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A189176
Row sums of the Riordan matrix (1+x/sqrt(1-4*x),(1-sqrt(1-4*x))/2) (A189175).
3
1, 2, 5, 15, 49, 168, 594, 2145, 7865, 29172, 109174, 411502, 1560090, 5943200, 22732740, 87253605, 335897865, 1296447900, 5015206350, 19439895090, 75487384830, 293595204240, 1143532045500, 4459774977450, 17413705988874, 68067249620328, 266326619546204, 1043003263740060
OFFSET
0,2
FORMULA
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.
G.f.: (1-5*x+4*x^2-(1-5*x)*sqrt(1-4x))/(2*x*(1-4*x))
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
From Gary W. Adamson, Nov 14 2011: (Start)
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:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
2, 1, 1, 1, 0, 0, ...
3, 1, 1, 1, 1, 0, ...
5, 1, 1, 1, 1, 1, ...
which means the top row of M^n is the n-th row in A189175. (End)
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
a(n) = Sum_{k=0..n} (k+1) * A090181(n,k). - Alois P. Heinz, Apr 04 2024
From Amiram Eldar, Sep 26 2025: (Start)
a(n) = (n+3)*A000108(n)/2 for n >= 1.
a(n) = A007851(n) + 1.
a(n) ~ n^(2*n-1) / sqrt(Pi*n).
Sum_{n>=0} 1/a(n) = 133 - 1304*Pi/(9*sqrt(3)) + 40*Pi^2/3.
Sum_{n>=0} (-1)^n/a(n) = 1679/15 - 5168*log(phi)/(5*sqrt(5)) + 480*log(phi)^2, where phi is the golden ratio (A001622). (End)
EXAMPLE
a(3) = 15 since the top row of M^3 = (6, 5, 3, 1, 0, 0, 0, ...)
MATHEMATICA
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}]
a[n_] := (n+3) * CatalanNumber[n] / 2; a[0] = 1; Array[a, 30, 0] (* Amiram Eldar, Sep 26 2025 *)
PROG
(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));
makelist(sum(T(n, k), k, 0, n), n, 0, 22);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Emanuele Munarini, Apr 18 2011
STATUS
approved