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%I #22 Jan 22 2017 21:53:56
%S 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,
%T 11,1,0,13068,8028,2754,638,107,13,1,0,109584,69264,24552,5944,1066,
%U 146,15,1,0,1026576,663696,241128,60216,11274,1650,191,17,1,0,10628640
%N A triangle of generalized Stirling numbers: sum of consecutive terms in the harmonic sequence multiplied by the product of their denominators.
%C 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)!.
%C 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
%C 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
%H G. C. Greubel, <a href="/A067176/b067176.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F 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.
%F a(n, k) = Sum_{i=1..n-k} i*k^(i-1)*abs(stirling1(n-k, i)). - _Vladeta Jovovic_, Feb 02 2003
%e 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.
%t 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 *)
%Y Columns are A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564, etc.
%Y Cf. A093905 and A165674. - _Johannes W. Meijer_, Oct 16 2009
%K nonn,tabl
%O 0,4
%A _Henry Bottomley_, Jan 09 2002
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