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A067176
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A triangle of generalized Stirling numbers: sum of consecutive terms in the harmonic sequence multiplied by the product of their denominators.
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6
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0, 1, 0, 3, 1, 0, 11, 5, 1, 0, 50, 26, 7, 1, 0, 274, 154, 47, 9, 1, 0, 1764, 1044, 342, 74, 11, 1, 0, 13068, 8028, 2754, 638, 107, 13, 1, 0, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 0, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1, 0, 10628640
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OFFSET
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0,4
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COMMENTS
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In the Coupon Collector's Problem with n types of coupon, the expected number of coupons required until there are only k types of coupon uncollected is a(n,k)*k!/(n-1)!.
If n+k is even, then a(n,k) is divisible by (n+k+1). For n>=k and k>= 0, a(n,k) = (n-k)!*H(k+1,n-k), where H(m,n) is a generalized harmonic number, i.e., H(0,n) = 1/n and H(m,n) = Sum_{j=1..n} H(m-1,j). - Leroy Quet, Dec 01 2006
This triangle is the same as triangle A165674, which is generated by the asymptotic expansion of the higher order exponential integral E(x,m=2,n), minus the first right hand column. - Johannes W. Meijer, Oct 16 2009
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LINKS
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FORMULA
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a(n, k) = (n!/k!)*Sum_{j=k+1..n} 1/j = (A000254(n) - A000254(k)*A008279(n, n-k))/A000142(k) = a(n-1, k)*n + (n-1)!/k! = (a(n, k-1)-n!/k!)/k.
a(n, k) = Sum_{i=1..n-k} i*k^(i-1)*abs(stirling1(n-k, i)). - Vladeta Jovovic, Feb 02 2003
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EXAMPLE
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Rows start 0; 1,0; 3,1,0; 11,5,1,0; 50,26,7,1,0; 274,154,47,9,1,0 etc. a(5,2) = 3*4*5*(1/3 + 1/4 + 1/5) = 4*5 + 3*5 + 3*4 = 20 + 15 + 12 = 47.
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MATHEMATICA
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T[0, k_] := 1; T[n_, k_] := T[n, k] = Sum[ i*k^(i - 1)*Abs[StirlingS1[n - k, i]], {i, 1, n - k}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* G. C. Greubel, Jan 21 2017 *)
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CROSSREFS
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Columns are A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564, etc.
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KEYWORD
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AUTHOR
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STATUS
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approved
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