sum_{k=0..n-1} binomial(n,k) * binomial(2*n+k,n-1-k) for n > 0. - Seiichi Manyama, Aug 05 2023
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * binomial(n,k) * binomial(4*n-2*k,n-1-k) for n > 0. - Seiichi Manyama, Aug 06 2023
G.f.: A(x) = sqrt(B(x)) where B(x) is the g.f. of A366400. - Seiichi Manyama, Mar 31 2024
a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} binomial(n,k) * binomial(3*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 01 2024
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+3*k/2+1/2,n)/(2*n+3*k+1). - Seiichi Manyama, Apr 04 2024
G.f.: Sum_{k>=0} binomial(5*k/2 + 1/2, k)*x^k/((5*k + 1)*(1 - x)^((5*k + 1)/2)). - Miles Wilson, Feb 02 2025

A219534 G.f. satisfies A(x) = 1 + x*(A(x)^2 + A(x)^4).

Original entry on oeis.org

1, 2, 12, 100, 968, 10208, 113792, 1318832, 15732064, 191878592, 2381917824, 29995598208, 382257383168, 4920505410816, 63882881030656, 835554927932160, 10999486798112256, 145626782310460416, 1937772463214168064, 25901381584638605312, 347618773649248088064

Offset: 0

Paul D. Hanna, Nov 21 2012

nonn

			G.f.: A(x) = 1 + 2*x + 12*x^2 + 100*x^3 + 968*x^4 + 10208*x^5 +...
Related expansions:
A(x)^2 = 1 + 4*x + 28*x^2 + 248*x^3 + 2480*x^4 + 26688*x^5 +...
A(x)^4 = 1 + 8*x + 72*x^2 + 720*x^3 + 7728*x^4 + 87104*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)^2) and G(x) = A(x/G(x)^2) where
G(x) = 1 + 2*x + 4*x^2 + 12*x^3 + 40*x^4 + 144*x^5 + 544*x^6 +...+ A025227(n+1)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..850

Cf. A025227, A219535, A219536, A219537, A219538.

Cf. A006318, A364167.

nmax=20; aa=ConstantArray[0,nmax]; aa[[1]]=2; Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[1+x*(AGF^2+AGF^4)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 10 2013 *)

/* Formula A(x) = 1 + x*(A(x)^2 + A(x)^4): */
{a(n)=local(A=1);for(i=1,n,A=1+x*(A^2+A^4) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion: */
{a(n)=local(A=1,G=(1-sqrt(1-4*x-4*x^2+x^3*O(x^n)))/(2*x));A=sqrt((1/x)*serreverse(x/G^2));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

Let G(x) = (1 - sqrt(1-4*x-4*x^2))/(2*x), then g.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion(x/G(x)^2) ),
(2) A(x) = G(x*A(x)^2) and G(x) = A(x/G(x)^2),
where x*G(x) is the g.f. of A025227.
Recurrence: 3*n*(3*n-1)*(3*n+1)*(131*n^3 - 666*n^2 + 1075*n - 558)*a(n) = 2*(26200*n^6 - 172500*n^5 + 431572*n^4 - 521613*n^3 + 316327*n^2 - 89058*n + 8640)*a(n-1) - 12*(n-2)*(1441*n^5 - 8767*n^4 + 19186*n^3 - 18172*n^2 + 6930*n - 810)*a(n-2) + 8*(n-3)*(n-2)*(2*n-5)*(131*n^3 - 273*n^2 + 136*n - 18)*a(n-3). - Vaclav Kotesovec, Sep 10 2013
a(n) ~ c*d^n/n^(3/2), where d = 2/81*(7217783 + 10611 * sqrt(786))^(1/3) + 74654/(81*(7217783 + 10611 * sqrt(786))^(1/3)) + 400/81 = 14.48001092254652246... is the root of the equation -16 + 132*d - 400*d^2 + 27*d^3 = 0 and c = 1/2358*sqrt(262)*sqrt((213070976 + 3034746 * sqrt(786))^(1/3) * ((213070976 + 3034746 * sqrt(786))^(2/3) + 336670 + 1310*(213070976 + 3034746 * sqrt(786))^(1/3)))/((213070976 + 3034746 * sqrt(786))^(1/3)*sqrt(Pi)) = 0.1929450901182412149... - Vaclav Kotesovec, Sep 10 2013
a(n) = (1/n) * Sum_{k=0..floor(n-1)/2} 2^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 01 2024
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(2*n+2*k+1,n)/(2*n+2*k+1). - Seiichi Manyama, Apr 03 2024

A336575 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} 3^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.

Original entry on oeis.org

1, 1, 3, 1, 3, 3, 1, 3, 12, 3, 1, 3, 21, 57, 3, 1, 3, 30, 192, 300, 3, 1, 3, 39, 408, 2001, 1686, 3, 1, 3, 48, 705, 6402, 22539, 9912, 3, 1, 3, 57, 1083, 14799, 109137, 267276, 60213, 3, 1, 3, 66, 1542, 28488, 338430, 1964010, 3287496, 374988, 3, 1, 3, 75, 2082, 48765, 817743, 8181597, 36718680, 41556585, 2381322, 3

Offset: 0

Seiichi Manyama, Jul 26 2020

			Square array begins:
  1,    1,     1,      1,      1,      1, ...
  3,    3,     3,      3,      3,      3, ...
  3,   12,    21,     30,     39,     48, ...
  3,   57,   192,    408,    705,   1083, ...
  3,  300,  2001,   6402,  14799,  28488, ...
  3, 1686, 22539, 109137, 338430, 817743, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-4 give: A122553, A047891, A219535, A336538, A336540.
Main diagonal gives A336578.
Cf. A336534, A336573, A336574.

T[0, k_] := 1; T[n_, k_] := Sum[3^j * Binomial[n, j] * Binomial[k*n, j - 1], {j, 1, n}]/n; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 27 2020 *)

T(n, k) = if(n==0, 1, sum(j=1, n, 3^j*binomial(n, j)*binomial(k*n, j-1))/n);

T(n, k) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k*(2+A)); polcoeff(A, n);

T(n, k) = sum(j=0, n, 2^(n-j)*binomial(n, j)*binomial(k*n+j+1, n)/(k*n+j+1));

T(n, k) = sum(j=0, n, 2^j*binomial(k*n+1, j)*binomial((k+1)*n-j, n-j))/(k*n+1);

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k * (2 + A_k(x)).
T(n,k) = Sum_{j=0..n} 2^(n-j) * binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} 2^j * binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = (1/n) * Sum_{j=0..n-1} (-2)^j * 3^(n-j) * binomial(n,j) * binomial((k+1)*n-j,n-1-j) for n > 0. - Seiichi Manyama, Aug 10 2023
T(n,k) = 3*hypergeom([1-n, -k*n], [2], 3) for n > 0. - Stefano Spezia, Aug 09 2025

A219536 G.f. satisfies A(x) = 1 + x(A(x)^2 + 2A(x)^3).

Original entry on oeis.org

1, 3, 24, 255, 3102, 40854, 566934, 8164263, 120864390, 1827982362, 28122626760, 438720097638, 6923868098820, 110346550539780, 1773394661610258, 28707809007278775, 467677404522668742, 7661583171651546786, 126137791939032756960, 2085923447593966281378

Offset: 0

Paul D. Hanna, Nov 21 2012

			G.f.: A(x) = 1 + 3*x + 24*x^2 + 255*x^3 + 3102*x^4 + 40854*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 57*x^2 + 654*x^3 + 8310*x^4 + 112560*x^5 +...
A(x)^3 = 1 + 9*x + 99*x^2 + 1224*x^3 + 16272*x^4 + 227187*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 645*x^4 + 4791*x^5 +...+ A103210(n)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..790
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.

Column k=2 of A336574.

Cf. A003168, A103210, A219534, A219535.

CoefficientList[1/x*InverseSeries[Series[4*x^2/(1-x-Sqrt[1-10*x+x^2]), {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 28 2013 *)

/* Formula A(x) = 1 + x*(A(x)^2 + 2*A(x)^3): */
{a(n)=my(A=1);for(i=1,n,A=1+x*(A^2+2*A^3) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion: */
{a(n)=my(A=1,G=(1-x-sqrt(1-10*x+x^2+x^3*O(x^n)))/(4*x));A=(1/x)*serreverse(x/G);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 26 2020

a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 26 2020

Let G(x) = (1-x - sqrt(1 - 10*x + x^2)) / (4*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where G(x) is the g.f. of A103210.
Recurrence: 4*n*(2*n+1)*(19*n-26)*a(n) = (2717*n^3 - 6435*n^2 + 4342*n - 840)*a(n-1) + 2*(n-2)*(2*n-3)*(19*n-7)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ (3/19)^(1/4) * (5+sqrt(57)) * ((143 + 19*sqrt(57))/16)^n / (16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
From Seiichi Manyama, Jul 26 2020: (Start)
a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
From Seiichi Manyama, Aug 10 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 3^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * 2^(n-k) * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)
a(n) = (-1)^(n+1) * (3/n) * Jacobi_P(n-1, 1, n+1, -5) for n >= 1. - Peter Bala, Sep 08 2024

A372214 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 2x - sqrt(1 - 8x + 4x^2))/(2x).

Original entry on oeis.org

1, 4, 40, 487, 6376, 86629, 1203823, 16984678, 242274280, 3484593028, 50444222665, 734066291974, 10728052396111, 157349171819155, 2314894133906086, 34145661019248487, 504810905195542504, 7478066502444399439, 110972913533524676080, 1649407167353221551706, 24549982881130265421001

Offset: 0

Peter Bala, Apr 23 2024

x*G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/2 is the o.g.f. of A047891.
The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. Examples of these supercongruences are given below.
More generally, for each integer m, we conjecture that the sequence {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 3*x + O(x^2)
  n = 2: G(x)^2 = 1 + 6*x + 33*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 9*x + 63*x^2 + 414*x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 12*x + 102*x^2 + 768*x^3 + 5493*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 33 = 40, a(3) = 1 + 9 + 63 + 414 = 487 and a(4) = 1 + 12 + 102 + 768 + 5493 = 6376.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
                                             row sums
  n = 0 |    1                                   1
  n = 1 |    3    1                              4
  n = 2 |   33    6     1                       40
  n = 3 |  414   63     9    1                 487
  n = 4 | 5493  768   102   12   1            6376
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 157349171819155 - 4 = (3^3)*(13^3)*269*9860941 == 0 (mod 13^3).
a(2*7) - a(2) = 2314894133906086 - 40 = 2*(3^4)*(7^3)*11*12119*312509 == 0 (mod 7^3).

Cf. A047891, A219535, A333090, A333093, A372215.

G := x -> (1/2)*(1 - 2*x - sqrt(1 - 8*x + 4*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 41):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);

Table[SeriesCoefficient[(2*(1 + x)^2/(1 - x + Sqrt[1 - 6*x - 3*x^2]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)

G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024

a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + ... is the o.g.f. of A219535.
Row sums of the Riordan array ( 1 + x*F'(x)/F(x), x*F(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ sqrt(1 + 17/sqrt(33)) * (59 + 11*sqrt(33))^n / (sqrt(3*Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, May 02 2024

cp's OEIS Frontend

A219537 G.f. satisfies A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4).

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A219534 G.f. satisfies A(x) = 1 + x*(A(x)^2 + A(x)^4).

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A336575 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} 3^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.

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A219536 G.f. satisfies A(x) = 1 + x*(A(x)^2 + 2*A(x)^3).

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A372214 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x).

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A219536 G.f. satisfies A(x) = 1 + x(A(x)^2 + 2A(x)^3).

A372214 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 2x - sqrt(1 - 8x + 4x^2))/(2x).