A116364 Row squared sums of Catalan triangle A033184.
1, 2, 9, 60, 490, 4534, 45689, 489920, 5508000, 64276492, 773029466, 9531003552, 119990158054, 1537695160070, 20009930706137, 263883333450760, 3521003563829212, 47470845904561648, 645960472314074400
Offset: 0
Keywords
Examples
The dot product of Catalan row 4 with itself equals
a(4) = [14,14,9,4,1]*[14,14,9,4,1] = 490
which is equivalent to obtaining the final term in these repeated partial sums of Catalan row 4:
14, 14, 9, 4, 1
28, 37, 41, 42
65, 106, 148
171, 319
490
Links
- G. C. Greubel, Table of n, a(n) for n = 0..830
Programs
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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
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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
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Maple
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
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Mathematica
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 *) -
PARI
a(n)=sum(k=0,n,((k+1)*binomial(2*n-k+1,n-k)/(2*n-k+1))^2)
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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
Formula
a(n) = Sum_{k=0..n} (C(2*n-k+1,n-k)*(k+1)/(2*n-k+1))^2.
Comments