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)

A120972 G.f. A(x) satisfies A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).

Original entry on oeis.org

1, 1, 3, 21, 217, 2814, 42510, 718647, 13270944, 263532276, 5567092665, 124143735663, 2905528740060, 71058906460091, 1809695198254281, 47861102278428198, 1311488806252697283, 37164457324943708739, 1087356593493807164289, 32801308084353988297404
Offset: 0

Views

Author

Paul D. Hanna, Jul 20 2006

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 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
Related expansions.
A(x)^3 = 1 + 3*x + 12*x^2 + 82*x^3 + 813*x^4 + 10212*x^5 + 150699*x^6 +...
A(A(x)-1) = 1 + x + 6*x^2 + 60*x^3 + 776*x^4 + 11802*x^5 + 201465*x^6 +...
A(A(x)-1)^3 = 1 + 3*x + 21*x^2 + 217*x^3 + 2814*x^4 + 42510*x^5 +...
x/A(x)^3 = x - 3*x^2 - 3*x^3 - 37*x^4 - 420*x^5 - 5823*x^6 -...
Series_Reversion(x/A(x)^3) = x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(3*n)/(3*n),
form a table of coefficients in A(x)^(3*n) as follows:
  A^3:  [(1), 3,  12,   82,    813,   10212,   150699,   2503233, ...];
  A^6:  [ 1, (6), 33,  236,   2262,   27270,   388906,   6289080, ...];
  A^9:  [ 1,  9, (63), 489,   4671,   54684,   756012,  11904813, ...];
  A^12: [ 1, 12, 102, (868),  8445,   97260,  1310040,  20112516, ...];
  A^15: [ 1, 15, 150, 1400, (14070), 161343,  2130505,  31961175, ...];
  A^18: [ 1, 18, 207, 2112,  22113, (255060), 3324003,  48876264, ...];
  A^21: [ 1, 21, 273, 3031,  33222,  388563, (5030529), 72769014, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[3/3*(1), 3/6*(6), 3/9*(63), 3/12*(868), 3/15*(14070), 3/18*(255060), ...].

Crossrefs

Cf. A120973; variants: A120970, A120974, A120976, A030266, A067145, A107096.
Cf. A381603.

Programs

  • Mathematica
    terms = 18; A[] = 1; Do[A[x] = 1 + x*A[A[x] - 1]^3 + 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)^3))[ #A]);A[n+1]}
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*subst(A^3,x,A-1+x*O(x^n)));polcoeff(A,n)}
  • PARI
    /* This sequence is generated when k=3, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=3, 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))}
for(n=0,25,print1(a(n),", "))
  • PARI
    b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(3*n+k, j)/(3*n+k)*b(n-j, 3*j)));
a(n) = if(n==0, 1, b(n-1, 3)); \\ Seiichi Manyama, Jun 04 2025

Formula

G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^3.
a(n) = [x^(n-1)] A(x)^(3*n)/n for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in A(x)^(3*n)/n for n>=1 (see comment).
Let B(x) be the g.f. of A120973, 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)^3),
(c) A(x) = B(x/A(x)^3),
(d) A(x) = 1 + x*B(x)^3,
(e) B(x) = 1 + x*B(x)^3*B(A(x)-1)^3,
(f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^3).
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120973.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(3*n+k,j)/(3*n+k) * b(n-j,3*j).
a(n) = b(n-1,3) for n > 0. (End)

A145349 G.f. satisfies: A(x/A(x)^3) = 1 + x*A(x).

Original entry on oeis.org

1, 1, 4, 34, 416, 6319, 111124, 2177346, 46440184, 1061938195, 25762345804, 658072997702, 17600573291712, 490770914734054, 14219015899154068, 426904437068035200, 13252855203929697200, 424634035832800883743, 14020984603588221319072, 476460483567807053745450
Offset: 0

Views

Author

Paul D. Hanna, Nov 11 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 + 4*x^2 + 34*x^3 + 416*x^4 + 6319*x^5 +...
A(x)^3 = 1 + 3*x + 15*x^2 + 127*x^3 + 1512*x^4 + 22419*x^5 +...
A(x/A(x)^3) = 1 + x + x^2 + 4*x^3 + 34*x^4 + 416*x^5 + 6319*x^6 +...
A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)^3):
G(x) = 1 + x + 7*x^2 + 82*x^3 + 1239*x^4 + 21942*x^5 + 434746*x^6 +...
G(x)^3 = 1 + 3*x + 24*x^2 + 289*x^3 + 4377*x^4 + 77097*x^5 +...
To illustrate the formula a(n) = [x^(n-1)] 4*A(x)^(3*n+1)/(3*n+1),
form a table of coefficients in A(x)^(3*n+1) as follows:
A^4: [(1), 4, 22, 188, 2217, 32516, 555972, ...];
A^7: [1, (7), 49, 441, 5131, 73248, 1220457, ...];
A^10: [1, 10, (85), 820, 9590, 134482, 2191060, ...];
A^13: [1, 13, 130, (1352), 16107, 223886, 3582072, ...];
A^16: [1, 16, 184, 2064, (25276), 351072, 5541912, ...];
A^19: [1, 19, 247, 2983, 37772, (527839), 8260174, ...];  ...
in which the main diagonal forms the initial terms of this sequence:
[4/4*(1), 4/7*(7), 4/10*(85), 4/13*(1352), 4/16*(25276), 4/19*(527839), ...].
ALTERNATE GENERATING METHOD.
This sequence forms column zero in the follow array.
Let A denote this sequence, and A^3 the self-convolution cube 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^3 and the remaining terms in a given row to obtain the next row:
[1, 1, 1, 4, 34, 416, 6319, 111124, 2177346, 46440184, 1061938195, ...];
[1, 4, 22, 188, 2217, 32516, 555972, 10655628, 223313220, 5034249556, ...];
[4, 34, 314, 3619, 50829, 833591, 15417781, 313704516, 6900409869, ...];
[34, 416, 5071, 70714, 1131649, 20377616, 404581945, 8712077584, ...];
[416, 6319, 92167, 1472688, 26106282, 508663862, 10756749655, ...];
[6319, 111124, 1843974, 32709364, 630332565, 13142389012, 294045605744, ...];
[111124, 2177346, 39908146, 770233015, 15926231353, 352361228127, ...];
[46440184, 1061938195, 22576531219, 497127293708, 11532665230502, ...]; ...

Crossrefs

Cf. variants: A002293, A145347, A145348, A120974, A145345, A145350, A147664.

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^4);polcoeff(A,n)}
  • PARI
    /* This sequence is generated when k=3, m=1: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=3, 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)^3*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)^4 where G(x) = A(x*G(x)^3) and A(x) = G(x/A(x)^3).
a(n) = [x^(n-1)] 4*A(x)^(3*n+1)/(3*n+1) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 4*A(x)^(3*n+1)/(3*n+1) for n>=1 (see comment).

A120975 G.f. A(x) satisfies A(x) = 1 + x*A(x)^4 * A(x*A(x)^4)^4.

Original entry on oeis.org

1, 1, 8, 108, 1888, 38798, 894308, 22517256, 609112756, 17507219813, 530495478900, 16850219461706, 558608940038072, 19263089278722726, 689119527976265884, 25519081467271687938, 976447764170903902364
Offset: 0

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Crossrefs

Cf. A120974; variants: A110447, A120971, A120973, A120977.

Programs

  • PARI
    {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)^4))[ #G]); A=Vec(((Ser(G)-1)/x)^(1/4));A[n+1]}
  • PARI
    a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, binomial(4*n+k, j)/(4*n+k)*a(n-j, 4*j))); \\ Seiichi Manyama, Mar 01 2025

Formula

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)^4) and A(x/G(x)^4) = G(x), where G(x) is the g.f. of A120974 and satisfies G(x/G(x)^4) = 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(4*n+k,j)/(4*n+k) * a(n-j,4*j). (End)

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

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Crossrefs

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

Programs

  • 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)^5))[ #A]);A[n+1]}
  • PARI
    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

Formula

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)
Showing 1-5 of 5 results.

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