A001711 Generalized Stirling numbers.

1, 7, 47, 342, 2754, 24552, 241128, 2592720, 30334320, 383970240, 5231113920, 76349105280, 1188825724800, 19675048780800, 344937224217600, 6386713749964800, 124548748102195200, 2551797512248320000, 54804198761303040000, 1231237843834521600000

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=3) ~ exp(-x)/x^2*(1 - 7/x + 47/x^2 - 342/x^3 + 2754/x^4 - 24552/x^5 + 241128/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

For n > 4, a(n) mod n = 0 for n composite, = n-3 for n prime. - Gary Detlefs, Jul 18 2011

From Petros Hadjicostas, Jun 11 2020: (Start)

For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) using slightly different notation. They were further examined by Mitrinovic and Mitrinovic (1962).

These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.

As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m. (Because an empty product is by definition 1, we may let R_0^0(a,b) = 1.)

With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m). (Array A008275 is the same as array A048994 but with no zero row and no zero column.)

We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.

For the current sequence, a(n) = R_{n+1}^1(a=-3, b=-1) for n >= 0. (End)

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..440[Terms 0 to 100 computed by T. D. Noe; terms 101 to 440 computed by G. C. Greubel, Jan 15 2017]
• D. S. Mitrinovic, Sur une classe de nombres reliés aux nombres de Stirling, Comptes rendus de l'Académie des sciences de Paris, t. 252 (1961), 2354-2356. [The numbers R_n^m(a,b) are introduced.]
• D. S. Mitrinovic and M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 (1962), 1-77.
• D. S. Mitrinovic and R. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., No. 77 (1962), 1-77 [jstor stable version].
• Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., 1 (No. 3, 1926), 44-49. [Annotated scanned copy]
• J. Riordan, Letter of 04/11/74.

Crossrefs

Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k=2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

Maple

a := n-> add(1/2*((n+3)!/(k+3)), k=0..n): seq(a(n), n=0..19); # Zerinvary Lajos, Jan 22 2008
a := n -> (n+1)!*hs2(n+1): hs2 := n-> add(hs(k), k=0..n): hs := n-> add(h(k), k=0..n): h := n-> add(1/k, k=1..n): seq(a(n), n=0..19); # Gary Detlefs, Jan 01 2011

Mathematica

f[k_] := k + 2; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}]; (* Clark Kimberling, Dec 29 2011 *)
Table[(n + 3)!*Sum[1/(2*k + 4), {k, 1, n + 1}], {n,0,100}] (* G. C. Greubel, Jan 15 2017 *)

PARI

for(n=0, 19, print1((n+1)! * sum(k=0, n, binomial(k + 2, 2) / (n + 1 - k)),", ")) \\ Indranil Ghosh, Mar 13 2017

PARI

R(n,m,a,b) =  sum(k=0, n-m, (-1)^k*a^k*b^(n-m-k)*binomial(m+k,k)*stirling(n, m+k,1));
aa(n) = R(n+1,1,-3,-1);
for(n=0, 19, print1(aa(n), ",")) \\ Petros Hadjicostas, Jun 11 2020

Formula

E.g.f.: -log(1 - x)/(1 - x)^3 if offset 1. With offset 0: (d/dx)(-log(1 - x)/(1 - x)^3) = (1 - 3*log(1 - x))/(1 - x)^4.

a(n) = Sum_{k=0..n} ((-1)^(n+k)*(k+1)*3^k*Stirling1(n+1, k+1)). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-3,k)/(n-k)). - Milan Janjic, Dec 14 2008

a(n) = ( A000254(n+3) - 3*A001710(n+3) )/2. - Gary Detlefs, May 24 2010

a(n) = ((n+3)!/4) * (2*h(n+3) - 3), where h(n) = Sum_{k=1..n} (1/k) is the n-th harmonic number. - Gary Detlefs, Aug 15 2010

a(n) = n!*[2]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. With offset 1. - Gary Detlefs, Jan 04 2011

a(n) = (n+3)! * Sum_{k=1..n+1} (1/(2*k+4)). - Gary Detlefs, Sep 14 2011

a(n) = (n+1)! * Sum_{k=0..n} (binomial(k+2,2)/(n+1-k)). - Gary Detlefs, Dec 01 2011

a(n) = A001705(n+2) - A182541(n+4). - Anton Zakharov, Jul 02 2016

a(n) ~ n^(n+7/2) * exp(-n) * sqrt(Pi/2) * log(n) * (1 + (gamma - 3/2)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jul 12 2016

Conjectural D-finite with recurrence: a(n) + (-2*n-5)*a(n-1) + (n+2)^2*a(n-2)=0. - R. J. Mathar, Feb 16 2020

From Petros Hadjicostas, Jun 11 2020: (Start)

Since a(n) = R_{n+1}^1(a=-3, b=-1), it follows from Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) that:

a(n) = [x] Product_{r=0}^n (x + 3 + r) = (Product_{r=0}^n (3 + r)) * Sum_{s=0}^n 1/(3 + s).

a(n) = (n + 2)!/2 + (n + 3)*a(n-1) for n >= 1. [This can be used to prove R. J. Mathar's recurrence above.] (End)

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

Maple programs corrected and edited by Johannes W. Meijer, Nov 28 2012