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A076000
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a(n) = Product_{k=1..n} k/floor(n/k).
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1
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1, 1, 2, 3, 12, 20, 120, 315, 1680, 6048, 60480, 138600, 1663200, 9266400, 69189120, 340540200, 5448643200, 22870848000, 411675264000, 2111894104320, 24135932620800, 230388447744000, 5068545850368000
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OFFSET
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1,3
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COMMENTS
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Sketch of proof that a(n) is an integer from Paul R. Pudaite, 9/28/2002: 1. n! = Product{p^([n/p]+[n/p^2]+...): prime p <= n}. 2. Product{[n/k]: k = 1...n} = Product{i^([n/i]-[n/i+1]): i=2...n}. 3. = Product{Product{Product{p^([n/i]-[n/i+1]): i such that p^k|i}: k such that p^k <= n}: prime p <= n}. 4. Reorganizing the exponents in the innermost product: ([n/p^k] - [n/(p^k+1)]) + ([n/(2 p^k)] - [n/(2 p^k + 1)] + ... = [n/p^k] - ([n/(p^k+1)] - [n/(2 p^k)]) - ... <= [n/p^k].
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LINKS
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FORMULA
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EXAMPLE
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a(6) = 6*5*4*3*2*1/([6/1]*[6/2]*[6/3]*[6/4]*[6/5]*[6/6]) = 6!/(6*3*2*1*1*1) = 20, where [x] denotes the greatest integer <= x.
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MATHEMATICA
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Table[Product[k/Floor[n/k], {k, n}], {n, 30}] (* Harvey P. Dale, Feb 27 2013 *)
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PROG
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(PARI) a(n) = prod(k=1, n, k/(n\k)); \\ Michel Marcus, Jun 24 2021
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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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