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 1-4 of 4 results.

A120971 G.f. A(x) satisfies A(x) = 1 + x*A(x)^2 * A( x*A(x)^2 )^2.

Original entry on oeis.org

1, 1, 4, 26, 218, 2151, 23854, 289555, 3783568, 52624689, 772928988, 11918181144, 192074926618, 3224153299106, 56213565222834, 1015694652332437, 18982833869517376, 366384235565593176, 7292660345274942402
Offset: 0

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 26*x^3 + 218*x^4 + 2151*x^5 + 23854*x^6 +...
From _Paul D. Hanna_, Apr 16 2007: G.f. A(x) is the unique solution to variable A in the infinite system of simultaneous equations:
A = 1 + x*B^2;
B = A*(1 + x*C^2);
C = B*(1 + x*D^2);
D = C*(1 + x*E^2);
E = D*(1 + x*F^2); ...
The above series begin:
B(x) = 1 + 2*x + 11*x^2 + 87*x^3 + 841*x^4 + 9288*x^5 + 113166*x^6 +...
C(x) = 1 + 3*x + 21*x^2 + 198*x^3 + 2204*x^4 + 27431*x^5 + 371102*x^6 +...
D(x) = 1 + 4*x + 34*x^2 + 374*x^3 + 4747*x^4 + 66350*x^5 + 996943*x^6 +...
E(x) = 1 + 5*x + 50*x^2 + 630*x^3 + 9015*x^4 + 140510*x^5 + 2334895*x^6 +...
F(x) = 1 + 6*x + 69*x^2 + 981*x^3 + 15658*x^4 + 270016*x^5 + 4933294*x^6 +...
		

Crossrefs

Programs

  • Mathematica
    m = 19; A[] = 0; Do[A[x] = 1 + x A[x]^2 A[x A[x]^2]^2 + O[x]^m, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Nov 07 2019 *)
  • 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)^2))[ #G]); A=Vec(((Ser(G)-1)/x)^(1/2));A[n+1]}
    
  • PARI
    a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, binomial(2*n+k, j)/(2*n+k)*a(n-j, 2*j))); \\ Seiichi Manyama, Mar 01 2025

Formula

G.f. A(x) satisfies:
(1) A(x) = G(G(x)-1),
(2) A(G(x)-1) = G(A(x)-1),
(3) A(x) = G(x*A(x)^2),
(4) A(x/G(x)^2) = G(x),
where G(x) is the g.f. of A120970 and satisfies G(x/G(x)^2) = 1 + x.
G.f. A(x) = F(x,1) where F(x,n) satisfies: F(x,n) = F(x,n-1)*(1 + x*F(x,n+1)^2) for n>0 with F(x,0)=1. - Paul D. Hanna, Apr 16 2007
Let B(x) = Sum_{n>=0} a(n)*x^(2*n+1), then B( x/(1+B(x)^2) ) = x. - Paul D. Hanna, Oct 30 2013
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(2*n+k,j)/(2*n+k) * a(n-j,2*j). (End)

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

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Crossrefs

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)^3))[ #G]); A=Vec(((Ser(G)-1)/x)^(1/3));A[n+1]}
    
  • PARI
    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

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)^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

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Crossrefs

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • 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)^4))[ #A]);A[n+1]}
    
  • PARI
    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

Formula

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)

A120977 G.f. A(x) satisfies A(x) = 1 + x*A(x)^5 * A(x*A(x)^5)^5.

Original entry on oeis.org

1, 1, 10, 170, 3745, 96960, 2814752, 89221360, 3037327145, 109825686370, 4185287088735, 167139924222426, 6964610755602495, 301800832258018835, 13564159649547824735, 630916661388096564620, 30316241123672291911875
Offset: 0

Views

Author

Paul D. Hanna, Jul 20 2006

Keywords

Crossrefs

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)^5))[ #G]); A=Vec(((Ser(G)-1)/x)^(1/5));A[n+1]}
    
  • PARI
    a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, binomial(5*n+k, j)/(5*n+k)*a(n-j, 5*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)^5) and A(x/G(x)^5) = G(x), where G(x) is the g.f. of A120976 and satisfies G(x/G(x)^5) = 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(5*n+k,j)/(5*n+k) * a(n-j,5*j). (End)
Showing 1-4 of 4 results.