cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Original entry on oeis.org

1, 1, 2, 9, 60, 504, 4946, 54430, 655362, 8496454, 117311198, 1711459903, 26228829200, 420370445830, 7021029571856, 121859518887327, 2192820745899978, 40831103986939664, 785429260324068156, 15585831041632684997, 318649154587152781210, 6704504768568697046504
Offset: 0

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Comments

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)

Examples

			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), ...].

Links

Crossrefs

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.

Programs

  • Mathematica
    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 *)
  • PARI
    {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),", "))
  • PARI
    /* 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),", "))
  • PARI
    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

Formula

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

Views

Author

Paul D. Hanna, Nov 12 2008

Keywords

Comments

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.

Examples

			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, ...]; ...

Links

Crossrefs

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

Programs

  • PARI
    {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)}
  • PARI
    /* 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),", "))
  • PARI
    /* 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

Formula

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

Views

Author

Paul D. Hanna, Nov 09 2008

Keywords

Comments

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.

Examples

			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, ...]; ...

Links

Crossrefs

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

Programs

  • PARI
    {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)}
  • PARI
    /* 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),", "))
  • PARI
    /* 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

Formula

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).

A145345 G.f. satisfies: A(x/A(x)) = 1 + x*A(x).

Original entry on oeis.org

1, 1, 2, 7, 34, 203, 1398, 10706, 89120, 794347, 7502170, 74511150, 773864654, 8368430208, 93905460014, 1090519614152, 13077315637592, 161643281777801, 2056306418177832, 26887064722265250, 360939404438509866
Offset: 0

Views

Author

Paul D. Hanna, Nov 05 2008

Keywords

Comments

From Paul D. Hanna, Nov 15 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)

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 34*x^4 + 203*x^5 + 1398*x^6 + ...
A(x/A(x)) = 1 + x + x^2 + 2*x^3 + 7*x^4 + 34*x^5 + 203*x^6 + 1398*x^7 + ...
A(x) = 1 + x*G(x)^2 where
G(x) = 1 + x + 3*x^2 + 14*x^3 + 83*x^4 + 574*x^5 + 4432*x^6 + ...
is the g.f. of A121687.
ALTERNATE GENERATING METHOD.
This sequence forms column zero in the following array.
Let A denote this sequence.
Start in row zero with this sequence, A, 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, 1, 2, 7, 34, 203, 1398, 10706, 89120, 794347, 7502170, ...];
[1, 2, 5, 18, 86, 502, 3387, 25496, 209242, 1843134, 17235671, ...];
[2, 7, 27, 128, 727, 4763, 34912, 280006, 2418537, 22240055, ...];
[7, 34, 169, 958, 6173, 44364, 349152, 2965098, 26864357, ...];
[34, 203, 1195, 7707, 54792, 425216, 3560600, 31842929, ...];
[203, 1398, 9308, 66310, 510689, 4231188, 37425922, ...];
[1398, 10706, 78414, 605401, 4987185, 43742924, 406387957, ...];
[10706, 89120, 705227, 5824356, 50853813, 469182452, ...];
[89120, 794347, 6707823, 58712463, 539651646, 5211277285, ...];
[794347, 7502170, 67008980, 617340184, 5942316416, 59827126712, ...]; ...

Links

Crossrefs

Cf. A383563, A121687, A145350, A145349, A147664, A302703.

Programs

  • PARI
    {a(n)=local(F=1+x);for(i=0,n,G=serreverse(x/(F+x*O(x^n)))/x;F=1+x*subst(F,x,x*G)^2);polcoeff(F,n)}
  • PARI
    {a(n)=local(F=1+x);for(i=0,n,G=serreverse(x/(F+x*O(x^n)))/x;F=1+x*G^2);polcoeff(F,n)} \\ Paul D. Hanna, Nov 08 2008
  • PARI
    /* This sequence is generated when k=1, m=1: A(x/A(x)^k) = 1 + x*A(x)^m */ {a(n,k=1,m=1)=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 15 2008
  • PARI
    /* Prints terms 0..30 */
{A=[1];
for(m=1,30,
  B=A;
  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
  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(m+1)))[m+1] - Vec(Ser(A))[m+1])/(m+1)); A[n+1]}
for(n=0, 30, print1(2^n*a(n), ", ")) \\ Vaclav Kotesovec, Jan 31 2023

Formula

G.f. satisfies: A(x) = 1 + x*G(x)^2 where G(x) = g.f. of A121687.
G.f. satisfies: A(x) = G(x/A(x)) where G(x) = A(x*G(x)) = g.f. of A121687. - Paul D. Hanna, Nov 08 2008
a(n) = [x^(n-1)] (2/(n+1))*A(x)^(n+1) for n>=1 with a(0)=1; i.e., a(n) equals 2/(n+1) times the coefficient of x^(n-1) in A(x)^(n+1) for n>=1. - Paul D. Hanna, Nov 15 2008

A217794 G.f. satisfies: A(x/A(x)^3) = 1 + x*A(x)^2.

Original entry on oeis.org

1, 1, 5, 50, 705, 12225, 244076, 5408850, 130087875, 3345893240, 91094645450, 2606026116783, 77908655011205, 2423758771361300, 78209895085897845, 2610730390260224350, 89962745874096081282, 3194452580744306873560, 116712765368638543260825
Offset: 0

Views

Author

Paul D. Hanna, Oct 11 2012

Keywords

Comments

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.

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 50*x^3 + 705*x^4 + 12225*x^5 + 244076*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 11*x^2 + 110*x^3 + 1535*x^4 + 26360*x^5 + 522152*x^6 +...
A(x)^3 = 1 + 3*x + 18*x^2 + 181*x^3 + 2505*x^4 + 42630*x^5 + 837968*x^6 +...
A(x/A(x)^3) = 1 + x + 2*x^2 + 11*x^3 + 110*x^4 + 1535*x^5 + 26360*x^6 +...
Note that A(x) = 1 + x*G(x)^5 where G(x) = A(x*G(x)^3):
G(x) = 1 + x + 8*x^2 + 107*x^3 + 1840*x^4 + 36973*x^5 + 828944*x^6 +...
G(x)^3 = 1 + 3*x + 27*x^2 + 370*x^3 + 6378*x^4 + 127608*x^5 +...
G(x)^5 = 1 + 5*x + 50*x^2 + 705*x^3 + 12225*x^4 + 244076*x^5 +...
To illustrate the formula a(n) = [x^(n-1)] 5*A(x)^(3*n+2)/(3*n+2),
form a table of coefficients in A(x)^(3*n+2) as follows:
A^5: [(1),  5, 35, 360, 4930, 82576, 1599555, 34576120, ...];
A^8: [1,(8), 68, 736, 10050, 165336, 3139976, 66675504, ...];
A^11: [1, 11,(110), 1265, 17435, 283712, 5299448, 110675125, ...];
A^14: [1, 14, 161,(1974), 27706, 448882, 8275827, 170250124, ...];
A^17: [1, 17, 221, 2890,(41565), 674373, 12316568, 250050450, ...];
A^20: [1, 20, 290, 4040, 59795,(976304), 17727230, 355906600, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[5/5*(1), 5/8*(8), 5/11*(110), 5/14*(1974), 5/17*(41565), 5/20*(976304), ...].

Crossrefs

Cf. A145349, A145350.

Programs

  • PARI
    {a(n)=local(A=1+x, G); for(i=0, n, G=(serreverse(x/(A+x*O(x^n))^3)/x)^(1/3); A=1+x*G^5); polcoeff(A, n)}
  • PARI
    /* This sequence is generated when k=3, m=2: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=3, 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), ", "))

Formula

G.f.: A(x) = 1 + x*G(x)^5 where G(x) = A(x*G(x)^3) and A(x) = G(x/A(x)^3).
a(n) = [x^(n-1)] 5*A(x)^(3*n+2)/(3*n+2) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 5*A(x)^(3*n+2)/(3*n+2) for n>=1 (see comment).
Showing 1-5 of 5 results.

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