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 A016779 a(n) = (3n + 1)^3. 10

%I

%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) = (3n + 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^2 / (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

%e 343 = (3*2+1)^3, 6859 = (3*6+1)^3.

%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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Last modified April 18 12:42 EDT 2019. Contains 322209 sequences. (Running on oeis4.)