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 A227053 a(n) = (n^3)! / (n^3-n)! = number of ways of placing n labeled balls into n^3 labeled boxes with at most one ball in each box. 0

%I #9 Jul 04 2013 03:14:04

%S 1,1,56,17550,15249024,28143753000,94689953674560,525169893772283760,

%T 4469844204191484518400,55337211594165488805417600,

%U 955860613004397508326213120000,22282564877342299983672172489536000,682182070950002359574696677978908672000,26812831292465310201469047550286967518976000

%N a(n) = (n^3)! / (n^3-n)! = number of ways of placing n labeled balls into n^3 labeled boxes with at most one ball in each box.

%o (Python)

%o import math

%o for n in range(20):

%o print str(math.factorial(n**3)/math.factorial(n**3-n))+',',

%Y Cf. A107444, A158362.

%K nonn,easy

%O 0,3

%A _Alex Ratushnyak_, Jun 29 2013

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Last modified August 13 01:27 EDT 2024. Contains 375113 sequences. (Running on oeis4.)