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%I #24 Apr 04 2024 15:00:09
%S 1,2,5,15,49,168,594,2145,7865,29172,109174,411502,1560090,5943200,
%T 22732740,87253605,335897865,1296447900,5015206350,19439895090,
%U 75487384830,293595204240,1143532045500,4459774977450,17413705988874,68067249620328,266326619546204
%N Row sums of the Riordan matrix (1+x/sqrt(1-4*x),(1-sqrt(1-4*x))/2) (A189175).
%F a(n) = Sum_{k=0..n} binomial(2*n-k,n-k)*(n^2+n*k-k^2-k)/((2*n-k)*(2*n-k-1)), for n>=2.
%F G.f.: (1-5*x+4*x^2-(1-5*x)*sqrt(1-4x))/(2*x*(1-4*x))
%F a(n) = Sum_{k=1..n} (3-k)*binomial(2*n-k-1,n-1), n>0, a(0)=1. - _Vladimir Kruchinin_, Oct 18 2011
%F From _Gary W. Adamson_, Nov 14 2011: (Start)
%F a(n) is the sum of top row terms in M^n, M = an infinite square production matrix as follows, with the Fibonacci sequence as the left border:
%F 1, 1, 0, 0, 0, 0, ...
%F 1, 1, 1, 0, 0, 0, ...
%F 2, 1, 1, 1, 0, 0, ...
%F 3, 1, 1, 1, 1, 0, ...
%F 5, 1, 1, 1, 1, 1, ...
%F which means the top row of M^n is the n-th row in A189175. (End)
%F Conjecture: (n+1)*a(n) + 3*(1-3*n)*a(n-1) + 10*(2*n-3)*a(n-2) = 0. - _R. J. Mathar_, Nov 15 2011
%F a(n) = Sum_{k=0..n} (k+1) * A090181(n,k). - _Alois P. Heinz_, Apr 04 2024
%e a(3) = 15 since the top row of M^3 = (6, 5, 3, 1, 0, 0, 0, ...)
%t T[n_,k_] := If[n==k,1,Binomial[2n-k,n-k](n^2+n k-k^2-k)/((2n-k)(2n-k-1))]; Table[Sum[T[n,k], {k,0,n}], {n,0,22}]
%o (Maxima) T(n,k):=if n=k then 1 else binomial(2*n-k,n-k)*(n^2+n*k-k^2-k)/((2*n-k)*(2*n-k-1));
%o makelist(sum(T(n,k),k,0,n),n,0,22);
%Y Cf. A090181, A189175.
%K nonn,easy
%O 0,2
%A _Emanuele Munarini_, Apr 18 2011
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