A072350 E.g.f. A(x) satisfies A(A(x)) = tan(x), where A(x) = Sum_{n>=1} a(n)*x^(2n-1)/(2n-1)!.
1, 1, 3, 17, 225, 3613, -42997, 8725357, 2116966081, -549193907111, -114757574954509, 117893333517545097, 14433599120070484321, -65568697910890921624715, 2968238619232726100394235, 86999609037195113208781248165
Offset: 1
Keywords
Examples
a(x) = x/1!+x^3/3!+3*x^5/5!+17*x^7/7!+225*x^9/9!+3613*x^11/11!-42997*x^13/13!+...
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..40
- Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation A^{2^n}(x)=F(x), arXiv:1302.1986 [math.CO], 2013.
Programs
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Mathematica
a[n_] := Module[{A, B, F}, F = Tan[x + O[x]^(2n+1)]; A = F; For[i = 0, i <= 2n-1, i++, B = InverseSeries[A, x]; A = (A + (B /. x -> F))/2]; If[n<1, 0, (2n-1)!*SeriesCoefficient[A, {x, 0, 2n-1}]]]; Table[a[n], {n, 1, 16}] (* Jean-François Alcover, Oct 29 2015, adapted from PARI *) -
PARI
{a(n)=local(A,B,F);F=tan(x+O(x^(2*n+1)));A=F; for(i=0,2*n-1,B=serreverse(A);A=(A+subst(B,x,F))/2); if(n<1,0,(2*n-1)!*polcoeff(A,2*n-1,x))} \\ Paul D. Hanna, Dec 09 2004
Formula
a(n)=(2*n-1)!*T(2*n-1,1), T(n,k)=if n=k then 1 else 1/2*(T059419(n,k)*k!/n!-sum(i=k+1..n-1, T(n,i)*T(i,k))). [Vladimir Kruchinin, Nov 11 2011]
Extensions
More terms from Paul D. Hanna, Dec 09 2004
Comments