A016791 - OEIS

%I #46 Dec 29 2024 21:01:42

%S 8,125,512,1331,2744,4913,8000,12167,17576,24389,32768,42875,54872,

%T 68921,85184,103823,125000,148877,175616,205379,238328,274625,314432,

%U 357911,405224,456533,512000,571787,636056,704969,778688,857375,941192,1030301,1124864,1225043

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

%C 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

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

%H Harry J. Smith, <a href="/A016791/b016791.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = A016789(n)^3. - _Nathaniel Johnston_, May 04 2011

%F 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

%F 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

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

%e a(4) = (3*4 + 2)^3 = 2744.

%e a(8) = (3*8 + 2)^3 = 17576.

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

%o (PARI) a(n) = { (3*n + 2)^3 } \\ _Harry J. Smith_, Jul 18 2009

%Y Cf. A054966, A016779, A016789, A073636.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_

%E More terms from _Harry J. Smith_, Jul 18 2009

%E First digital root in proof in comment line corrected. - _Ant King_, May 01 2013