A165674 -id:A165674 - cp's OEIS frontend

A093344 a(n) = n! * Sum_{i=1..n} (1/i)*Sum_{j=0..i-1} 1/j!.

0, 1, 4, 17, 84, 485, 3236, 24609, 210572, 2004749, 21033900, 241237001, 3003349124, 40345599957, 581765196884, 8963453118065, 146969877361116, 2555361954692189, 46963373856864092, 909707559383702169, 18524816853636447380, 395634467245613474981

Offset: 0

Ralf Stephan, Apr 26 2004

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..400

Cf. A000254, A000774.

Equals for n=>1 the row sums of A165674 and A093905. - Johannes W. Meijer, Oct 16 2009

f:= gfun:-rectoproc({a(0) = 0, a(1) = 1, a(2) = 4, a(n) = 2*n*a(n-1) + (2-n^2)*a(n-2) + (n-2)^2*a(n-3)},a(n),remember):
seq(f(n),n=0..50); # Robert Israel, Oct 28 2015

Round@Table[E n! Sum[Gamma[k, 1]/k!, {k, 1, n}], {n, 0, 20}]
Round@Table[E ((HarmonicNumber[n] + ExpIntegralEi[-1] - EulerGamma) n! + HypergeometricPFQ[{n+1,n+1},{n+2,n+2},-1]/(n+1)^2), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)

a(n) = n!*sum(i=1,n,1/i*sum(j=0,i-1,1/j!))

E.g.f.: exp(1)*(Ei(1,1-x)-Ei(1,1))/(1-x). - Vladeta Jovovic, May 05 2007
a(n) = Integral_{x=1..oo} exp(1-x)*(x^n*log(x) - n!/x) dx. - Groux Roland, Mar 12 2011
From Vladimir Reshetnikov, Oct 28 2015: (Start)
a(n) = exp(1)*(H(n)*n! + (Ei(-1)-gamma)*n! + hypergeom([n+1,n+1],[n+2,n+2],-1)/(n+1)^2), where H(n)*n! = A000254(n), -Ei(-1) is A099285, gamma is A001620.
Recurrence: a(0) = 0, a(1) = 1, a(2) = 4, a(n) = 2*n*a(n-1) + (2-n^2)*a(n-2) + (n-2)^2*a(n-3).
(End)
a(n) = n!*e*Sum_{k=1..n} Gamma(k,1)/k!. - Robert Israel, Oct 28 2015

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

Original entry on oeis.org

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

Henry Bottomley, Jan 09 2002

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

			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.

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

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

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

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 *)

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

A093905 Triangle read by rows: for 0 <= k < n, a(n, k) is the sum of the products of all subsets of {n-k, n-k+1, ..., n} with k members.

Original entry on oeis.org

1, 1, 3, 1, 5, 11, 1, 7, 26, 50, 1, 9, 47, 154, 274, 1, 11, 74, 342, 1044, 1764, 1, 13, 107, 638, 2754, 8028, 13068, 1, 15, 146, 1066, 5944, 24552, 69264, 109584, 1, 17, 191, 1650, 11274, 60216, 241128, 663696, 1026576, 1, 19, 242, 2414, 19524, 127860

Offset: 1

Amarnath Murthy, Apr 24 2004

Triangle A165674, which is the reversal of this triangle, is generated by the asymptotic expansion of the higher order exponential integral E(x,m=2,n). - Johannes W. Meijer, Oct 16 2009

			Triangle begins:
1
1 3
1 5 11
1 7 26 50
1 9 47 154 274
...
a(5, 3) = 4*3*2+5*3*2+5*4*2+5*4*3 = 154.

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

The leading diagonal is given by A000254, Stirling numbers of first kind. The next nine diagonals are A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562 and A051564, generalized Stirling numbers.
Cf. A001705, A001711, A067176, A093344, A105954.
A165674 is the reversal of this triangle. - Johannes W. Meijer, Oct 16 2009

T[n_, 0] := 1; T[n_, k_]:= Product[i, {i, n - k, n}]*Sum[1/i, {i, n - k, n}]; Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] (* G. C. Greubel, Jan 21 2017 *)

a(n, k) = prod(i=n-k, n, i)*sum(i=n-k,n,1/i);
tabl(nn) = for (n=1, nn, for (k=0, n-1, print1(a(n,k), ", ")); print()); \\ Michel Marcus, Jan 21 2017

a(n, k) = (Product_{i=n-k..n} i)*(Sum_{i=n-k..n} 1/i), where a(n, 0) = 1.

a(n, k) = A067176(n, n-k-1) = A105954(k+1, n-k). Row sums are given by A093344.

Edited and extended by David Wasserman, Apr 24 2007

cp's OEIS Frontend

A093344 a(n) = n! * Sum_{i=1..n} (1/i)*Sum_{j=0..i-1} 1/j!.

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A067176 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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A093905 Triangle read by rows: for 0 <= k < n, a(n, k) is the sum of the products of all subsets of {n-k, n-k+1, ..., n} with k members.

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