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A199823 G.f. A(x) satisfies A(A(x)) = (x+2*x^2)/(1-2*x-4*x^2). 1
0, 1, 2, 2, 6, 2, 20, 4, -10, 334, -864, 236, 8356, 22252, -450052, -347224, 30131822, -54733650, -2300158256, 10315132412, 221235097892, -1741162058468, -25836213977848, 319542298035704, 3570412755518780, -66526876971287412, -567916947391285280 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..200

Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation A^{2^n}(x)=F(x), arXiv:1302.1986 [math.CO], 2013.

FORMULA

a(n) = T(n,1), T(n,k) = 1 if n=k, else T(n,k) = 1/2*((sum(i=k..n, C(i-1,k-1) * C(i,n-i))) * 2^(n-k) - sum(i=k+1..n-1, T(n,i)*T(i,k))).

MAPLE

a:= n-> T(n, 1):

T:= proc(n, k) option remember;

      `if`(n=k, 1, (add(binomial(i-1, k-1) *binomial(i, n-i), i=k..n)

                   *2^(n-k) -add(T(n, i)*T(i, k), i=k+1..n-1))/2)

    end:

seq(a(n), n=0..30); # Alois P. Heinz, Nov 11 2011

MATHEMATICA

a[n_] := T[n, 1]; T[n_, k_] := T[n, k] = If[n == k, 1, 1/2*(Sum[Binomial[i-1, k-1] * Binomial[i, n-i], {i, k, n}]*2^(n-k) - Sum[T[n, i]*T[i, k], {i, k+1, n-1}]) ]; Table[a[n], {n, 0, 30}] (* Jean-Fran├žois Alcover, Apr 21 2015, from formula *)

CROSSREFS

Sequence in context: A061057 A038667 A304104 * A013608 A324158 A196441

Adjacent sequences:  A199820 A199821 A199822 * A199824 A199825 A199826

KEYWORD

sign

AUTHOR

Vladimir Kruchinin, Nov 11 2011

STATUS

approved

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Last modified April 14 12:11 EDT 2021. Contains 342949 sequences. (Running on oeis4.)