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.

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A294784 E.g.f. A(x) satisfies: A'(x) = (1 + A(x)^2)^2/4.

Original entry on oeis.org

1, 1, 2, 8, 46, 346, 3212, 35468, 453976, 6607936, 107781992, 1947158168, 38592660016, 832595731696, 19422479520992, 487137028505408, 13072025077208416, 373697069074031776, 11338183238037941312, 363881995144694554688, 12316073980019762824576, 438441199984650577010176, 16376568508223695174746752, 640396538780869661656846208, 26164698834332206196492375296, 1114866540340266230645081994496
Offset: 0

Views

Author

Paul D. Hanna, Nov 09 2017

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 2*x^2/2! + 8*x^3/3! + 46*x^4/4! + 346*x^5/5! + 3212*x^6/6! + 35468*x^7/7! + 453976*x^8/8! + 6607936*x^9/9! + 107781992*x^10/10! + 1947158168*x^11/11! + 38592660016*x^12/12! + 832595731696*x^13/13! + 19422479520992*x^14/14! + 487137028505408*x^15/15! +...
such that A'(x) = (1 + A(x)^2)^2/4.
RELATED SERIES.
Series_Reversion( Integral sqrt(1-x^2)/(1+x) dx ) = x + x^2/2! + 2*x^3/3! + 8*x^4/4! + 46*x^5/5! + 346*x^6/6! + 3212*x^7/7! + 35468*x^8/8! +...
which equals Integral A(x) dx.
(A(x)^2 - 1)/(A(x)^2 + 1) = x + x^2/2! + 2*x^3/3! + 8*x^4/4! + 46*x^5/5! + 346*x^6/6! + 3212*x^7/7! + 35468*x^8/8! +...
which equals Integral A(x) dx.
Note that asin( Integral A(x) dx ) = Series_Reversion(x + cos(x) - 1), the e.g.f. of A200317.
A(x)^2 = 1 + 2*x + 6*x^2/2! + 28*x^3/3! + 180*x^4/4! + 1472*x^5/5! + 14616*x^6/6! + 170728*x^7/7! + 2293320*x^8/8! + 34822592*x^9/9! + 589761216*x^10/10! +...
(A(x) - 1/A(x))/2 = x + x^2/2! + 5*x^3/3! + 26*x^4/4! + 196*x^5/5! + 1786*x^6/6! + 19550*x^7/7! + 248156*x^8/8! + 3588916*x^9/9! + 58220416*x^10/10! +...
(A(x) + 1/A(x))/2 = 1 + x^2/2! + 3*x^3/3! + 20*x^4/4! + 150*x^5/5! + 1426*x^6/6! + 15918*x^7/7! + 205820*x^8/8! + 3019020*x^9/9! + 49561576*x^10/10! +...
where (A(x) - 1/A(x))/2  =  (A(x) + 1/A(x))/2 * Integral A(x) dx.

Links

Crossrefs

Cf. A294785, A200317.

Programs

  • PARI
    {a(n) = my(A=1); A = deriv( sin( serreverse( x + cos(x +x^2*O(x^n)) - 1 ) ) ); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n) = my(A=1); A = deriv( serreverse( intformal( sqrt( (1-x)/(1+x +x*O(x^n)) ) ) ) ); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n) = my(A=1); for(i=1, n+1, A = 1 + intformal( (1 + A^2)^2/4 +x*O(x^n)) ); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

E.g.f. A(x) satisfies:
(1) A(x) = d/dx sin( Series_Reversion( x + cos(x) - 1 ) ).
(2) A(x) = d/dx Series_Reversion( asin(x) + sqrt(1-x^2) - 1 ).
(3) A(x) = d/dx Series_Reversion( Integral sqrt(1-x^2)/(1+x) dx ).
(4) A(x) = sqrt(1 - B(x)^2) / (1 - B(x)) where B(x) = Integral A(x) dx.
(5) Integral A(x) dx = (A(x)^2 - 1)/(A(x)^2 + 1).
(6) A(x) = (A(x) + 1/A(x))/2 * ( 1 + Integral A(x) dx ).
(7) exp( Integral (A(x) + 1/A(x))/2 dx ) = 1 + Integral A(x) dx.
(8) A(x) = 1 + Integral (1 + A(x)^2)^2/4 dx.
a(n) ~ c * (2/(Pi-2))^n * n^(n-1/6) / exp(n), where c = 1.2415... - Vaclav Kotesovec, Nov 11 2017

A200318 E.g.f. satisfies: A(x) = x-1 + cosh(A(x)).

Original entry on oeis.org

1, 1, 3, 16, 120, 1156, 13608, 189316, 3039060, 55291336, 1124309208, 25268818576, 622008616320, 16642670404816, 480923246983728, 14926731083999296, 495243684302520000, 17491488288340789696, 655224017429959987968, 25947019896579324410176, 1083050878686674070676800
Offset: 1

Views

Author

Paul D. Hanna, Nov 15 2011

Keywords

Comments

a(n) is the number of leaf labeled rooted trees with n leaves in which the outdegrees of the root and all internal nodes are positive even integers. - Geoffrey Critzer, Jul 31 2016

Examples

			E.g.f.: A(x) = x + x^2/2! + 3*x^3/3! + 16*x^4/4! + 120*x^5/5! +...
where A(1+x - cosh(x)) = x and A(x) = x-1 + cosh(A(x)).
The e.g.f. satisfies:
A(x) = x + (cosh(x)-1) + d/dx (cosh(x)-1)^2/2! + d^2/dx^2 (cosh(x)-1)^3/3! + d^3/dx^3 (cosh(x)-1)^4/4! +...
as well as the logarithmic series:
log(A(x)/x) = (cosh(x)-1)/x + d/dx (cosh(x)-1)^2/x/2! - d^2/dx^2 (cosh(x)-1)^3/x/3! + d^3/dx^3 (cosh(x)-1)^4/x/4! +...

Links

  • Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, page 128,(labeled hierarchies).

Crossrefs

Cf. A200317, A000311, A052526.

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[1 + x - Cosh[x],{x,0,20}],x],x] * Range[0,20]!] (* Vaclav Kotesovec, Jan 10 2014 *)
  • PARI
    {a(n)=n!*polcoeff(serreverse(1+x-cosh(x+x^2*O(x^n))),n)}
for(n=1, 21, print1(a(n), ", "))
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=x+x^2+x*O(x^n)); for(i=1, n, A=x+sum(m=1, n, Dx(m-1, (cosh(x+x*O(x^n))-1)^m)/m!)+x*O(x^n)); n!*polcoeff(A, n)}
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=x+x^2+x*O(x^n)); for(i=1, n, A=x*exp(sum(m=1, n, Dx(m-1, (cosh(x+x*O(x^n))-1)^m/x)/m!)+x*O(x^n))); n!*polcoeff(A, n)}

Formula

E.g.f. satisfies:
(1) A(x) = Series_Reversion(1+x - cosh(x)).
(2) A(x) = x + Sum_{n>=1} d^(n-1)/dx^(n-1) (cosh(x) - 1)^n / n!.
(3) A(x) = x*exp( Sum_{n>=1} d^(n-1)/dx^(n-1) (cosh(x) - 1)^n/x / n! ).
a(n) ~ n^(n-1) / (2^(1/4) * exp(n) * (1-sqrt(2)+log(1+sqrt(2)))^(n-1/2)). - Vaclav Kotesovec, Jan 10 2014
Showing 1-2 of 2 results.

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