login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A001712
Generalized Stirling numbers.
(Formerly M4861 N2077)
7
1, 12, 119, 1175, 12154, 133938, 1580508, 19978308, 270074016, 3894932448, 59760168192, 972751628160, 16752851775360, 304473528961920, 5825460745532160, 117070467915075840, 2465958106403712000, 54336917746726272000, 1250216389189281024000
OFFSET
0,2
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
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]
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)
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
EXTENSIONS
More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
STATUS
approved