

A000578


The cubes: a(n) = n^3.
(Formerly M4499 N1905)


955



0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000
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OFFSET

0,3


COMMENTS

a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the nth group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the nth partial sum = (n(n + 1)/2)^2.  Amarnath Murthy, Sep 14 2002
Total number of triangles resulting from crisscrossing cevians within a triangle so that two of its sides are each npartitioned.  Lekraj Beedassy, Jun 02 2004
Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal antidiamond numbers (vertex structure 7) (cf. A100188 = structured antidiamonds). Cf. A100145 for more on structured polyhedral numbers.  James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square.  Amarnath Murthy, Sep 09 2005
Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total.  Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2  X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k  XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k  1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1)  XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1.  Mohamed Bouhamida, Oct 04 2007
Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1).  K.V.Iyer, Mar 16 2009
Number of units of a(n) belongs to a periodic sequence: 0, 1, 8, 7, 4, 5, 6, 3, 2, 9.  Mohamed Bouhamida, Sep 04 2009
Totally multiplicative sequence with a(p) = p^3 for prime p.  Jaroslav Krizek, Nov 01 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012).  Daniel Forgues, May 14 2010
Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3  n is t = 2.  Artur Jasinski, Jun 30 2010
The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646.  R. J. Mathar, Mar 10 2011
The number of atoms in a bcc (bodycentered cubic) rhombic hexahedron with n shells is n^3 (T. P. Martin, Shells of atoms, eq.(8)).  Brigitte Stepanov, Jul 02 2011
Twice the area of a triangle with vertices at (0, 0), (t(n  1), t(n)), and (t(n), t(n  1)), where t = A000217 are triangular numbers.  J. M. Bergot, Jun 25 2013
For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)).  J. M. Bergot, Jun 14 2014
27, 64, 343, and 1331 are conjectured to be the only cubes not divisible by 10 with 2 distinct digits. See A155146 for cubes with 3 distinct digits and A155147 for cubes with 4 distinct digits.  Derek Orr, Sep 23 2014
Determinants of the spiral knots S(4,k,(1,1,1)). a(k) = det(S(4,k,(1,1,1))).  Ryan Stees, Dec 14 2014
One of the oldestknown examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform.  Charles R Greathouse IV, Jan 21 2015
We construct a number triangle from the integers 1, 2, 3, ... 2*n1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
1
2 2
3 3 3
4 4 4 4
5 5 5
6 6
7
(End)
All terms are == {0,1,8} (mod 9).  Zak Seidov, Jul 13 2015
The sequence is the third partial sum of (0, 1, 5, 6, 6, 6, ...).  Gary W. Adamson, Sep 27 2015
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3.  Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Ross, 2012].  N. J. A. Sloane, Feb 08 2017
Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice.  David Nacin, Feb 22 2017
Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n1).  Gregory L. Simay, Jul 27 2018
By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k.  Felix Fröhlich, Jul 27 2018


REFERENCES

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. AddisonWesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439  445. See http://www.mhikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find.  N. J. A. Sloane, Feb 18 2019]
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).
D. Wells, You Are A Mathematician, pp. 238241, Penguin Books 1995.


LINKS

N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010).
M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/1(mod m), Involve, Vol. 8 (2015), No. 3, 361384.
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199241, eq. (8).
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]


FORMULA

G.f.: x*(1+4*x+x^2)/(1x)^4.  Simon Plouffe in his 1992 dissertation
a(n) = lcm(n, (n  1)^2)  (n  1)^2. E.g.: lcm(1, (1  1)^2)  (1  1)^2 = 0, lcm(2, (2  1)^2)  (2  1)^2 = 1, lcm(3, (3  1)^2)  (3  1)^2 = 8, ...  Mats Granvik, Sep 24 2007
Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...].  Gary W. Adamson, Nov 21 2007
a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3).[Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37).  Wolfdieter Lang, Jul 17 2019]
a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3).  Ron Knott, Jun 10 2019
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k1)*k/6)/((k+3)!/6) at k=0.  Alexander R. Povolotsky, May 17 2008
a(n) = 3*a(n1)  3*a(n2) + a(n3) + 6.  Ant King Apr 29 2013
a(k) = det(S(4,k,(1,1,1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k1)  b(k2) = b(2)*b(k1)  b(k2).  Ryan Stees, Dec 14 2014
a(n) = 4*a(n1)  6*a(n2) + 4*a(n3)  a(n4).  Jon Tavasanis, Feb 21 2016
a(n) = n + Sum_{j=0..n1} Sum_{k=1..2} binomial(3,k)*j^(3k).  Patrick J. McNab, Mar 28 2016
a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3).  Tony Foster III, Nov 14 2017
Sum_{n>=1} 1/a(n) = zeta(3) (A002117).
Sum_{n>=1} (1)^(n+1)/a(n) = 3*zeta(3)/4 (A197070). (End)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1  1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)


EXAMPLE

For k=3, b(3) = 2 b(2)  b(1) = 41 = 3, so det(S(4,3,(1,1,1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27.  Patrick J. McNab, Mar 28 2016


MAPLE

isA000578 := proc(r)
local p;
if r = 0 or r =1 then
true;
else
for p in ifactors(r)[2] do
if op(2, p) mod 3 <> 0 then
return false;
end if;
end do:
true ;
end if;


MATHEMATICA

CoefficientList[Series[x (1 + 4 x + x^2)/(1  x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *)
Accumulate[Table[3n^2+3n+1, {n, 0, 20}]] (* or *) LinearRecurrence[{4, 6, 4, 1}, {1, 8, 27, 64}, 20](* Harvey P. Dale, Aug 18 2018 *)


PROG

(Haskell)
a000578 = (^ 3)
a000578_list = 0 : 1 : 8 : zipWith (+)
(map (+ 6) a000578_list)
(map (* 3) $ tail $ zipWith () (tail a000578_list) a000578_list)
(Magma) I:=[0, 1, 8, 27]; [n le 4 select I[n] else 4*Self(n1)6*Self(n2)+4*Self(n3)Self(n4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014
(Python)
A000578_list, m = [], [6, 6, 1, 0]
for _ in range(10**2):
for i in range(3):


CROSSREFS

(1/12)*t*(n^3n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.


KEYWORD

nonn,core,easy,nice,mult


AUTHOR



STATUS

approved



