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%I M4159 #190 Jul 09 2023 08:33:12
%S 0,0,0,6,24,60,120,210,336,504,720,990,1320,1716,2184,2730,3360,4080,
%T 4896,5814,6840,7980,9240,10626,12144,13800,15600,17550,19656,21924,
%U 24360,26970,29760,32736,35904,39270,42840,46620,50616,54834,59280,63960,68880
%N a(n) = n*(n-1)*(n-2) (or n!/(n-3)!).
%C _Ed Pegg Jr_ conjectures that n^3 - n = k! has a solution if and only if n is 2, 3, 5 or 9 (when k is 3, 4, 5 and 6).
%C Three-dimensional promic (or oblong) numbers, cf. A002378. - _Alexandre Wajnberg_, Dec 29 2005
%C Doubled first differences of tritriangular numbers A050534(n) = (1/8)n(n + 1)(n - 1)(n - 2). a(n) = 2*(A050534(n+1) - A050534(n)). - _Alexander Adamchuk_, Apr 11 2006
%C If Y is a 4-subset of an n-set X then, for n >= 6, a(n-4) is the number of (n-5)-subsets of X having exactly two elements in common with Y. - _Milan Janjic_, Dec 28 2007
%C Convolution of A005843 with A008585. - _Reinhard Zumkeller_, Mar 07 2009
%C a(n) = A000578(n) - A000567(n). - _Reinhard Zumkeller_, Sep 18 2009
%C For n > 3: a(n) = A173333(n, n-3). - _Reinhard Zumkeller_, Feb 19 2010
%C Let H be the n X n Hilbert matrix H(i, j) = 1/(i+j-1) for 1 <= i, j <= n. Let B be the inverse matrix of H. The sum of the elements in row 2 of B equals (-1)^n a(n+1). - _T. D. Noe_, May 01 2011
%C a(n) equals 2^(n-1) times the coefficient of log(3) in 2F1(n-2, n-2, n, -2). - _John M. Campbell_, Jul 16 2011
%C For n > 2 a(n) = 1/(Integral_{x = 0..Pi/2} (sin(x))^5*(cos(x))^(2*n-5)). - _Francesco Daddi_, Aug 02 2011
%C a(n) is the number of functions f:[3] -> [n] that are injective since there are n choices for f(1), (n-1) choices for f(2), and (n-2) choices for f(3). Also, a(n+1) is the number of functions f:[3] -> [n] that are width-2 restricted (that is, the pre-image under f of any element in [n] is of size 2 or less). See "Width-restricted finite functions" link below. - _Dennis P. Walsh_, Mar 01 2012
%C This sequence is produced by three consecutive triangular numbers t(n-1), t(n-2) and t(n-3) in the expression 2*t(n-1)*(t(n-2)-t(n-3)) for n = 0, 1, 2, ... - _J. M. Bergot_, May 14, 2012
%C For n > 2: A020639(a(n)) = 2; A006530(a(n)) = A093074(n-1). - _Reinhard Zumkeller_, Jul 04 2012
%C Number of contact points between equal spheres arranged in a tetrahedron with n - 1 spheres in each edge. - _Ignacio Larrosa Cañestro_, Jan 07 2013
%C Also for n >= 3, area of Pythagorean triangle in which one side differs from hypotenuse by two units. Consider any Pythagorean triple (2n, n^2-1, n^2+1) where n > 1. The area of such a Pythagorean triangle is n(n^2-1). For n = 2, 3, 4,.. the areas are 6, 24, 60, .... which are the given terms of the series. - _Jayanta Basu_, Apr 11 2013
%C Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices (chromatic polynomial) of the complete graph K_3. - _Tom Copeland_, Apr 05 2014
%C Starting with 6, 24, 60, 120, ..., a(n) is the number of permutations of length n>=3 avoiding the partially ordered pattern (POP) {1>2} of length 5. That is, the number of length n permutations having no subsequences of length 5 in which the first element is larger than the second element. - _Sergey Kitaev_, Dec 11 2020
%D R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
%D L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 40.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A007531/b007531.txt">Table of n, a(n) for n = 0..1000</a>
%H G. D. Birkhoff, <a href="https://www.jstor.org/stable/1967597">A determinant formula for the number of ways of coloring a map</a>, Ann. Math., 14:42-4. See 1st polynomial p. 5.
%H Alice L. L. Gao and Sergey Kitaev, <a href="https://arxiv.org/abs/1903.08946">On partially ordered patterns of length 4 and 5 in permutations</a>, arXiv:1903.08946 [math.CO], 2019.
%H Alice L. L. Gao and Sergey Kitaev, <a href="https://doi.org/10.37236/8605">On partially ordered patterns of length 4 and 5 in permutations</a>, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%H Milan Janjic and B. Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv:1301.4550 [math.CO], 2013.
%H Michael Penn, <a href="https://www.youtube.com/watch?v=FqaDCoXwQlc">A natural nested root</a>, YouTube video, 2022.
%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv:1406.3081 [math.CO], 2014-2015.
%H Michelle Rudolph-Lilith, <a href="http://arxiv.org/abs/1508.07894">On the Product Representation of Number Sequences, with Application to the Fibonacci Family</a>, arXiv:1508.07894 [math.NT], 2015.
%H Dennis Walsh, <a href="http://capone.mtsu.edu/dwalsh/WIDTH_2.pdf">Width-restricted finite functions</a>
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = 6*A000292(n-2).
%F a(n) = Sum_{i=1..n} polygorial(3,i) where polygorial(3,i) = A028896(i-1). - Daniel Dockery (peritus(AT)gmail.com), Jun 16 2003
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6, n > 2. - _Zak Seidov_, Feb 09 2006
%F G.f.: 6*x^2/(1-x)^4.
%F a(-n) = -a(n+2).
%F 1/6 + 3/24 + 5/60 + ... = Sum_{k>=1} (2*k-1)/(k*(k+1)*(k+2)) = 3/4. [Jolley Eq. 213]
%F a(n+1) = n^3 - n. - _Mohammad K. Azarian_, Jul 26 2007
%F E.g.f.: x^3*exp(x). - _Geoffrey Critzer_, Feb 08 2009
%F If the first 0 is eliminated, a(n) = floor(n^5/(n^2+1)). - _Gary Detlefs_, Feb 11 2010
%F 1/6 + 1/24 + 1/60 + ... = Sum_{n>=1} 1/(n*(n+1)*(n+2)) = 1/4. - _Mohammad K. Azarian_, Dec 29 2010
%F a(0) = 0, a(n) = a(n-1) + 3*(n-1)*(n-2). - _Jean-François Alcover_, Jan 08 2013
%F (a(n+1) - a(n))/6 = A000217(n-2) for n > 0. - _J. M. Bergot_, Jul 30 2013
%F Partial sums of A028896. - _R. J. Mathar_, Aug 28 2014
%F 1/6 + 1/24 + 1/60 + ... + 1/(n*(n+1)*(n+2)) = n*(n+3)/(4*(n+1)*(n+2)). - _Christina Steffan_, Jul 20 2015
%F a(n+2)^2 = A005563(n)^3 + A005563(n)^2. - _Bruno Berselli_, May 03 2018
%F a(n)*a(n+1) + A000096(n-3)^2 = m^2 (a perfect square), m = ((a(n)+a(n+1))/2)-n. - _Ezhilarasu Velayutham_, May 21 2019
%F Sum_{n>=3} (-1)^(n+1)/a(n) = 2*log(2) - 5/4. - _Amiram Eldar_, Jul 02 2020
%F For n >= 3, (a(n) + (a(n) + (a(n) + ...)^(1/3))^(1/3))^(1/3) = n - 1. - _Paolo Xausa_, Apr 09 2022
%p [seq(6*binomial(n,3),n=0..41)]; # _Zerinvary Lajos_, Nov 24 2006
%t Table[n^3 - 3n^2 + 2n, {n, 0, 42}]
%t Table[FactorialPower[n, 3], {n, 0, 42}] (* _Arkadiusz Wesolowski_, Oct 29 2012 *)
%o (PARI) a(n)=n*(n-1)*(n-2)
%o (Magma) [n*(n-1)*(n-2): n in [0..40]]; // _Vincenzo Librandi_, May 02 2011
%o (Haskell)
%o a007531 n = product [n-2..n] -- _Reinhard Zumkeller_, Jul 04 2012
%o (Sage) [n*(n-1)*(n-2) for n in range(40)] # _G. C. Greubel_, Feb 11 2019
%Y Cf. A002378, A005563, A084939, A084940, A084941, A084942, A084943, A084944.
%Y Cf. A028896.
%K nonn,easy
%O 0,4
%A _N. J. A. Sloane_, _Robert G. Wilson v_
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