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%I #31 Apr 21 2024 16:17:58
%S 1,2,8,56,616,9856,216832,6288128,232660736,10702393856,599334055936,
%T 40155381747712,3172275158069248,291849314542370816,
%U 30936027341491306496,3743259308320448086016,512826525239901387784192,78975284886944813718765568,13583749000554507959627677696,2594496059105911020288886439936
%N a(0)=1, a(n)= Product_{k=1..n} k*(k+1)/2+1.
%C Partial products of A000124.
%H Harvey P. Dale, <a href="/A128814/b128814.txt">Table of n, a(n) for n = 0..269</a>
%H Spencer J. Franks, Pamela E. Harris, Kimberly Harry, Jan Kretschmann, and Megan Vance, <a href="https://arxiv.org/abs/2301.10830">Counting Parking Sequences and Parking Assortments Through Permutations</a>, arXiv:2301.10830 [math.CO], 2023.
%F a(n) grows roughly like n*(n!)^2/2^n. [Corrected by _Vaclav Kotesovec_, Mar 18 2023]
%F G.f.: G(0)/(2*x^2) - 1/x^2 - 1/x, where G(k)= 1 + 1/(1 - x*(k^2-k+2)/(x*(k^2-k+2) + 2/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jul 18 2013
%F a(n) = -2^(-n-1)*Gamma(n+3/2+sqrt(-7)/2)*Gamma(n+3/2-sqrt(-7)/2)*sin((3+sqrt(-7))*Pi/2)/Pi. - _Robert Israel_, May 19 2014
%F a(n) ~ cosh(sqrt(7)*Pi/2) * n^(2*(n+1)) / (2^n * exp(2*n)). - _Vaclav Kotesovec_, Mar 18 2023
%p a[0]:=1:for n from 1 to 20 do a[n]:=product(k*(k+1)/2+1,k=1..n) od: seq(a[n],n=0..20);
%t FoldList[Times,Accumulate[Range[0,20]]+1] (* _Harvey P. Dale_, Apr 21 2024 *)
%o (PARI) a(n) = if (n, prod(k=1, n, k*(k+1)/2+1), 1); \\ _Michel Marcus_, Mar 18 2023
%Y Cf. A000124.
%K easy,nonn,changed
%O 0,2
%A _Miklos Kristof_, Apr 10 2007
%E More terms from _Michel Marcus_, Mar 18 2023
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