

A007531


a(n) = n*(n1)*(n2) (or n!/(n3)!).
(Formerly M4159)


91



0, 0, 0, 6, 24, 60, 120, 210, 336, 504, 720, 990, 1320, 1716, 2184, 2730, 3360, 4080, 4896, 5814, 6840, 7980, 9240, 10626, 12144, 13800, 15600, 17550, 19656, 21924, 24360, 26970, 29760, 32736, 35904, 39270, 42840, 46620, 50616, 54834, 59280, 63960, 68880
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OFFSET

0,4


COMMENTS

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).
If Y is a 4subset of an nset X then, for n >= 6, a(n4) is the number of (n5)subsets of X having exactly two elements in common with Y.  Milan Janjic, Dec 28 2007
Let H be the n X n Hilbert matrix H(i, j) = 1/(i+j1) 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
a(n) equals 2^(n1) times the coefficient of log(3) in 2F1(n2, n2, n, 2).  John M. Campbell, Jul 16 2011
For n > 2 a(n) = 1/(Integral_{x = 0..Pi/2} (sin(x))^5*(cos(x))^(2*n5)).  Francesco Daddi, Aug 02 2011
a(n) is the number of functions f:[3] > [n] that are injective since there are n choices for f(1), (n1) choices for f(2), and (n2) choices for f(3). Also, a(n+1) is the number of functions f:[3] > [n] that are width2 restricted (that is, the preimage under f of any element in [n] is of size 2 or less). See "Widthrestricted finite functions" link below.  Dennis P. Walsh, Mar 01 2012
This sequence is produced by three consecutive triangular numbers t(n1), t(n2) and t(n3) in the expression 2*t(n1)*(t(n2)t(n3)) for n = 0, 1, 2, ...  J. M. Bergot, May 14, 2012
Number of contact points between equal spheres arranged in a tetrahedron with n  1 spheres in each edge.  Ignacio Larrosa Cañestro, Jan 07 2013
Also for n >= 3, area of Pythagorean triangle in which one side differs from hypotenuse by two units. Consider any Pythagorean triple (2n, n^21, n^2+1) where n > 1. The area of such a Pythagorean triangle is n(n^21). For n = 2, 3, 4,.. the areas are 6, 24, 60, .... which are the given terms of the series.  Jayanta Basu, Apr 11 2013
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
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


REFERENCES

R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 40.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA

a(n) = Sum_{i=1..n} polygorial(3,i) where polygorial(3,i) = A028896(i1).  Daniel Dockery (peritus(AT)gmail.com), Jun 16 2003
a(n) = 3*a(n1)  3*a(n2) + a(n3) + 6, n > 2.  Zak Seidov, Feb 09 2006
G.f.: 6*x^2/(1x)^4.
a(n) = a(n+2).
1/6 + 3/24 + 5/60 + ... = Sum_{k>=1} (2*k1)/(k*(k+1)*(k+2)) = 3/4. [Jolley Eq. 213]
If the first 0 is eliminated, a(n) = floor(n^5/(n^2+1)).  Gary Detlefs, Feb 11 2010
1/6 + 1/24 + 1/60 + ... = Sum_{n>=1} 1/(n*(n+1)*(n+2)) = 1/4.  Mohammad K. Azarian, Dec 29 2010
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
Sum_{n>=3} (1)^(n+1)/a(n) = 2*log(2)  5/4.  Amiram Eldar, Jul 02 2020
For n >= 3, (a(n) + (a(n) + (a(n) + ...)^(1/3))^(1/3))^(1/3) = n  1.  Paolo Xausa, Apr 09 2022


MAPLE



MATHEMATICA

Table[n^3  3n^2 + 2n, {n, 0, 42}]


PROG

(PARI) a(n)=n*(n1)*(n2)
(Haskell)
(Sage) [n*(n1)*(n2) for n in range(40)] # G. C. Greubel, Feb 11 2019


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



