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A181783
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Array described in comments to A053482, here read by increasing antidiagonals. See comments below.
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3
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1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 4, 1, 1, 1, 16, 21, 7, 1, 1, 1, 65, 142, 63, 11, 1, 1, 1, 326, 1201, 709, 151, 16, 1, 1, 1, 1957, 12336, 9709, 2521, 311, 22, 1, 1, 1, 13700, 149989, 157971, 50045, 7186, 575, 29, 1, 1, 1, 109601, 2113546, 2993467, 1158871, 193765, 17536, 981, 37, 1
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OFFSET
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0,9
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COMMENTS
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We denote by a(n,k) the number in row number n >= 0 and column number k >= 0. The recurrence which defines the array is a(n,k) = n*(k-1)*a(n-1,k) + a(n,k-1). The initial values are given by a(n,0) = 1 = a(0,k) for all n >= 0 and k >= 0.
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LINKS
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FORMULA
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If we consider the e.g.f. Psi(k) of column number k we have: Psi(k)(z) = Psi(k-1)(z)/(1-(k-1)*z) with Psi(1)(z) = exp(z). Then Psi(k)(z) = exp(z)/Product_{j=0..k-1} (1 - j*z). We conclude that a(n,k) = n!*Sum_{m=0..n} Sum_{j=1..k-1} (-1)^(k-1-j)*j^(m+k-2)/((n-m)!*(j-1)!*(k-1-j)!). It seems after the recurrence (and its proof) in A053482 that:
a(n,k) = -Sum_{j=1..k-1} s1(k,k-j)*n*(n-1)*...*(n-k+1)*a(n-j,k) + 1 where s1(m,n) are the classical Stirling numbers of the first kind.
a(n,1) = 1 for every n.
a(1,k) = 1 + k*(k-1)/2 for every k.
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EXAMPLE
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Array read row after row:
1, 1, 1, 1, 1, 1, 1, ...
1, 1, 2, 4, 7, 11, 16, ...
1, 1, 5, 21, 63, 151, 311, ...
1, 1, 16, 142, 709, 2521, 7186, ...
1, 1, 65, 1201, 9709, 50045, 193765, ...
1, 1, 326, 12336, 157971, 1158871, 6002996, ...
1, 1, 1957, 149989, 2993467, 30806371, 210896251, ...
...
a(4,3) = 1201.
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MAPLE
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option remember;
if n =0 or k = 0 then
1;
else
n*(k-1)*procname(n-1, k)+procname(n, k-1) ;
end if;
end proc:
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MATHEMATICA
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T[n_, k_] := T[n, k] = If[n == 0 || k == 0, 1, n (k - 1) T[n - 1, k] + T[n, k - 1]];
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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