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a(n) = Sum_{k=0..n} k*binomial(4*n+2,2*k)
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%I #14 Jun 14 2024 11:58:00

%S 0,15,465,11102,236997,4751010,91474890,1712391420,31398038701,

%T 566621243642,10097483539038,178113001428004,3115342162844450,

%U 54103694774702292,933929099838928692,16037182307150776056,274132978890654857853,4667160114821964359530,79177297937966956038102,1338972240005810710258452

%N a(n) = Sum_{k=0..n} k*binomial(4*n+2,2*k)

%H Seiichi Manyama, <a href="/A306609/b306609.txt">Table of n, a(n) for n = 0..828</a>

%H MathOverflow, <a href="https://mathoverflow.net/questions/324341/specific-partial-sum-of-even-odd-binomial-coefficients/324343">Specific partial sum of even/odd binomial coefficients</a>

%F a(n) = (n+1/2)*(16^n - binomial(4*n,2*n)) = (2*n+1)*A000346(2*n-1).

%F -512*(4*n + 1)*(86*n + 213)*(3 + 4*n)*(n + 1)*a(n) + 32*(2336*n^4 + 8800*n^3 + 10524*n^2 + 11540*n + 9703)*a(n + 1) - 2*(n + 2)*(544*n^3 - 1072*n^2 + 1138*n + 8055)*a(n + 2) - (2*n + 5)*(26*n - 31)*(n + 3)*(n + 2)*a(n + 3) = 0.

%F a(n) ~ 16^n * (n - sqrt(n/(2*Pi)) + 1/2).

%p f:= n -> (n+1/2)*(16^n-binomial(4*n,2*n)):

%p map(f, [$0..30]);

%t Table[Sum[k Binomial[4n+2,2k],{k,0,n}],{n,0,20}] (* _Harvey P. Dale_, Jun 14 2024 *)

%o (GAP) List([0..30],n->Sum([0..n],k->k*Binomial(4*n+2,2*k))); # _Muniru A Asiru_, Mar 01 2019

%Y Cf. A000346.

%K nonn

%O 0,2

%A _Robert Israel_, Feb 28 2019