A067176 A triangle of generalized Stirling numbers: sum of consecutive terms in the harmonic sequence multiplied by the product of their denominators.

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

Offset

0,4

Comments

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

Links

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Formula

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

Example

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.

Mathematica

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 *)

Crossrefs

Columns are A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564, etc.

Cf. A093905 and A165674. - Johannes W. Meijer, Oct 16 2009

Sequence in context: A191578 A288385 A245667 * A249480 A271704 A307419

Adjacent sequences: A067173 A067174 A067175 * A067177 A067178 A067179

Keyword

nonn,tabl

Author

Henry Bottomley, Jan 09 2002

Status

approved