A001711 Generalized Stirling numbers. (Formerly M4429 N1873)

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

Offset

0,2

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.

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)

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 *)

Prog

(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

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.

Sequence in context: A098405 A104092 A024187 * A088057 A249477 A108434

Adjacent sequences: A001708 A001709 A001710 * A001712 A001713 A001714

Keyword

nonn

Author

N. J. A. Sloane

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

Status

approved