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A084915
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a(n) = (n!)^2*n.
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3
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0, 1, 8, 108, 2304, 72000, 3110400, 177811200, 13005619200, 1185137049600, 131681894400000, 17526860144640000, 2753310393630720000, 504085244567224320000
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OFFSET
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0,3
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COMMENTS
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Used to prove that Sum_{n>=1} 1/A002378(n) = 1. Examining Sum_{n=1..k} 1/A002378(n) gives 1/2, 1/2 + 1/6, 1/2 + 1/6 + 1/12. Simplifying gives 1/2, 8/12, 108/144, where the numerators are this sequence and the denominators are A010790. Therefore we have k!^2*k/k!(k+1)! = k*k!/(k+1)! = k/(k+1), which tends to 1 as k tends to infinity.
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LINKS
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FORMULA
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a(n) = n!*(n+1)! - n!^2.
a(n) = det(PS(i+2,j+1), 1 <= i,j <= n-1), where PS(n,k) are Legendre-Stirling numbers of the second kind (A071951) and n > 0. [Mircea Merca, Apr 06 2013]
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EXAMPLE
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a(3) = 3!^2*3 = 36*3 = 108.
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PROG
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(PARI) for(n=1, 50, print1(n!^2*n", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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