

A016791


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


14



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
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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



FORMULA

G.f.: (8+93*x+60*x^2+x^3)/(14*x+6*x^24*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(n1)6*a(n2)+4*a(n3)a(n4).  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



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS

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


STATUS

approved



