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.

A122940 L.g.f.: A(x) satisfies: A(x+x^2) = 2*A(x) - log(1+x) with A(0)=0; thus A(x) = log(B(x)), where B(x) is g.f. of A122938.

Original entry on oeis.org

1, 1, 4, 17, 106, 796, 7176, 75057, 894100, 11946906, 176939192, 2876683340, 50931297912, 975391344376, 20090039762944, 442830738561585, 10400937450758286, 259318357362882148, 6839990934297006668
Offset: 1

Views

Author

Paul D. Hanna, Sep 25 2006

Keywords

Comments

a(n) = n * Sum_{k=0..n-1} (-1)^(n-k-1)*A122941(n-k,k)/(n-k).

Examples

			To illustrate A(x+x^2) = 2*A(x) - log(1+x):
A(x) = x + 1*x^2/2 + 4*x^3/3 + 17*x^4/4 + 106*x^5/5 + 796*x^6/6 +...
A(x+x^2) = x + 3*x^2/2 + 7*x^3/3 + 35*x^4/4 + 211*x^5/5 + 1593*x^6/6 +...
		

Crossrefs

Cf. A122938; related tables: A122941, A122888.

Programs

  • PARI
    {a(n)=local(A=x+x*O(x^n)); for(i=0,n,A=-A+subst(A,x,x+x^2)+log(1+x+x*O(x^n)));n*polcoeff(A,n)}

Formula

L.g.f.: A(x) = Sum_{n>=1} a(n)*x^n/n = Sum_{n>=0} log(1 + F_n(x))/2^(n+1) where F_0(x)=x, F_{n+1}(x) = F_n(x+x^2); a sum involving self-compositions of x+x^2 (cf. A122888).

A122939 G.f. A(x) satisfies: A(x+x^2) = A(x)^2/(1+x)^2.

Original entry on oeis.org

1, 2, 3, 6, 17, 70, 390, 2776, 24042, 244864, 2862185, 37715474, 552685976, 8910951840, 156709821779, 2984589501562, 61188398397436, 1343410717573876, 31445844702847347, 781689483100388326, 20564696601659697997
Offset: 0

Views

Author

Paul D. Hanna, Sep 21 2006

Keywords

Comments

Self-convolution of A122938. See A122888 for the table of self-compositions of x+x^2.

Examples

			G.f.: A(x) = (1 + x) * (1 + x+x^2)^(1/2) * (1 + x+2x^2+2x^3+x^4)^(1/4) * (1 + x+3x^2+6x^3+9x^4+10x^5+8x^6+4x^7+x^8)^(1/8) *...
		

Crossrefs

Cf. A122938 (square-root), A122888 (table).

Programs

  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(i=0,n,A=-A+2*(1+x)*sqrt(subst(A,x,x+x^2+x*O(x^n))));polcoeff(A,n)}

Formula

G.f.: A(x) = Product_{n>=0} (1 + F_n(x) )^(1/2^n) where F_0(x)=x, F_{n+1}(x)=F_n(x+x^2); a product that involves the n-th self-compositions of x+x^2.
Showing 1-2 of 2 results.