

A007531


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


67



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).
Threedimensional promic (or oblong) numbers, cf. A002378.  Alexandre Wajnberg, Dec 29 2005
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
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
Convolution of A005843 with A008585.  Reinhard Zumkeller, Mar 07 2009
a(n) = A000578(n)  A000567(n).  Reinhard Zumkeller, Sep 18 2009
For n > 3: a(n) = A173333(n, n3).  Reinhard Zumkeller, Feb 19 2010
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
For n > 2: A020639(a(n)) = 2; A006530(a(n)) = A093074(n1).  Reinhard Zumkeller, Jul 04 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


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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
M. Janjic and B. Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv:1406.3081 [math.CO], 20142015.
Michelle RudolphLilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894, 2015
Dennis Walsh, Widthrestricted finite functions
Index entries for twoway infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

a(n) = 6*A000292(n2).
a(n) = Sum(Polygorial(3, i), i = 1..n) where Polygorial(3,i) = A028896(i1).  Daniel Dockery (peritus(AT)gmail.com), Jun 16 2003
a(n) = 3a(n1)  3a(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 + ... = 3/4. [Jolley Eq. 213]
a(n+1) = n^3  n.  Mohammad K. Azarian, Jul 26 2007
E.g.f.: x^3*exp(x).  Geoffrey Critzer, Feb 08 2009
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 + ... + 1/(n*(n+1)*(n+2))+... = 1/4.  Mohammad K. Azarian, Dec 29 2010
a(0) = 0, a(n) = a(n1) + 3*(n1)*(n2).  JeanFrançois Alcover, Jan 08 2013
(a(n+1)  a(n))/6 = A000217(n2) for n > 0.  J. M. Bergot, Jul 30 2013
Partial sums of A028896.  R. J. Mathar, Aug 28 2014
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


MAPLE

[seq(6*binomial(n, 3), n=0..41)]; # Zerinvary Lajos, Nov 24 2006


MATHEMATICA

Table[n^3  3n^2 + 2n, {n, 0, 42}]
Table[FactorialPower[n, 3], {n, 0, 42}] (* Arkadiusz Wesolowski, Oct 29 2012 *)


PROG

(PARI) a(n)=n*(n1)*(n2)
(MAGMA) [n*(n1)*(n2): n in [0..40]]; // Vincenzo Librandi, May 02 2011
(Haskell)
a007531 n = product [n2..n]  Reinhard Zumkeller, Jul 04 2012


CROSSREFS

Cf. A002378, A005563, A084939, A084940, A084941, A084942, A084943, A084944.
Cf. A007531, A028896.
Sequence in context: A086768 A160944 A160936 * A258345 A258351 A130669
Adjacent sequences: A007528 A007529 A007530 * A007532 A007533 A007534


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Robert G. Wilson v


STATUS

approved



