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A000399
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Unsigned Stirling numbers of first kind s(n,3).
(Formerly M4218 N1762)
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30
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1, 6, 35, 225, 1624, 13132, 118124, 1172700, 12753576, 150917976, 1931559552, 26596717056, 392156797824, 6165817614720, 102992244837120, 1821602444624640, 34012249593822720, 668609730341153280, 13803759753640704000
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OFFSET
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3,2
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COMMENTS
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Number of permutations of n elements with exactly 3 cycles.
The asymptotic expansion of the higher order exponential integral E(x,m=3,n=1) ~ exp(-x)/x^3*(1 - 6/x + 35/x^2 - 225/x^3 + 1624/x^4 - 13132/x^5 + ...) leads to the sequence given above. See A163931 and A163932 for more information. - Johannes W. Meijer, Oct 20 2009
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 217.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
Shanzhen Gao, Permutations with Restricted Structure (in preparation). - Shanzhen Gao, Sep 14 2010
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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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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FORMULA
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Let P(n-1,X) = (X+1)(X+2)(X+3)...(X+n-1); then a(n) is the coefficient of X^2; or a(n) = P''(n-1,0)/2!. - Benoit Cloitre, May 09 2002 [Edited by Petros Hadjicostas, Jun 29 2020 to agree with the offset 3]
E.g.f.: -log(1-x)^3/3!.
a(n) is the coefficient of x^(n+3) in (-log(1-x))^3, multiplied by (n+3)!/6.
a(n) = ((Sum_{i=1..n-1} 1/i)^2 - Sum_{i=1..n-1} 1/i^2)*(n-1)!/2 for n >= 3. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 18 2000
a(n) = det(|S(i+3,j+2)|, 1 <= i,j <= n-3), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 06 2013
a(n) = Gamma(n)*(HarmonicNumber(n-1)^2 + Zeta(2,n) - Zeta(2))/2. - Gerry Martens, Jul 05 2015
a(n) = (n-3)! + (2*n-3)*a(n-1) - (n-2)^2*a(n-2) for n >= 5.
a(n) = 3*(n-2)*a(n-1) - (3*n^2-15*n+19)*a(n-2) + (n-3)^3*a(n-3) for n >= 6. (End)
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EXAMPLE
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(-log(1-x))^3 = x^3 + 3/2*x^4 + 7/4*x^5 + 15/8*x^6 + ...
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MAPLE
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MATHEMATICA
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a=Log[1/(1-x)]; Range[0, 20]! CoefficientList[Series[a^3/3!, {x, 0, 20}], x]
f[n_] := Abs@ StirlingS1[n, 3]; Array[f, 19, 3]
f[n_] := Gamma[n]*(HarmonicNumber[n - 1]^2 + Zeta[2, n] - Zeta[2])/2; Array[f, 19, 3] (* Robert G. Wilson v, Jul 05 2015 *)
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PROG
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(MuPAD) f := proc(n) option remember; begin n^3*f(n-3)-(3*n^2+3*n+1)*f(n-2)+3*(n+1)*f(n-1) end_proc: f(0) := 1: f(1) := 6: f(2) := 35:
(PARI) for(n=2, 50, print1(polcoeff(prod(i=1, n, x+i), 2, x), ", "))
(Sage) [stirling_number1(i+2, 3) for i in range(1, 22)] # Zerinvary Lajos, Jun 27 2008
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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