A120972 -id:A120972 - cp's OEIS frontend

A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).

1, 1, 2, 9, 60, 504, 4946, 54430, 655362, 8496454, 117311198, 1711459903, 26228829200, 420370445830, 7021029571856, 121859518887327, 2192820745899978, 40831103986939664, 785429260324068156, 15585831041632684997, 318649154587152781210, 6704504768568697046504

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

From Paul D. Hanna, Nov 16 2008: (Start)
More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1. (End)

			G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ...
Related expansions.
A(x)^2 = 1 + 2*x + 5*x^2 + 22*x^3 + 142*x^4 + 1164*x^5 + 11221*x^6 + ...
A(A(x)-1) = 1 + x + 4*x^2 + 26*x^3 + 218*x^4 + 2151*x^5 + 23854*x^6 + ...
A(A(x)-1)^2 = 1 + 2*x + 9*x^2 + 60*x^3 + 504*x^4 + 4946*x^5 + ...
x/A(x)^2 = x - 2*x^2 - x^3 - 10*x^4 - 73*x^5 - 662*x^6 - 6842*x^7 - ...
Series_Reversion(x/A(x)^2) = x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ...
To illustrate the formula a(n) = [x^(n-1)] 2*A(x)^(2*n)/(2*n),
form a table of coefficients in A(x)^(2*n) as follows:
  A^2:  [(1), 2,   5,   22,   142,   1164,   11221,   121848, ...];
  A^4:  [ 1, (4), 14,   64,   397,   3116,   29002,   306468, ...];
  A^6:  [ 1,  6, (27), 134,   825,   6270,   56492,   580902, ...];
  A^8:  [ 1,  8,  44, (240), 1502,  11200,   98144,   983016, ...];
  A^10: [ 1, 10,  65,  390, (2520), 18672,  160115,  1565260, ...];
  A^12: [ 1, 12,  90,  592,  3987, (29676), 250730,  2399388, ...];
  A^14: [ 1, 14, 119,  854,  6027,  45458, (381010), 3582266, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[2/2*(1), 2/4*(4), 2/6*(27), 2/8*(240), 2/10*(2520), 2/12*(29676), ...].

Vaclav Kotesovec, Table of n, a(n) for n = 0..350

Cf. A120971; variants: A120972, A120974, A120976, A030266, A067145, A107096.
Cf. related variants: A145347, A145348, A147664, A145349, A145350. - Paul D. Hanna, Nov 16 2008
Cf. A381602.

terms = 21; A[] = 1; Do[A[x] = 1 + x*A[A[x] - 1]^2 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^2))[ #A]);A[n+1]}
for(n=0,30,print1(a(n),", "))

/* This sequence is generated when k=2, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */ {a(n,k=2,m=0)=local(A=sum(i=0,n-1,a(i,k,m)*x^i));if(n==0,1,polcoeff((m+k)/(m+k*n)*A^(m+k*n),n-1))} \\ Paul D. Hanna, Nov 16 2008
for(n=0,30,print1(a(n),", "))

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(2*n+k, j)/(2*n+k)*b(n-j, 2*j)));
a(n) = if(n==0, 1, b(n-1, 2)); \\ Seiichi Manyama, Jun 04 2025

G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^2.
Let B(x) be the g.f. of A120971, then B(x) and g.f. A(x) are related by:
(a) B(x) = A(A(x)-1),
(b) B(x) = A(x*B(x)^2),
(c) A(x) = B(x/A(x)^2),
(d) A(x) = 1 + x*B(x)^2,
(e) B(x) = 1 + x*B(x)^2*B(A(x)-1)^2,
(f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^2).
a(n) = [x^(n-1)] (1/n)*A(x)^(2n) for n>=1 with a(0)=1; i.e., a(n) equals 1/n times the coefficient of x^(n-1) in A(x)^(2n) for n>=1. [Paul D. Hanna, Nov 16 2008]
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120971.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(2*n+k,j)/(2*n+k) * b(n-j,2*j).
a(n) = b(n-1,2) for n > 0. (End)

A145350 G.f. satisfies: A(x/A(x)^2) = 1 + x*A(x).

Original entry on oeis.org

1, 1, 3, 18, 154, 1632, 20007, 273164, 4058556, 64628487, 1091488334, 19403175105, 361028420037, 7000932594042, 141010975529568, 2942134448306481, 63449975020918843, 1411787024678728344, 32360032648643379471, 763096191377494726161, 18491954778730596443088

Offset: 0

Paul D. Hanna, Nov 12 2008

nonn

More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

			G.f.: A(x) = 1 + x + 3*x^2 + 18*x^3 + 154*x^4 + 1632*x^5 +...
A(x)^2 = 1 + 2*x + 7*x^2 + 42*x^3 + 353*x^4 + 3680*x^5 + 44526*x^6+...
A(x/A(x)^2) = 1 + x + x^2 + 3*x^3 + 18*x^4 + 154*x^5 + 1632*x^6 +...
A(x) = 1 + x*G(x)^3 where G(x) = A(x*G(x)^2):
G(x) = 1 + x + 5*x^2 + 41*x^3 + 432*x^4 + 5329*x^5 + 73512*x^6 +...
G(x)^2 = 1 + 2*x + 11*x^2 + 92*x^3 + 971*x^4 + 11932*x^5 +...
To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(2*n+1)/(2*n+1),
form a table of coefficients in A(x)^(2*n+1) as follows:
A^3: [(1), 3, 12, 73, 606, 6225, 74370, 994668, ...];
A^5: [1, (5), 25, 160, 1315, 13191, 153930, 2017620, ...];
A^7: [1, 7, (42), 287, 2373, 23436, 267988, 3445835, ...];
A^9: [1, 9, 63, (462), 3888, 38106, 428637, 5414760, ...];
A^11: [1, 11, 88, 693, (5984), 58619, 651354, 8099410, ...];
A^13: [1, 13, 117, 988, 8801, (86697), 955656, 11723712, ...];
A^15: [1, 15, 150, 1355, 12495, 124398, (1365820), 16571385, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[3/3*(1), 3/5*(5), 3/7*(42), 3/9*(462), 3/11*(5984), 3/13*(86697), ...].
ALTERNATE GENERATING METHOD.
This sequence forms column zero in the follow array.
Let A denote this sequence, and A^2 the self-convolution square of A.
Start in row zero with this sequence, A, after prepending an initial '1', then repeat: drop the initial term and perform convolution with A^2 and the remaining terms in a given row to obtain the next row:
[1, 1, 1, 3, 18, 154, 1632, 20007, 273164, 4058556, 64628487, ...];
[1, 3, 12, 73, 606, 6225, 74370, 994668, 14535285, 228349287, ...];
[3, 18, 118, 962, 9511, 109404, 1415942, 20128565, 309001962, ...];
[18, 154, 1324, 13017, 146470, 1849625, 25701033, 386747469, ...];
[154, 1632, 16743, 188240, 2343654, 32006379, 473572975, ...];
[1632, 20007, 233150, 2905879, 39290669, 573813430, 8978918475, ...];
[20007, 273164, 3512228, 47574771, 689590692, 10679554646, ...];
[273164, 4058556, 56511375, 820798718, 12635699895, ...];
[4058556, 64628487, 962231360, 14843336308, 241004566025, ...]; ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..374

Cf. variants: A001764, A147664, A120972, A145345, A145349, A147664.

{a(n)=local(A=1+x,G);for(i=0,n,G=(serreverse(x/(A+x*O(x^n))^2)/x)^(1/2);A=1+x*G^3);polcoeff(A,n)}

/* This sequence is generated when k=2, m=1: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=2, m=1)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)+x*O(x^n)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0,20,print1(a(n),", "))

/* Prints terms 0..30 */
{A=[1];
for(m=1,30,
  B=A;
  for(i=1,m-1, C=Vec(Ser(A)^2*Ser(B)); B=vector(#C-1,n,C[n+1]) );
  A=concat(A,0);A[#A]=B[1]
);
A} \\ Paul D. Hanna, Jan 10 2016

G.f.: A(x) = 1 + x*G(x)^3 where G(x) = A(x*G(x)^2) and A(x) = G(x/A(x)^2).

a(n) = [x^(n-1)] 3*A(x)^(2*n+1)/(2*n+1) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 3*A(x)^(2*n+1)/(2*n+1) for n>=1.

A147664 G.f. satisfies: A(x/A(x)) = 1 + x*A(x)^2.

Original entry on oeis.org

1, 1, 3, 15, 100, 801, 7296, 73174, 791751, 9116613, 110640310, 1405349658, 18585016509, 254855278565, 3612425924919, 52793266545585, 793851646358364, 12261570084250926, 194260753173421656, 3153098224666860712

Offset: 0

Paul D. Hanna, Nov 09 2008

nonn

More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

			G.f.: A(x) = 1 + x + 3*x^2 + 15*x^3 + 100*x^4 + 801*x^5 + 7296*x^6 +...
A(x)^2 = 1 + 2*x + 7*x^2 + 36*x^3 + 239*x^4 + 1892*x^5 + 17019*x^6 +...
A(x/A(x)) = 1 + x + 2*x^2 + 7*x^3 + 36*x^4 + 239*x^5 + 1892*x^6 +...
A(x) = 1 + x*G(x)^3 where G(x) = A(x*G(x)) is the g.f. of A182953:
G(x) = 1 + x + 4*x^2 + 25*x^3 + 197*x^4 + 1797*x^5 + 18178*x^6 +...
To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(n+2)/(n+2),
form a table of coefficients in A(x)^(n+2) as follows:
A^3: [(1), 3, 12, 64, 426, 3345, 29766, 291999, ...];
A^4: [1, (4), 18, 100, 671, 5244, 46248, 449264, ...];
A^5: [1, 5, (25), 145, 985, 7686, 67305, 648085, ...];
A^6: [1, 6, 33, (200), 1380, 10782, 93922, 897402, ...];
A^7: [1, 7, 42, 266, (1869), 14658, 127246, 1207753, ...];
A^8: [1, 8, 52, 344, 2466, (19456), 168604, 1591496, ...];
A^9: [1, 9, 63, 435, 3186, 25335, (219522), 2063052, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[3/3*(1), 3/4*(4), 3/5*(25), 3/6*(200), 3/7*(1869), 3/8*(19456), ...].
ALTERNATE GENERATING METHOD.
This sequence forms column zero in the follow array.
Let A denote this sequence, and A^2 the self-convolution square of A.
Start in row zero with A^2, after prepending an initial '1', then repeat: drop the initial term and perform convolution with A and the remaining terms in a given row to obtain the next row:
[1, 1, 2, 7, 36, 239, 1892, 17019, 168746, 1807656, 20634852, 248560373, ...];
[1, 3, 12, 64, 426, 3345, 29766, 291999, 3097746, 35059659, 419160576,...];
[3, 15, 85, 571, 4443, 38952, 376090, 3930156, 43875903, 518191486, ...];
[15, 100, 701, 5494, 47883, 457451, 4724372, 52138050, 609215321, ...];
[100, 801, 6495, 56980, 542331, 5558082, 60793521, 704009142, ...];
[801, 7296, 65878, 630811, 6448878, 70124397, 806356465, 9749112990, ...];
[7296, 73174, 718577, 7386763, 80183478, 917721557, 11031656810, ...];
[73174, 791751, 8324862, 90823582, 1038161379, 12431200320, 155525155360, ...]; ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..385

Cf. A182953, A145348, A145350, A120972, A145345, A145349.

{a(n)=local(F=1+x,G);for(i=0,n,G=serreverse(x/(F+x*O(x^n))^1)/x;F=1+x*G^3);polcoeff(F,n)}

/* This sequence is generated when k=1, m=2: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=1, m=2)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)+x*O(x^n)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0,20,print1(a(n),", "))

/* Prints terms 0..30 */
{A=[1];
for(m=1,30,
  B=Vec(Ser(A)^2);
  for(i=1,m-1, C=Vec(Ser(A)*Ser(B)); B=vector(#C-1,n,C[n+1]) );
  A=concat(A,0); A[#A]=B[1];
);
A} \\ Paul D. Hanna, Jan 10 2016

G.f.: A(x) = 1 + x*G(x)^3 where G(x) = A(x*G(x)) and A(x) = G(x/A(x)) is the g.f. of A182953.

a(n) = [x^(n-1)] 3*A(x)^(n+2)/(n+2) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 3*A(x)^(n+2)/(n+2) for n>=1 (see comment).

A120973 G.f. A(x) satisfies A(x) = 1 + xA(x)^3 A(x*A(x)^3)^3.

Original entry on oeis.org

1, 1, 6, 60, 776, 11802, 201465, 3759100, 75404151, 1608036861, 36172106112, 853346084343, 21021015647574, 538868533164995, 14336235065928966, 394957784033440194, 11246848201518516044, 330520280036501809758

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

Cf. A120972; variants: A110447, A120971, A120975, A120977.

{a(n)=local(A,G=[1,1]);for(i=1,n,G=concat(G,0); G[ #G]=-Vec(subst(Ser(G),x,x/Ser(G)^3))[ #G]); A=Vec(((Ser(G)-1)/x)^(1/3));A[n+1]}

a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, binomial(3*n+k, j)/(3*n+k)*a(n-j, 3*j))); \\ Seiichi Manyama, Mar 01 2025

G.f. A(x) satisfies: A(x) = G(G(x)-1), A(G(x)-1) = G(A(x)-1), A(x) = G(x*A(x)^3) and A(x/G(x)^3) = G(x), where G(x) is the g.f. of A120972 and satisfies G(x/G(x)^3) = 1 + x.
From Seiichi Manyama, Mar 01 2025: (Start)
Let a(n,k) = [x^n] A(x)^k.
a(n,0) = 0^n; a(n,k) = k * Sum_{j=0..n} binomial(3*n+k,j)/(3*n+k) * a(n-j,3*j). (End)

A120974 G.f. A(x) satisfies A(x/A(x)^4) = 1 + x; thus A(x) = 1 + series_reversion(x/A(x)^4).

Original entry on oeis.org

1, 1, 4, 38, 532, 9329, 190312, 4340296, 108043128, 2890318936, 82209697588, 2467155342740, 77676395612884, 2554497746708964, 87449858261161216, 3107829518797739032, 114399270654847628768, 4353537522757357068296, 171010040645759712226048

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

Robert Israel, Table of n, a(n) for n = 0..250

Cf. A120975; variants: A120970, A120972, A120976, A030266, A067145, A107096.

A:= x -> 1:
for m from 1 to 30 do
  Ap:= unapply(A(x)+c*x^m,x);
  S:= series(Ap(x/Ap(x)^4)-1-x, x, m+1);
  cs:= solve(convert(S,polynom),c);
  A:= subs(c=cs, eval(Ap));
od:
seq(coeff(A(x),x,m),m=0..30);# Robert Israel, Oct 25 2019

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[ A[x/A[x]^4] - 1 - x + 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 02 2019 *)

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^4))[ #A]);A[n+1]}

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(4*n+k, j)/(4*n+k)*b(n-j, 4*j)));
a(n) = if(n==0, 1, b(n-1, 4)); \\ Seiichi Manyama, Jun 04 2025

G.f. satisfies: A(x) = 1 + x*B(x)^4 = 1 + (1 + x*C(x)^4 )^4 where B(x) and C(x) satisfy: C(x) = B(x)*B(A(x)-1), B(x) = A(A(x)-1), B(A(x)-1) = A(B(x)-1), B(x/A(x)^4) = A(x), B(x) = A(x*B(x)^4) and B(x) is g.f. of A120975.
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120975.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(4*n+k,j)/(4*n+k) * b(n-j,4*j).
a(n) = b(n-1,4) for n > 0. (End)

A145348 G.f. satisfies: A(x/A(x)^2) = 1 + x*A(x)^2.

Original entry on oeis.org

1, 1, 4, 30, 312, 3965, 57824, 933998, 16346728, 305539046, 6037780164, 125227212342, 2711254371568, 61021656441091, 1423063422363676, 34297379607790288, 852463916004336464, 21812807282389353798

Offset: 0

Paul D. Hanna, Nov 09 2008

nonn

More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

			G.f.: A(x) = 1 + x + 4*x^2 + 30*x^3 + 312*x^4 + 3965*x^5 +...
A(x)^2 = 1 + 2*x + 9*x^2 + 68*x^3 + 700*x^4 + 8794*x^5 + 126974*x^6+..
A(x/A(x)^2) = 1 + x + 2*x^2 + 9*x^3 + 68*x^4 + 700*x^5 + 8794*x^6 +...
A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)^2):
G(x) = 1 + x + 6*x^2 + 59*x^3 + 742*x^4 + 10877*x^5 + 177612*x^6 +...
G(x)^2 = 1 + 2*x + 13*x^2 + 130*x^3 + 1638*x^4 + 23946*x^5 +...
To illustrate the formula a(n) = [x^(n-1)] 2*A(x)^(2*n+2)/(n+1),
form a table of coefficients in A(x)^(2*n+2) as follows:
A^4: [(1), 4, 22, 172, 1753, 21612, 306348, ...];
A^6: [1, (6), 39, 320, 3267, 39756, 554595, ...];
A^8: [1, 8, (60), 520, 5366, 64816, 892308, ...];
A^10: [1, 10, 85, (780), 8190, 98702, 1344920,  ...];
A^12: [1, 12, 114, 1108, (11895), 143676, 1943488, ...];
A^14: [1, 14, 147, 1512, 16653, (202384), 2725541, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[2/2*(1), 2/3*(6), 2/4*(60), 2/5*(780), 2/6*(11895), 2/7*(202384), ...].

Vaclav Kotesovec, Table of n, a(n) for n = 0..230

Cf. A145350, A147664, A120972.

{a(n)=local(F=1+x,G);for(i=0,n,G=serreverse(x/(F+x*O(x^n))^2)/x;F=1+x*G^2);polcoeff(F,n)}

/* This sequence is generated when k=2, m=2: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=2, m=2)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)+x*O(x^n)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0,20,print1(a(n),", "))

G.f.: A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)^2) and A(x) = G(x/A(x)^2).

a(n) = [x^(n-1)] 2*A(x)^(2*n+2)/(n+1) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 2*A(x)^(2*n+2)/(n+1) for n>=1 (see comment).

A120976 G.f. A(x) satisfies A(x/A(x)^5) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^5).

Original entry on oeis.org

1, 1, 5, 60, 1060, 23430, 602001, 17281760, 541258390, 18210836060, 651246905140, 24566101401035, 971933892729980, 40156993344526860, 1726753293393763625, 77065076699967844390, 3561820706538663354320, 170162336673835615653925, 8389644485709060522744640

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

Cf. A120977; variants: A120970, A120972, A120974, A030266, A067145, A107096.

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^5))[ #A]);A[n+1]}

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(5*n+k, j)/(5*n+k)*b(n-j, 5*j)));
a(n) = if(n==0, 1, b(n-1, 5)); \\ Seiichi Manyama, Jun 04 2025

G.f. satisfies: A(x) = 1 + x*B(x)^5 = 1 + (1 + x*C(x)^5 )^5 where B(x) and C(x) satisfy: C(x) = B(x)*B(A(x)-1), B(x) = A(A(x)-1), B(A(x)-1) = A(B(x)-1), B(x/A(x)^5) = A(x), B(x) = A(x*B(x)^5) and B(x) is g.f. of A120977.
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120977.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(5*n+k,j)/(5*n+k) * b(n-j,5*j).
a(n) = b(n-1,5) for n > 0. (End)

A145347 G.f. satisfies: A(x/A(x)) = 1 + x*A(x)^3.

Original entry on oeis.org

1, 1, 4, 26, 220, 2203, 24836, 306104, 4047988, 56713521, 834286612, 12801754120, 203889888832, 3357619794321, 56999146850380, 995081586539016, 17830012791062632, 327376145842252333, 6151225530281186372, 118142009771446643592, 2317165307900630229384

Offset: 0

Paul D. Hanna, Nov 09 2008

nonn

More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

			G.f.: A(x) = 1 + x + 4*x^2 + 26*x^3 + 220*x^4 + 2203*x^5 + 24836*x^6 +...
A(x)^3 = 1 + 3*x + 15*x^2 + 103*x^3 + 876*x^4 + 8679*x^5 + 96382*x^6 +...
A(x/A(x)) = 1 + x + 3*x^2 + 15*x^3 + 103*x^4 + 876*x^5 + 8679*x^6 +...
A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)):
G(x) = 1 + x + 5*x^2 + 39*x^3 + 381*x^4 + 4284*x^5 + 53163*x^6 +...
To illustrate the formula a(n) = [x^(n-1)] 4*A(x)^(n+3)/(n+3),
form a table of coefficients in A(x)^(n+3) as follows:
A^4: [(1), 4, 22, 156, 1337, 13220, 145988, 1759876, ...];
A^5: [1, (5), 30, 220, 1905, 18836, 207100, 2481740, ...];
A^6: [1, 6, (39), 296, 2595, 25704, 281727, 3358488, ...];
A^7: [1, 7, 49, (385), 3423, 34020, 372141, 4416658, ...];
A^8: [1, 8, 60, 488, (4406), 44000, 480900, 5686480, ...];
A^9: [1, 9, 72, 606, 5562, (55881), 610872, 7202268, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[4/4*(1), 4/5*(5), 4/6*(39), 4/7*(385), 4/8*(4406), 4/9*(55881), ...].

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A384265, A145348, A145350, A147664, A120972.

{a(n) = my(F=1+x); for(i=0,n, G=serreverse(x/(F+x*O(x^n))^1)/x; F=1+x*G^4); polcoef(F,n)}

/* This sequence is generated when k=1, m=3: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=1, m=3) = my(A=sum(i=0, n-1, a(i, k, m)*x^i) +x*O(x^n)); if(n==0, 1, polcoef((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0,20,print1(a(n),", "))

G.f.: A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)) and A(x) = G(x/A(x)).

a(n) = [x^(n-1)] 4*A(x)^(n+3)/(n+3) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 4*A(x)^(n+3)/(n+3) for n>=1 (see comment).

A381570 G.f. A(x) satisfies A(x) = (1 + xA(xA(x)))^3.

Original entry on oeis.org

1, 3, 12, 82, 732, 7944, 99156, 1381464, 21065853, 346932822, 6112226961, 114383442888, 2261347164766, 47025363829497, 1025005545866361, 23349137897005296, 554467427766694440, 13696046757037152183, 351231525904387758222, 9335221780768641038952

Offset: 0

Seiichi Manyama, Feb 28 2025

nonn

Column k=1 of A381569.

Cf. A087949, A120972, A212029, A381568, A381574.

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

See A381569.

G.f.: B(x)^3, where B(x) is the g.f. of A212029.

cp's OEIS Frontend

A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A145350 G.f. satisfies: A(x/A(x)^2) = 1 + x*A(x).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

PARI

Formula

A147664 G.f. satisfies: A(x/A(x)) = 1 + x*A(x)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

PARI

Formula

A120973 G.f. A(x) satisfies A(x) = 1 + x*A(x)^3 * A(x*A(x)^3)^3.

Views

Author

Keywords

Crossrefs

Programs

PARI

PARI

Formula

A120974 G.f. A(x) satisfies A(x/A(x)^4) = 1 + x; thus A(x) = 1 + series_reversion(x/A(x)^4).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

A145348 G.f. satisfies: A(x/A(x)^2) = 1 + x*A(x)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A120976 G.f. A(x) satisfies A(x/A(x)^5) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^5).

Views

Author

Keywords

Crossrefs

Programs

A120973 G.f. A(x) satisfies A(x) = 1 + xA(x)^3 A(x*A(x)^3)^3.

A381570 G.f. A(x) satisfies A(x) = (1 + xA(xA(x)))^3.