A001713 Generalized Stirling numbers.
1, 18, 245, 3135, 40369, 537628, 7494416, 109911300, 1698920916, 27679825272, 474957547272, 8572072384512, 162478082312064, 3229079010579072, 67177961946534528, 1460629706845766400, 33139181950164806400, 783398920650352012800, 19268391564147377318400
Offset: 0
Keywords
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
- 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 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].
- 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.
Programs
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Mathematica
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 *)
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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
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PARI
b(n) = prod(r=0, n+2, r+3); 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))))); for(n=0, 18, print1(b(n)*c(n), ", ")) \\ Petros Hadjicostas, Jun 12 2020
Formula
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]
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
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
From Petros Hadjicostas, Jun 12 2020: (Start)
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)).
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:
(i) a(n) = A001712(n) + (n+5)*a(n-1) for n >= 1.
(ii) a(n) = A001711(n) + (2*n+9)*a(n-1) - (n+4)^2*a(n-2) for n >= 2.
(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.
(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)
Extensions
More terms from Vladeta Jovovic, May 05 2003
Comments