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 1-2 of 2 results.

A182954 G.f. satisfies: A(x) = 1 + x*A(x) * A( x*A(x) )^4.

Original entry on oeis.org

1, 1, 5, 39, 381, 4284, 53163, 710810, 10085621, 150326044, 2336828792, 37687170215, 628069684439, 10782885724300, 190248852445782, 3442896376032300, 63804661588968521, 1209314277690837796
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2010

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 39*x^3 + 381*x^4 + 4284*x^5 + ...
Related expansions:
A(x*A(x)) = 1 + x + 6*x^2 + 54*x^3 + 592*x^4 + 7331*x^5 + 98870*x^6 + ...
A(x*A(x))^4 = 1 + 4*x + 30*x^2 + 292*x^3 + 3305*x^4 + 41420*x^5 + ...
The g.f. satisfies:
log(A(x)) = A(x)^4*x + {d/dx x*A(x)^8}*x^2/2! + {d^2/dx^2 x^2*A(x)^12}*x^3/3! + {d^3/dx^3 x^3*A(x)^16}*x^4/4! + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A = 1 + sum(i=1,n-1,a(i)*x^i+x*O(x^n)));
    for(i=1,n, A = exp( sum(m=1,n, sum(k=0,n-m, binomial(m+k-1,k)*polcoef(A^(4*m),k)*x^k) * x^m/m ) + x*O(x^n))); polcoef(A,n)}
    
  • PARI
    {a(n, m=1) = if(n==0, 1, if(m==0, 0^n, sum(k=0, n, m*binomial(n+m, k)/(n+m)*a(n-k, 4*k))))}

Formula

G.f. A(x) satisfies:
* A(x) = exp( Sum_{m>=0} {d^m/dx^m x^m*A(x)^(4m+4)} * x^(m+1)/(m+1)! );
* A(x) = exp( Sum_{m>=1} [Sum_{k>=0} C(m+k-1,k)*{[y^k] A(y)^(4m)}*x^k]*x^m/m);
which are equivalent.
Recurrence:
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n with a(0,m)=1, then
a(n,m) = Sum_{k=0..n} m*C(n+m,k)/(n+m) * a(n-k,4k).

A182953 G.f. satisfies: A(x) = 1 + x*A(x) * A( x*A(x) )^3.

Original entry on oeis.org

1, 1, 4, 25, 197, 1797, 18178, 198937, 2318858, 28487593, 366129764, 4896068759, 67843403960, 971032668429, 14319735032679, 217136949146091, 3379973833321141, 53936100582832901, 881318215466710693, 14731508761600217914
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2010

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 25*x^3 + 197*x^4 + 1797*x^5 +...
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 100*x^3 + 801*x^4 + 7296*x^5 + 73174*x^6 +...
A(x*A(x)) = 1 + x + 5*x^2 + 37*x^3 + 333*x^4 + 3389*x^5 + 37634*x^6 +...
A(x*A(x))^3 = 1 + 3*x + 18*x^2 + 142*x^3 + 1311*x^4 + 13461*x^5 +...
The g.f. satisfies:
log(A(x)) = A(x)^3*x + {d/dx x*A(x)^6}*x^2/2! + {d^2/dx^2 x^2*A(x)^9}*x^3/3! + {d^3/dx^3 x^3*A(x)^12}*x^4/4! +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+sum(i=1,n-1,a(i)*x^i+x*O(x^n)));
    for(i=1,n,A=exp(sum(m=1,n,sum(k=0,n-m,binomial(m+k-1,k)*polcoeff(A^(3*m),k)*x^k)*x^m/m)+x*O(x^n)));polcoeff(A,n)}
    
  • PARI
    {a(n, m=1)=if(n==0, 1, if(m==0, 0^n, sum(k=0, n, m*binomial(n+m, k)/(n+m)*a(n-k, 3*k))))}

Formula

G.f. A(x) satisfies:
* A(x) = exp( Sum_{m>=0} {d^m/dx^m x^m*A(x)^(3m+3)} * x^(m+1)/(m+1)! );
* A(x) = exp( Sum_{m>=1} [Sum_{k>=0} C(m+k-1,k)*{[y^k] A(y)^(3m)}*x^k]*x^m/m);
which are equivalent.
Recurrence:
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n with a(0,m)=1, then
a(n,m) = Sum_{k=0..n} m*C(n+m,k)/(n+m) * a(n-k,3k).
Given g.f. A(x), then G(x) = 1 + x*A(x)^3 satisfies G(x/G(x)) = 1 + x*G(x)^2 and G(x) is the g.f. of A147664.
Showing 1-2 of 2 results.