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.

A276358 G.f. A(x) satisfies: A(x - x*A(x)) = x + x*A(x).

Original entry on oeis.org

1, 2, 8, 46, 324, 2608, 23136, 221370, 2252872, 24153284, 270922880, 3163154736, 38291322000, 479133266432, 6181998751808, 82084129578414, 1119798740473788, 15674024566862424, 224843628257016920, 3302256609111585300, 49613275311027132672, 761926428688868584400, 11952618573953745931536, 191418290850831848697272, 3127755564602007721663352, 52118116918762815035493760, 885205781290692080951844800, 15318116453244882343710519680, 269953482313408263924956600000
Offset: 1

Views

Author

Paul D. Hanna, Sep 04 2016

Keywords

Examples

			G.f.: A(x) = x + 2*x^2 + 8*x^3 + 46*x^4 + 324*x^5 + 2608*x^6 + 23136*x^7 + 221370*x^8 + 2252872*x^9 + 24153284*x^10 + 270922880*x^11 + 3163154736*x^12 +...
such that A(x - x*A(x)) = x + x*A(x).
RELATED SERIES.
A(x - x*A(x)) = x + x^2 + 2*x^3 + 8*x^4 + 46*x^5 + 324*x^6 + 2608*x^7 +...
which equals x + x*A(x).
Series_Reversion( x - x*A(x) ) = x + x^2 + 4*x^3 + 23*x^4 + 162*x^5 + 1304*x^6 + 11568*x^7 + 110685*x^8 + 1126436*x^9 + 12076642*x^10 + 135461440*x^11 + 1581577368*x^12 +...
which equals (x + A(x))/2.
A( (x + A(x))/2 ) = x + 3*x^2 + 16*x^3 + 111*x^4 + 898*x^5 + 8068*x^6 + 78400*x^7 + 810875*x^8 + 8832804*x^9 + 100592970*x^10 + 1191393144*x^11 + 14616198024*x^12 +...
which equals (A(x) - x)/(A(x) + x).
		

Crossrefs

Cf. A275765.

Programs

  • PARI
    {a(n) = my(A=x); for(i=1,n, A = 2*serreverse( x - x*A +x*O(x^n) ) - x ); polcoeff(A,n)}
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    {a(n) = my(A=x, B); for(i=1,n, B = (x + A)/2 +x*O(x^n); A = x*(1 + subst(A,x,B))/(1 - subst(A,x,B)) ); polcoeff(A,n)}
    for(n=1,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = 2 * Series_Reversion( x - x*A(x) ) - x.
(2) A(x) = x * (1 + A(B(x))) / (1 - A(B(x))), where B(x) = (x + A(x))/2.
(3) A( (x + A(x))/2 ) = (A(x) - x) / (A(x) + x).