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

%I #18 Feb 11 2019 01:42:42

%S 1,9,28,121,496,2100,9017,38969,169975,744984,3282005,14513236,

%T 64394500,286519305,1277975053,5712392313,25581765122,114754116351,

%U 515530099946,2319115721576,10445215621547,47096725844837,212569226371737,960306310551860,4341968468524371

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

%H G. C. Greubel, <a href="/A198059/b198059.txt">Table of n, a(n) for n = 1..1000</a>

%F Logarithmic derivative of A197601.

%F L.g.f.: Sum_{n>=1} [Sum_{k=0..2*n} C(2*n,k)^2 *x^k] *x^n/n.

%F L.g.f.: Sum_{n>=1} (1-x)^(4*n+1) *[Sum_{k>=0} C(2*n+k,k)^2 *x^k] *x^n/n.

%F G.f.: sqrt((1 + (15*x^3+2*x^2-x+3)/w - (x^3-2*x^2+x-4)/sqrt(w))/2) - 2 where w = (x^3-2*x^2-3*x-1)*(x^3-2*x^2+5*x-1). - _Mark van Hoeij_, May 06 2013

%e L.g.f.: L(x) = x + 9*x^2/2 + 28*x^3/3 + 121*x^4/4 + 496*x^5/5 + 2100*x^6/6 + ...

%e where

%e exp(L(x)) = 1 + x + 5*x^2 + 14*x^3 + 52*x^4 + 187*x^5 + 708*x^6 + ... + A197601(n)*x^n + ...

%e The l.g.f. equals the series:

%e L(x) = (1 + 2^2*x + x^2)*x

%e + (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)*x^2/2

%e + (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)*x^3/3

%e + (1 + 8^2*x + 28^2*x^2 + 56^2*x^3 + 70^2*x^4 + 56^2*x^5 + 28^2*x^6 + 8^2*x^7 + x^8)*x^4/4

%e + (1 + 10^2*x + 45^2*x^2 + 120^2*x^3 + 210^2*x^4 + 252^2*x^5 + 210^2*x^6 + 120^2*x^7 + 45^2*x^8 + 10^2*x^9 + x^10)*x^5/5 + ...

%e which involves the squares of the coefficients in even powers of (1+x).

%e Also,

%e L(x) = (1-x)^5*(1 + 3^2*x + 6^2*x^2 + 10^2*x^3 + 15^2*x^4 + ...)*x

%e + (1-x)^9*(1 + 5^2*x + 15^2*x^2 + 35^2*x^3 + 70^2*x^4 + ...)*x^2/2

%e + (1-x)^13*(1 + 7^2*x + 28^2*x^2 + 84^2*x^3 + 210^2*x^4 + ...)*x^3/3

%e + (1-x)^17*(1 + 9^2*x + 45^2*x^2 + 165^2*x^3 + 495^2*x^4 + ...)*x^4/4

%e + (1-x)^21*(1 + 11^2*x + 66^2*x^2 + 286^2*x^3 + 1001^2*x^4 + ...)*x^5/5 + ...

%e which involves the squares of the coefficients in odd powers of 1/(1-x).

%p w := (x^3-2*x^2-3*x-1)*(x^3-2*x^2+5*x-1);

%p sqrt((1 + (15*x^3+2*x^2-x+3)/w - (x^3-2*x^2+x-4)/sqrt(w))/2) - 2;

%p series(%,x=0,30); # _Mark van Hoeij_, May 06 2013

%t Table[Sum[Binomial[2k,n-k]^2 n/k,{k,n}],{n,30}] (* _Harvey P. Dale_, Oct 25 2011 *)

%o (PARI) {a(n)=n*sum(k=1,n,binomial(2*k,n-k)^2/k)}

%o (PARI) {a(n)=n*polcoeff(sum(m=1, n, sum(k=0, n, binomial(2*m, k)^2 *x^k)*x^m/m)+x*O(x^n), n)}

%o (PARI) {a(n)=n*polcoeff(sum(m=1, n, (1-x+x*O(x^n))^(4*m+1) *sum(k=0, n-m+1, binomial(2*m+k, k)^2 *x^k)*x^m/m+x*O(x^n)), n)}

%Y Cf. A197601 (exp).

%K nonn

%O 1,2

%A _Paul D. Hanna_, Oct 20 2011