0,3

a(n) = sum of the next n odd numbers; i.e. group the odd numbers so that the n-th 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 n-th partial sum = {n(n+1)/2}^2. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 14 2002

Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 02 2004

n^3 is the sum of the first n centered hexagonal numbers (A003215). - Alonso Delarte (alonso.delarte(AT)gmail.com), Jul 29 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 anti-diamond numbers (vertex structure 7) (Cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers . - James A. Record (james.record(AT)gmail.com), Nov. 7, 2004.

Schlaefli symbol for this polyhedron: {4,3}

Least multiple of n such that every partial sum is a square. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), 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

a(n) = {least common multiple of n and (n-1)^2}-(n-1)^2. E.g.: {least common multiple of 1 and (1-1)^2}-(1-1)^2 = 0, {least common multiple of 2 and (2-1)^2}-(2-1)^2 = 1, {least common multiple of 3 and (3-1)^2}-(3-1)^2 = 8, ... - Mats Granvik (mgranvik(AT)abo.fi), Sep 24 2007

The solutions of the Diophantine equation: (X/Y)^2 - XY = 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 (bhmd95(AT)yahoo.fr), Oct 04 2007

Excepting for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). [From Kailasam Viswanathan Iyer (kvi(AT)nitt.edu), Mar 16 2009]

Number of units of a(n) belongs to a periodic sequence: 0, 1, 8, 7, 4, 5, 6, 3, 2, 9. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009]

a(n) = A007531(n) + A000567(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 18 2009]

Totally multiplicative sequence with a(p) = p^3 for prime p. [From Jaroslav Krizek (jaroslav.krizek(AT)atlas.cz), Nov 01 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (8).

D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000

Index entries for sequences related to linear recurrences with constant coefficients

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

H. Bottomley, Illustration of initial terms

H. Bottomley, Some Smarandache-type multiplicative sequences

Hyun Kwang Kim, On Regular Polytope Numbers

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Ronald Yannone, Hilbert Matrix Analyses

Multiplicative with a(p^e) = p^(3e). - David W. Wilson (davidwwilson(AT)comcast.net), Aug 01, 2001.

G.f.: x(1+4x+x^2)/(1-x)^4. - Michael Somos, May 06 2003

Dirichlet generating function: zeta(s-3). - Franklin T. Adams-Watters, Sep 11 2005. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 09 2005

E.g.f.: (x+3x^2+x^3)*e^x. - Franklin T. Adams-Watters, Sep 11 2005. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 09 2005

a(n)=sum(sum(n, j=1..n),k=1..n), n>=0. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 11 2007

a(n) = Sum(Sum(A002024(j,i): i<=j<n+i): 1<=i<=n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 24 2007

Starting (1, 8, 27, 64, 125,...), = binomial tansform of [1, 7, 12, 6, 0, 0, 0,...]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 21 2007

a(n) = C(n+2,3) + 4 C(n+1,3) + C(n,3)

This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=0. - Alexander R. Povolotsky (pevnev(AT)juno.com), May 17 2008

G.f.: sage: taylor( mul( x*(x^2+4*x+1)/(x-1)^4 for i in xrange(1,2)),x,0,40)# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 03 2009]

A000578 := n->n^3;

a:=n->sum(sum(n, j=1..n), k=1..n): seq(a(n), n=0..40); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 11 2007

with(combinat):a:=n->sum(sum(sum(binomial(5, 2)/10, j=0..n), k=0..n), m=0..n): seq(a(n), n=-1..39); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 30 2007

A000578:=(1+4*z+z**2)/(z-1)^4; [S. Plouffe in his 1992 dissertation if sequence starts at a(1).]

Table[n^3, {n, 0, 30}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Apr 01 2006

(PARI) A000578(n)=n^3 - M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 12 2008

isA000578(n)={n==round(sqrtn(n, 3))^3} - M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 12 2008

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

Cf. A065876.

a(n)= sum (A003215)

Cf. A030078(n)=A000578(A000040(n)): cubes of primes; sums of cubes: A003325, A024670 and references therein: A003072, ...

Cf. A101102, A101097, A101094, A024166, A000537.

Subsequence of A145784. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 19 2008]

Sequence in context: A069939 A118880 A048390 this_sequence A062292 A030295 A052045

Adjacent sequences: A000575 A000576 A000577 this_sequence A000579 A000580 A000581

nonn,core,easy,nice,mult,new

N. J. A. Sloane (njas(AT)research.att.com).

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Jun 20 2000

Removed attribute "conjectured" from Plouffe g.f R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 11 2009