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Search: id:A067176
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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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4
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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
(list; table; graph; listen)
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OFFSET
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0,4
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COMMENT
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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, ie H(0,n) =1/n, and H(m,n) = sum{j=1 to n} H(m-1,j). - Leroy Quet (qq-quet(AT)mindspring.com), Dec 01 2006
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FORMULA
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a(n, k) =(n!/k!)*sum_{k<j<=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 (vladeta(AT)Eunet.yu), 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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CROSSREFS
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Columns are A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564, etc.
Sequence in context: A135871 A126178 A094753 this_sequence A137431 A131222 A114151
Adjacent sequences: A067173 A067174 A067175 this_sequence A067177 A067178 A067179
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KEYWORD
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nonn,tabl
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AUTHOR
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Henry Bottomley (se16(AT)btinternet.com), Jan 09 2002
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