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A001714
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Generalized Stirling numbers.
(Formerly M5184 N2252)
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3
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1, 25, 445, 7140, 111769, 1767087, 28699460, 483004280, 8460980836, 154594537812, 2948470152264, 58696064973000, 1219007251826064, 26390216795274288, 594982297852020288, 13955257961738192448, 340154857108405040256, 8606960634143667938688
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OFFSET
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0,2
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COMMENTS
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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
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)
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REFERENCES
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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).
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LINKS
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FORMULA
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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
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)
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
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STATUS
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approved
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