0,4

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 Dec 01 2006

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Oct 16 2009: (Start)

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.

(End)

Leroy Quet, Home Page (listed in lieu of email address)

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.rs), Feb 02 2003

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.

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

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Oct 16 2009: (Start)

Cf. A093905 and A165674.

(End)

Sequence in context: A126178 A094753 A143398 this_sequence A137431 A131222 A114151

Adjacent sequences: A067173 A067174 A067175 this_sequence A067177 A067178 A067179

nonn,tabl

Henry Bottomley (se16(AT)btinternet.com), Jan 09 2002