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A089664
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a(n) = S2(n,1), where S2(n, t) = Sum_{k=0..n} k^t *(Sum_{j=0..k} binomial(n,j))^2.
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10
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0, 4, 41, 306, 1966, 11540, 63726, 336700, 1720364, 8562024, 41718190, 199753004, 942561636, 4392660376, 20253510956, 92519626200, 419201709976, 1885719209936, 8428262686254, 37453751742604, 165575219275700, 728534225415864, 3191850894862564
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OFFSET
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0,2
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LINKS
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FORMULA
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a(n) = (1/8)*(n*(3*n+5)*4^n - 2*n*(n-1)*binomial(2*n, n)). (see Wang and Zhang, p. 338)
a(n) = (1/2)*( n*(3*n+5)*4^(n-1) - 3*binomial(n+1, 3)*Catalan(n) ).
G.f.: x*(4*(1-x) - 3*x*sqrt(1-4*x))/(1-4*x)^3.
E.g.f.: 2*x*(2 + 3*x)*exp(4*x) - (x^2/2)*(3*BesselI(0, 2*x) + 4*BesselI(1, 2*x) + BesselI(2, 2*x))*exp(2*x)). (End)
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MATHEMATICA
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Table[(n*(3*n+5)*4^n -2*n*(n-1)*Binomial[2*n, n])/8, {n, 0, 40}] (* G. C. Greubel, May 25 2022 *)
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PROG
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(SageMath) [(1/2)*(n*(3*n+5)*4^(n-1) -3*binomial(n+1, 3)*catalan_number(n)) for n in (0..40)] # G. C. Greubel, May 25 2022
(PARI) a(n)=n*(3*n+5)*2^(2*n-3) - 3*binomial(n+1, 3)*binomial(2*n, n)/(n+1)/2 \\ Charles R Greathouse IV, Oct 23 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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