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A220113 E.g.f. A(x)=sum{n>0, a(n)x^(2*n-1)/(2*n-1)!} satisfies A(A(x))=sin(2*x)/2. 0
1, -2, -12, -424, -29808, -2966816, -237449920, 76118167936, 84317834342656, 53499781544238592, 20080969948883956736, -10740526073453596649472, -31099457241702481710116864 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$, arXiv:1302.1986
FORMULA
a(n)=T(2*n-1,1), T(n,m)=1/2*(2^(n-2*m)*(((-1)^(n-m)+1)*sum(i=0..m/2, (2*i-m)^n*binomial(m,i)*(-1)^((n+m)/2-i)))/m!-sum(i=m+1..n-1, T(n,i)*T(i,m))), n>m, T(n,n)=1.
MATHEMATICA
t[n_, m_] := t[n, m] = 1/2*(2^(n - 2*m)*(((-1)^(n-m) + 1)* Sum[(2*i - m)^n*Binomial[m, i]*(-1)^((n+m)/2 - i), {i, 0, m/2}])/m! - Sum[t[n, i]*t[i, m], {i, m+1, n-1}]); t[n_, n_] = 1; Table[ t[2*n-1, 1], {n, 1, 13}] (* Jean-François Alcover, Feb 22 2013 *)
PROG
(Maxima)
T(n, m):=if n=m then 1 else 1/2*(2^(n-2*m)*(((-1)^(n-m)+1)*sum((2*i-m)^n*binomial(m, i)*(-1)^((n+m)/2-i), i, 0, m/2))/m!-sum(T(n, i)*T(i, m), i, m+1, n-1));
makelist(((T3(2*n-1, 1))), n, 1, 7);
CROSSREFS
Cf. A048602.
Sequence in context: A324616 A060942 A072446 * A015181 A012378 A012383
KEYWORD
sign
AUTHOR
Dmitry Kruchinin, Dec 05 2012
STATUS
approved

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Last modified March 29 09:59 EDT 2024. Contains 371268 sequences. (Running on oeis4.)