OFFSET
0,2
COMMENTS
The asymptotic expansion of the higher-order exponential integral E(x,m=5,n=3) ~ exp(-x)/x^5*(1 - 25/x + 445/x^2 - 7140/x^3 + 111769/x^4 - ...) leads to the sequence given above. See A163931 for E(x,m,n) information and A163932 for a Maple procedure for the asymptotic expansion. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 13 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) and Mitrinovic and Mitrinovic (1962) using slightly different notation.
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_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.
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.
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+4}^4(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
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.
FORMULA
a(n) = Sum_{k=0..n} (-1)^(n+k) * binomial(k+4, 4) * 3^k * Stirling1(n+4, k+4). - 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-4) = |f(n,4,3)| for n >= 4. - Milan Janjic, Dec 21 2008
From Petros Hadjicostas, Jun 14 2020: (Start)
a(n) = [x^4] Product_{r=0}^{n+3} (x + 3 + r) = (Product_{r=0}^{n+3} (r+3)) * Sum_{0 <= i < j < k < m <= n+3} 1/((3+i)*(3+j)*(3+k)*(3+m)).
E.g.f.: Sum_{n>=0} a(n)*x^(n+4)/(n+4)! = (log(1 - x))^4/(1 - x)^3/24.
Since a(n) = R_{n+4}^4(a=-3, b=-1), A001713(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+4}^4(a=-3,b=-1) = R_{n+3}^3(a=-3,b=-1) + (n+6)*R_{n+3}^4(a=-3,b=-1) implies the following:
(i) a(n) = A001713(n) + (n+6)*a(n-1) for n >= 1.
(ii) a(n) = A001712(n) + (2*n+11)*a(n-1) - (n+5)^2*a(n-2) for n >= 2.
(iii) a(n) = A001711(n) + 3*(n+5)*a(n-1) - (3*n^2+27*n+61)*a(n-2) + (n+4)^3*a(n-3) for n >= 3.
(iv) a(n) = (n+2)!/2 + 2*(2*n+9)*a(n-1) - (6*n^2+48*n+97)*a(n-2) + (2*n+7)*(2*n^2+14*n+25)*a(n-3) - (n+3)^4*a(n-4) for n >= 4.
(v) By taking the difference a(n) - (n+2)*a(n-1), and using (iv) above, we get a 5th-order linear recurrence with polynomial coefficients of degree at most 5. We omit the details. (End)
MATHEMATICA
nn = 24; t = Range[0, nn]! CoefficientList[Series[Log[1 - x]^4/(24*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 4] (* T. D. Noe, Aug 09 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
STATUS
approved