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Sums of lists produced by a variant of the iteration that produces the Catalan numbers: start with 0 and at each iteration replace each integer k with the list 0,1,...,k-1,k,k+1,k,k-1,...,1,0 and let a(n) be the sum of the resulting (flattened) list after n iterations.
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%I #47 Sep 01 2023 14:13:19

%S 0,1,6,31,156,785,3978,20335,104856,545073,2854350,15046383,79787700,

%T 425360481,2278586898,12259138975,66216193968,358941938849,

%U 1952111592342,10648449309823,58245727453260,319406931168241,1755674399021466,9671384910586511

%N Sums of lists produced by a variant of the iteration that produces the Catalan numbers: start with 0 and at each iteration replace each integer k with the list 0,1,...,k-1,k,k+1,k,k-1,...,1,0 and let a(n) be the sum of the resulting (flattened) list after n iterations.

%C Number of diagonals emanating from a fixed vertex of a convex (n+3)-gon in all of its dissections. Example: a(1)=1 because in the three dissections of a convex quadrilateral ABCD (namely: empty, {AC}, {BD}) there is only one diagonal emanating from A.

%H Fung Lam, <a href="/A065096/b065096.txt">Table of n, a(n) for n = 0..1000</a>

%H S. B. Ekhad and M. Yang, <a href="http://sites.math.rutgers.edu/~zeilberg/tokhniot/oMathar1maple12.txt">Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences</a>, (2017).

%F G.f.: (1-3*z-sqrt(1-6*z+z^2))^2/(16*z^3).

%F a(n) = (1/Pi)*Integral_{x=3-2*sqrt(2)..3+2*sqrt(2)} x^n*sqrt(-x^2+6x-1)*(x-3)/8. - _Paul Barry_, Sep 16 2006

%F a(0) = 0 and, for n > 0, a(n) = Sum_{k=1..n} A001003(k)*A001003(n+1-k). - _Philippe Deléham_, Jan 27 2004

%F Conjecture: (n+3)*a(n) + 3*(-3*n-4)*a(n-1) + (19*n-9)*a(n-2) + 3*(-n+2)*a(n-3) = 0. - _R. J. Mathar_, Nov 24 2012

%F Recurrence: (n+3)*a(n) = -9*(n-3)*a(n-4) + 30*(2*n-3)*a(n-3) - 46*n*a(n-2) + 6*(2*n+3)*a(n-1). - _Fung Lam_, Jan 29 2014

%F a(n) ~ (3*sqrt(2)-4)^(3/2) * (3+2*sqrt(2))^(n+3) / (4 * sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Feb 13 2014

%F From _Peter Bala_, Aug 30 2023: (Start)

%F a(n) = Sum_{k = 0..n-1} 2^(k+1)/(n+1) * binomial(n+1, k)*binomial(n+1, k+2).

%F (n+3)*(n-1)*a(n) = 3*n*(2*n+1)*a(n-1) - n*(n-1)*a(n-2) with a(0) = 0 and a(1) = 1.

%F G.f. A(x) satisfies the algebraic equation 4*x^3*A(x)^2 - (5*x^2 - 6*x + 1)*A(x) + x = 0 and the differential equation

%F (3*x^4 - 19*x^3 + 9*x^2 - x)*dA/dx + (3*x^3 - 29*x^2 + 21*x - 3)*A(x) + 4*x = 0 with A(0) = 0. (End)

%p a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 else (3*n*(2*n+1)*a(n-1) - n*(n-1)*a(n-2))/((n+3)*(n-1)) end if; end:

%p seq(a(n), n = 0..20); # _Peter Bala_, Aug 30 2023

%t Table[Plus@@Flatten[Nest[ #/.a_Integer:> Join[Range[0, a+1], Range[a, 0, -1]]&, {0}, n]], {n, 0, 10}]

%t Table[Range[n, 0, -1].Table[a[n, k], {k, 0, n}], {n, 0, 36}] (* with a[n, k] as defined in A033877 *)

%Y Cf. A000108, A001003.

%K nonn,easy

%O 0,3

%A _Wouter Meeussen_, Nov 11 2001