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A005286
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a(n) = (n + 3)*(n^2 + 6*n + 2)/6.
(Formerly M4109)
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10
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1, 6, 15, 29, 49, 76, 111, 155, 209, 274, 351, 441, 545, 664, 799, 951, 1121, 1310, 1519, 1749, 2001, 2276, 2575, 2899, 3249, 3626, 4031, 4465, 4929, 5424, 5951, 6511, 7105, 7734, 8399, 9101, 9841, 10620, 11439, 12299, 13201, 14146, 15135, 16169, 17249
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OFFSET
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0,2
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COMMENTS
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Number of permutations of [n+3] with three inversions. - Michael Somos, Jun 25 2002
For n >= 2, a(n) is also the number of multiplications between two nonzero matrix elements involved in calculating the product of an (n+1) X (n+1) Hessenberg matrix and an (n+1) X (n+1) upper triangular matrix. The formula for n X n matrices is (n+2)(n^2+4n-3)/6 multiplications, n >= 3. - John M. Coffey, Jul 18 2016
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REFERENCES
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 255, #2, b(n,3).
R. K. Guy, personal communication.
E. Netto, Lehrbuch der Combinatorik. 2nd ed., Teubner, Leipzig, 1927, p. 96.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1999; see Exercise 1.30, p. 49.
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LINKS
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FORMULA
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G.f.: (1+2*x-3*x^2+x^3)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
(m^3-7*m)/6 for m >= 3 gives the same sequence. - N. J. A. Sloane, Jul 15 2011
a(0)=1, a(1)=6, a(2)=15, a(3)=29, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Mar 07 2012
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MATHEMATICA
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LinearRecurrence[{4, -6, 4, -1}, {1, 6, 15, 29}, 50] (* Harvey P. Dale, Mar 07 2012 *)
Table[Binomial[n, 3] + Binomial[n, 2] - n, {n, 3, 47}] (* or *)
CoefficientList[Series[(1 + 2 x - 3 x^2 + x^3)/(1 - x)^4, {x, 0, 44}], x] (* Michael De Vlieger, Jul 09 2016 *)
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PROG
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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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