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%I #59 Feb 07 2024 09:43:14
%S 1,64,343,1000,2197,4096,6859,10648,15625,21952,29791,39304,50653,
%T 64000,79507,97336,117649,140608,166375,195112,226981,262144,300763,
%U 343000,389017,438976,493039,551368,614125,681472,753571,830584,912673,1000000,1092727,1191016
%N a(n) = (3*n + 1)^3.
%C The inverse binomial transform is 1, 63, 216, 162, 0, 0, 0 (0 continued). _R. J. Mathar_, May 07 2008
%C Perfect cubes with digital root 1 in base 10. Proof: perfect cubes are one of (3*s)^3, (3*s+1)^3 or (3*s+2)^3. Digital roots of (3*s)^3 are 0, digital roots of (3*s+1)^3 are 1, and digital roots of (3*s+2)^3 are 8, using trinomial expansion and the multiplicative property of digits roots. - _R. J. Mathar_, Jul 31 2010
%D S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.3.
%D Amarnath Murthy, Fabricating a perfect cube with a given valid digit sum (to be published)
%H Harry J. Smith, <a href="/A016779/b016779.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 Sum_{n>=0} 1/a(n) = 2*Pi^3 / (81*sqrt(3)) + 13*zeta(3)/27.
%F O.g.f.: (1+60*x+93*x^2+8*x^3)/(1-x)^4. - _R. J. Mathar_, May 07 2008
%F E.g.f.: (1 + 63*x + 108*x^2 + 27*x^3)*exp(x). - _Ilya Gutkovskiy_, Jun 16 2016
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Wesley Ivan Hurt_, Oct 02 2020
%F Sum_{n>=1) (-1)^n/a(n) = A226735. - _R. J. Mathar_, Feb 07 2024
%e a(2) = (3*2+1)^3 = 343.
%e a(6) = (3*6+1)^3 = 6859.
%t Table[(3n+1)^3,{n,0,100}] (* _Mohammad K. Azarian_, Jun 15 2016 *)
%t LinearRecurrence[{4,-6,4,-1},{1,64,343,1000},40] (* _Harvey P. Dale_, Oct 31 2016 *)
%o (PARI) { b=0; for (n=0, 1000, until (s==1, b++; s=b^3; s-=9*(s\9)); write("b016779.txt", n, " ", b^3) ) } \\ _Harry J. Smith_, Jul 18 2009
%o (Magma) [(3*n+1)^3: n in [0..30]]; // _Vincenzo Librandi_, May 09 2011
%o (PARI) a(n)=(3*n+1)^3 \\ _Charles R Greathouse IV_, Jan 02 2012
%Y Cf. A016791, A054966.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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