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A116363 a(n) = dot product of row n in Catalan triangle A033184 with row n in Pascal's triangle.

%I A116363 #24 Jun 22 2025 00:17:19
%S A116363 1,2,7,30,141,698,3571,18686,99385,535122,2908863,15932766,87809541,
%T A116363 486421770,2706138987,15110359038,84637982961,475381503266,
%U A116363 2676447372535,15100548901790,85357620588541,483304834607322
%N A116363 a(n) = dot product of row n in Catalan triangle A033184 with row n in Pascal's triangle.
%H A116363 G. C. Greubel, <a href="/A116363/b116363.txt">Table of n, a(n) for n = 0..1000</a>
%H A116363 D. Drake, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Drake/drake.html">Bijections from Weighted Dyck Paths to Schröder Paths</a>, J. Int. Seq. 13 (2010) # 10.9.2.
%F A116363 a(n) = Sum_{k=0..n} C(n,k)*C(2*n-k+1,n-k)*(k+1)/(2*n-k+1).
%F A116363 G.f. A(x) satisfies: d/dx[log(1 - 4*x*A(x))] = -4*(1-5*x)/(1-13*x+43*x^2-7*x^3).
%F A116363 O.g.f.: 2*(R+x)/(R*(R+x+1)), where R = sqrt(x^2+6*x+1). [_Dan Drake_, May 19 2010]
%F A116363 Conjecture: +(2*n+5)*(n+1)*a(n) +4*(-3*n^2-9*n+5)*a(n-1) +(2*n+7)*(n-1)*a(n-2)=0. - _R. J. Mathar_, Jun 22 2016
%e A116363 The dot product of Catalan row 4 and Pascal row 4 equals a(4) = [14,14,9,4,1]*[1,4,6,4,1] = 141
%e A116363 which is equivalent to obtaining the final term in these repeated partial sums of Pascal row 4:
%e A116363   1,4, 6, 4, 1
%e A116363    5,11,15,16
%e A116363     16,31,47
%e A116363      47,94
%e A116363       141
%t A116363 Table[Sum[Binomial[n, j]*Binomial[2*n-j+1, n-j]*(j+1)/(2*n-j+1), {j,0,n} ], {n,0,30}] (* _G. C. Greubel_, May 12 2019 *)
%o A116363 (PARI) a(n)=sum(k=0,n,binomial(n,k)*binomial(2*n-k+1,n-k)*(k+1)/(2*n-k+1))
%o A116363 for(n=0,30,print1(a(n),", "))
%o A116363 (Magma) [(&+[Binomial(n,j)*Binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1): j in [0..n]]): n in [0..30]]; // _G. C. Greubel_, May 12 2019
%o A116363 (Sage) [sum(binomial(n,j)*binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1) for j in (0..n)) for n in (0..30)] # _G. C. Greubel_, May 12 2019
%o A116363 (GAP) List([0..30], n-> Sum([0..n], j-> Binomial(n,j)*Binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1))); # _G. C. Greubel_, May 12 2019
%Y A116363 Cf. A033184.
%K A116363 nonn
%O A116363 0,2
%A A116363 _Paul D. Hanna_, Feb 04 2006