A001712 Generalized Stirling numbers.

1, 12, 119, 1175, 12154, 133938, 1580508, 19978308, 270074016, 3894932448, 59760168192, 972751628160, 16752851775360, 304473528961920, 5825460745532160, 117070467915075840, 2465958106403712000, 54336917746726272000, 1250216389189281024000

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=3,n=3) ~ exp(-x)/x^3*(1 - 12/x + 119/x^2 - 1175/x^3 + 12154/x^4 - 133938/x^5 + ...) leads to the sequence given above. See A163931 and A163932 for more information. - Johannes W. Meijer, Oct 20 2009

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+2}^2(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

• T. D. Noe, Table of n, a(n) for n = 0..100
• Matt Davis, Quadrant Marked Mesh Patterns and the r-Stirling Numbers, arXiv preprint arXiv:1412.0345 [math.CO], 2014.
• Matt Davis, Quadrant Marked Mesh Patterns and the r-Stirling Numbers, J. Int. Seq. 18 (2015), #15.10.1.
• Sergey Kitaev and Jeffrey Remmel, Simple marked mesh patterns, arXiv preprint arXiv:1201.1323 [math.CO], 2012.
• Sergey Kitaev and Jeffrey Remmel, Quadrant Marked Mesh Patterns, J. Int. Seq. 15 (2012), #12.4.7.
• 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]

Crossrefs

Cf. A001711, A008275, A048994, A163931, A163932.

Programs

Maple

A001712 := proc(n)
    add((-1)^(n+k)*binomial(k+2, 2)*3^k*Stirling1(n+2, k+2), k=0..n) ;
end proc:
seq(A001712(n), n=0..10) ; # R. J. Mathar, Jun 09 2018

Mathematica

nn = 22; t = Range[0, nn]! CoefficientList[Series[Log[1 - x]^2/(2*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 2] (* T. D. Noe, Aug 09 2012 *)

PARI

a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+2, 2)*3^k*stirling(n+2, k+2, 1)) \\ Michel Marcus, Jan 20 2016

PARI

b(n) = prod(r=0, n+1, r+3);
c(n) = sum(i=0, n+1, sum(j=i+1, n+1, 1/((3+i)*(3+j))));
for(n=0, 18, print1(b(n)*c(n),",")) \\ Petros Hadjicostas, Jun 11 2020

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(k+2, 2)*3^k*Stirling1(n+2, k+2). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

E.g.f.: (1 - 7*log(1 - x) + 6*log(1 - x)^2)/(1 - x)^5. - Vladeta Jovovic, Mar 01 2004

If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k, i)*Product_{j=0..k-1} (-a-j), then a(n-2) = |f(n,2,3)|, for n >= 2. [Milan Janjic, Dec 21 2008]

Conjecture: a(n) + 3*(-n-3)*a(n-1) + (3*n^2 + 15*n + 19)*a(n-2) - (n+2)^3*a(n-3)=0. - R. J. Mathar, Jun 09 2018

From Petros Hadjicostas, Jun 11 2020: (Start)

a(n) = [x^2] Product_{r=0}^{n+1} (x + 3 + r) = (Product_{r=0}^{n+1} (r+3)) * Sum_{0 <= i < j <= n+1} 1/((3+i)*(3+j)).

Since a(n) = R_{n+2}^2(a=-3, b=-1) and A001711(n) = R_{n+1}^1(a=-3,b=-1), the equation R_{n+2}^2(a=-3,b=-1) = R_{n+1}^1(a=-3,b=-1) + (n+4)*R_{n+1}^2(a=-3,b=-1) implies the following:

(i) a(n) = A001711(n) + (n+4)*a(n-1) for n >= 1.

(ii) a(n) = (n+2)!/2 + (2*n+7)*a(n-1) - (n+3)^2*a(n-2) for n >= 2.

(iii) R. J. Mathar's recurrence above. (End)

Extensions

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