A016791 a(n) = (3*n + 2)^3.

8, 125, 512, 1331, 2744, 4913, 8000, 12167, 17576, 24389, 32768, 42875, 54872, 68921, 85184, 103823, 125000, 148877, 175616, 205379, 238328, 274625, 314432, 357911, 405224, 456533, 512000, 571787, 636056, 704969, 778688, 857375, 941192

Offset

0,1

Comments

Also the perfect cubes with digital root 8. [Proof: perfect cubes are either of the form (3n)^3 or of the form (3n+1)^3 or of the form (3n+2)^3. These subsets have digital root 9, 1 and 8 respectively.] - R. J. Mathar, Oct 02 2008

References

Amarnath Murthy, Fabricating a perfect cube with a given valid digit sum (to be published)

Links

Harry J. Smith, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

a(n) = A016789(n)^3. - Nathaniel Johnston, May 04 2011

G.f.: (8+93*x+60*x^2+x^3)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 02 2012

a(0)=8, a(1)=125, a(2)=512, a(3)=1331, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Feb 20 2013

Sum_{n>=0} 1/a(n) = -2*Pi^3 / (81*sqrt(3)) + 13*zeta(3)/27. - Amiram Eldar, Oct 02 2020

Example

a(4) = (3*4 + 2)^3 = 2744.
a(8) = (3*8 + 2)^3 = 17576.

Mathematica

(3*Range[0, 40]+2)^3 (* or *) LinearRecurrence[{4, -6, 4, -1}, {8, 125, 512, 1331}, 40] (* Harvey P. Dale, Feb 20 2013 *)

Prog

(PARI) { for (n=0, 1000, write("b016791.txt", n, " ", (3*n + 2)^3) ) } \\ Harry J. Smith, Jul 18 2009
(PARI) { b=0; for (n=1, 1000, until (s==8, b++; s=b^3; s-=9*(s\9)); write("b016791.txt", n, " ", b^3) ) } \\ Harry J. Smith, Jul 18 2009

Crossrefs

Cf. A054966, A016779, A073636.

Sequence in context: A065082 A053058 A050803 * A061103 A264143 A110272

Adjacent sequences: A016788 A016789 A016790 * A016792 A016793 A016794

Keyword

nonn,easy

Author

N. J. A. Sloane

Extensions

More terms from Harry J. Smith, Jul 18 2009

First digital root in proof in comment line corrected. - Ant King, May 01 2013

Status

approved