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%I #14 Sep 08 2022 08:45:24
%S 1,2,9,60,490,4534,45689,489920,5508000,64276492,773029466,9531003552,
%T 119990158054,1537695160070,20009930706137,263883333450760,
%U 3521003563829212,47470845904561648,645960472314074400
%N Row squared sums of Catalan triangle A033184.
%C Number of 321-avoiding permutations in which the length of the longest increasing subsequence is n. Example: a(2)=9 because we have 12, 132, 312, 213, 231, 3142, 3412, 2143 and 2413. Column sums of triangle in A126217 (n >= 1). - _Emeric Deutsch_, Sep 07 2007
%H G. C. Greubel, <a href="/A116364/b116364.txt">Table of n, a(n) for n = 0..830</a>
%F a(n) = Sum_{k=0..n} (C(2*n-k+1,n-k)*(k+1)/(2*n-k+1))^2.
%e The dot product of Catalan row 4 with itself equals
%e a(4) = [14,14,9,4,1]*[14,14,9,4,1] = 490
%e which is equivalent to obtaining the final term in these repeated partial sums of Catalan row 4:
%e 14, 14, 9, 4, 1
%e 28, 37, 41, 42
%e 65, 106, 148
%e 171, 319
%e 490
%p a:=proc(k) options operator, arrow: sum((2*k-n+1)^2*binomial(n+1,k+1)^2/(n+1)^2,n=k..2*k) end proc: 1,seq(a(k),k=1..17); # _Emeric Deutsch_, Sep 07 2007
%t Table[Sum[(Binomial[2*n-j+1, n-j]*(j+1)/(2*n-j+1))^2, {j, 0, n}], {n, 0, 30}] (* _G. C. Greubel_, May 12 2019 *)
%o (PARI) a(n)=sum(k=0,n,((k+1)*binomial(2*n-k+1,n-k)/(2*n-k+1))^2)
%o (Magma) [(&+[(Binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1))^2: j in [0..n]]): n in [0..30]]; // _G. C. Greubel_, May 12 2019
%o (Sage) [sum(( binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1) )^2 for j in (0..n)) for n in (0..30)] # _G. C. Greubel_, May 12 2019
%o (GAP) List([0..30], n-> Sum([0..n], j-> (Binomial(2*n-j+1, n-j)* (j+1)/(2*n-j+1))^2 )) # _G. C. Greubel_, May 12 2019
%Y Cf. A033184, A116363.
%Y Cf. A126217.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Feb 04 2006
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