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A163176
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The n-th Minkowski number divided by the n-th factorial: a(n) = A053657(n)/n!.
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8
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1, 1, 4, 2, 48, 16, 576, 144, 3840, 768, 9216, 1536, 3870720, 552960, 442368, 55296, 26542080, 2949120, 2229534720, 222953472, 70071091200, 6370099200, 76441190400, 6370099200, 16694755983360, 1284211998720, 570760888320
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OFFSET
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1,3
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COMMENTS
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a(n) is an integer by Legendre's formula for the exponent of the highest power of a prime dividing n!.
See A053657 for additional comments, references, and links.
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LINKS
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FORMULA
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a(n) = (1/n!)*Product_{p prime} p^(Sum_{k>=0} ((n-1)/((p-1)p^k))).
a(n) = (1/n!)*denominator([y^(n-1)](y/(exp(y)-1))^x). - Peter Luschny, May 13 2019
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EXAMPLE
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MAPLE
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L := proc(n, p, r) local q, s; q := p-r; s := 0;
do if q > n then break fi; s := s+iquo(n, q);
q := q*p od; s end; mul(p^(L(n-1, p, 1)-L(n, p, 0)),
ser := series((y/(exp(y)-1))^x, y, 29):
c := n -> denom(coeff(ser, y, n-1)):
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MATHEMATICA
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a[n_] := (1/n!)*Product[ p^Sum[ Floor[ (n-1)/((p-1)*p^k) ], {k, 0, n}], {p, Select[ Range[2, n], PrimeQ]}]; Table[ a[n], {n, 1, 27}] (* Jean-François Alcover, Dec 07 2011 *)
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PROG
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(Julia)
using Primes
function L(n, p, r)
s, q = 0, p - r
while q <= n
s += div(n, q)
q *= p
end
s end
n < 2 && return 1
P = primes(n)
prod(p^(L(n-1, p, 1) - L(n, p, 0)) for p in P)
end
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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