A281070
a(n) is the numerator of 6 * Sum_{k=0..n} ((k+1)/(n-k+1)^2) * (Catalan(k)/(2^(2*k+1)))^2.
Original entry on oeis.org
3, 9, 109, 1037, 91027, 1540981, 447810157, 147053171, 503445581741, 16337573574319, 88973047698967, 3588920671411951, 2314594755016141847, 20685050199210758743, 2160689714871889935101, 121435710295138581181033, 16427863327419202412927713
Offset: 0
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a[n_]=6(Sum[(1/(n-k+1)^2)((CatalanNumber[k])/(2^(2k+1)))^2(k+1),{k,0, n}]); Numerator /@a/@ Range[0,10]
A282195
a(n) is the numerator of Sum_{m=0..n}(Sum_{k=0..m} ((k+1)/(m-k+1)^2) * (Catalan(k)/(2^(2*k)))^2)*(Sum_{k=0..n-m} ((k+1)/(n-m-k+1)^2) * (Catalan(k)/(2^(2*k)))^2).
Original entry on oeis.org
1, 3, 299, 1691, 4451729, 13446833, 16372396819, 208298035171, 1669160962863, 446401251163753, 6516008708737202119, 44233149340111747277, 5029067414956952883994601, 5810809342741928035310687, 46442062699559407155897191, 1018306138326248284055588777, 369103117042133718901423551221401
Offset: 0
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b[n_]=(Sum[((k+1)/(n-k+1)^2)((CatalanNumber[k])/(2^(2k)))^2, {k, 0, n}]); a[n_] = Sum[(b[k]*b[n - k]), {k, 0, n}]; Numerator /@a/@ Range[0, 10]
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C(n) = binomial(2*n,n)/(n+1);
b(n) = sum(k=0, n, ((k+1)/(n-k+1)^2) * (C(k)/(2^(2*k)))^2);
a(n) = numerator(sum(k=0, n, b(k)*b(n-k))); \\ Michel Marcus, Feb 11 2017
A282196
a(n) is the denominator of Sum_{m=0..n}(Sum_{k=0..m} ((k+1)/(m-k+1)^2) * (Catalan(k)/(2^(2*k)))^2)*(Sum_{k=0..n-m} ((k+1)/(n-m-k+1)^2) * (Catalan(k)/(2^(2*k)))^2).
Original entry on oeis.org
1, 4, 576, 4608, 16588800, 66355200, 104044953600, 1664719257600, 16441671680000, 5327101624320000, 92819418702151680000, 742555349617213440000, 98385613602882311946240000, 131180818137176415928320000, 1199367480111327231344640000, 29850923949437477757911040000, 12196892137874302548391671889920000
Offset: 0
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b[n_]=(Sum[((k+1)/(n-k+1)^2)((CatalanNumber[k])/(2^(2k)))^2, {k, 0, n}]); a[n_] = Sum[(b[k]*b[n - k]), {k, 0, n}]; Denominator /@a/@ Range[0, 10]
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C(n) = binomial(2*n,n)/(n+1);
b(n) = sum(k=0, n, ((k+1)/(n-k+1)^2) * (C(k)/(2^(2*k)))^2);
a(n) = denominator(sum(k=0, n, b(k)*b(n-k))); \\ Michel Marcus, Feb 11 2017
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