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A069322
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Square array read by antidiagonals of floor[(n+k)^(n+k)/(n^n*k^k)].
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1
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1, 1, 1, 1, 4, 1, 1, 6, 6, 1, 1, 9, 16, 9, 1, 1, 12, 28, 28, 12, 1, 1, 14, 45, 64, 45, 14, 1, 1, 17, 65, 119, 119, 65, 17, 1, 1, 20, 89, 198, 256, 198, 89, 20, 1, 1, 23, 117, 307, 484, 484, 307, 117, 23, 1, 1, 25, 149, 449, 837, 1024, 837, 449, 149, 25, 1, 1, 28, 184, 629
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OFFSET
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0,5
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COMMENTS
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T(n,k)*sqrt(3)/(n*k*Pi) provides a rough approximation for A067059.
a(n,k) is an analog of the binomial coefficients over transformations instead of permutations. - Chad Brewbaker, Nov 25 2013
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LINKS
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FORMULA
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EXAMPLE
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Rows start: 1,1,1,1,1,1,...; 1,4,6,9,12,14,...; 1,6,16,28,45,65,...; 1,9,28,64,119,198,...; etc. T(3,5)=floor[8^8/(3^3*5^5)]=floor[16777216 /84375]=floor[198.84...]=198.
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MATHEMATICA
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t[n_, 0] := 1; t[n_, n_] := 1; t[n_, k_] := Floor[(n^n)/((k^k)*((n - k)^(n - k)))]; Table[t[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 22 2018 *)
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PROG
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(Ruby)
def transitorial(n)
return n**n
end
def transnomial(n, k)
return transitorial(n)/(transitorial(k) *transitorial(n-k))
end
0.upto(15) do |i|
0.upto(i) do |j|
print transnomial(i, j).to_s + " "
end
puts ""
(PARI) for(n=0, 15, for(k=0, n, print1(if(k==0, 1, if(k==n, 1, floor((n^n)/(( k^k)*((n - k)^(n - k)))))), ", "))) \\ G. C. Greubel, Apr 22 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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