A001713 - OEIS

%I M5060 N2190 #63 Jun 26 2020 13:58:50

%S 1,18,245,3135,40369,537628,7494416,109911300,1698920916,27679825272,

%T 474957547272,8572072384512,162478082312064,3229079010579072,

%U 67177961946534528,1460629706845766400,33139181950164806400,783398920650352012800,19268391564147377318400

%N Generalized Stirling numbers.

%C The asymptotic expansion of the higher order exponential integral E(x,m=4,n=3) ~ exp(-x)/x^4*(1 - 18/x + 245/x^2 - 3135/x^3 + 40369/x^4 - 537628/x^5 + ...) leads to the sequence given above. See A163931 and A163934 for more information. - _Johannes W. Meijer_, Oct 20 2009

%C From _Petros Hadjicostas_, Jun 12 2020: (Start)

%C 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) and Mitrinovic and Mitrinovic (1962) using slightly different notation.

%C 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.

%C 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_0^0(a,b) = 1, R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m.

%C 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) for n, m >= 0.

%C 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.

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

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

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

%H T. D. Noe, <a href="/A001713/b001713.txt">Table of n, a(n) for n = 0..100</a>

%H D. S. Mitrinovic, <a href="https://gallica.bnf.fr/ark:/12148/bpt6k762d/f996.image.r=1961%20mitrinovic">Sur une classe de nombres reliés aux nombres de Stirling</a>, Comptes rendus de l'Académie des sciences de Paris, t. 252 (1961), 2354-2356. [The numbers R_n^m(a,b) are introduced.]

%H D. S. Mitrinovic and R. S. Mitrinovic, <a href="https://www.jstor.org/stable/43667130">Tableaux d'une classe de nombres reliés aux nombres de Stirling</a>, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., No. 77 (1962), 1-77 [jstor stable version].

%H D. S. Mitrinovic and M. S. Mitrinovic, <a href="http://pefmath2.etf.rs/files/47/77.pdf">Tableaux d'une classe de nombres reliés aux nombres de Stirling</a>, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 (1962), 1-77.

%F E.g.f.: Sum_{n>=0} a(n)*x^(n+3)/(n+3)! = (log(1 - x)/(x - 1))^3/6. - _Vladeta Jovovic_, May 05 2003 [Edited by _Petros Hadjicostas_, Jun 13 2020]

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

%F 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-3) = |f(n,3,3)| for n >= 3. - _Milan Janjic_, Dec 21 2008

%F From _Petros Hadjicostas_, Jun 12 2020: (Start)

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

%F Since a(n) = R_{n+3}^3(a=-3, b=-1), A001712(n) = R_{n+2}^2(a=-3,b=-1), and A001711(n) = R_{n+1}^1(a=-3, b=-1), the equation R_{n+3}^3(a=-3,b=-1) = R_{n+2}^2(a=-3,b=-1) + (n+5)*R_{n+2}^3(a=-3,b=-1) implies the following:

%F (i) a(n) = A001712(n) + (n+5)*a(n-1) for n >= 1.

%F (ii) a(n) = A001711(n) + (2*n+9)*a(n-1) - (n+4)^2*a(n-2) for n >= 2.

%F (iii) a(n) = (n+2)!/2 + 3*(n+4)*a(n-1) - (3*n^2+21*n+37)*a(n-2) + (n+3)^3*a(n-3) for n >= 3.

%F (iv) a(n) = 2*(2*n+7)*a(n-1) - (6*n^2+36*n+55)*a(n-2) + (2*n^2+10*n+13)*(2*n+5)*a(n-3) - (n+2)^4*a(n-4) for n >= 4. (End)

%t nn = 23; t = Range[0, nn]! CoefficientList[Series[-Log[1 - x]^3/(6*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 3] (* _T. D. Noe_, Aug 09 2012 *)

%o (PARI) a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+3, 3)*3^k*stirling(n+3, k+3, 1)); \\ _Michel Marcus_, Jan 20 2016

%o (PARI) b(n) = prod(r=0, n+2, r+3);

%o c(n) = sum(i=0, n+2, sum(j=i+1, n+2, sum(k=j+1, n+2, 1/((3+i)*(3+j)*(3+k)))));

%o for(n=0, 18, print1(b(n)*c(n), ", ")) \\ _Petros Hadjicostas_, Jun 12 2020

%Y Cf. A000254, A001706, A001711, A001712, A001719, A048994.

%K nonn

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Vladeta Jovovic_, May 05 2003