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A003955
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a(n) = (2*n + 4) * (1*3*5*...*(2*n+1))^2.
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1
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4, 54, 1800, 110250, 10716300, 1512784350, 292183491600, 73958946311250, 23749039426612500, 9430743556307823750, 4537044990907363935000, 2600104866872495148416250, 1750070583471871734510937500, 1366930130733208386919792968750, 1226227455943070136959515612500000
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OFFSET
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0,1
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LINKS
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FORMULA
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Equals (n+2)!*(n+1)!*binomial(2*n+2, n+1)^2/2^(2*n+1). - G. C. Greubel, Sep 24 2019
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MAPLE
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seq((n+2)!*(n+1)!*binomial(2*n+2, n+1)^2/2^(2*n+1), n=0..20); # G. C. Greubel, Sep 24 2019
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MATHEMATICA
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Table[(n+2)!*(n+1)!*Binomial[2*n+2, n+1]^2/2^(2*n+1), {n, 0, 20}] (* G. C. Greubel, Sep 24 2019 *)
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PROG
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(PARI) vector(21, n, (n+1)!*n!*binomial(2*n, n)^2/2^(2*n-1) ) \\ G. C. Greubel, Sep 24 2019
(Magma) F:=Factorial; [F(n+2)*F(n+1)*Binomial(2*n+2, n+1)^2/2^(2*n+1): n in [0..20]]; // G. C. Greubel, Sep 24 2019
(Sage) f=factorial; [f(n+2)*f(n+1)*binomial(2*n+2, n+1)^2/2^(2*n+1) for n in (0..20)] # G. C. Greubel, Sep 24 2019
(GAP) F:=Factorial;; List([0..20], n-> F(n+2)*F(n+1)*Binomial(2*n+2, n+1)^2/2^(2*n+1) ); # G. C. Greubel, Sep 24 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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