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

1, 64, 343, 1000, 2197, 4096, 6859, 10648, 15625, 21952, 29791, 39304, 50653, 64000, 79507, 97336, 117649, 140608, 166375, 195112, 226981, 262144, 300763, 343000, 389017, 438976, 493039, 551368, 614125, 681472, 753571, 830584, 912673, 1000000, 1092727, 1191016

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The inverse binomial transform is 1, 63, 216, 162, 0, 0, 0 (0 continued). R. J. Mathar, May 07 2008

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

Examples

			a(2) = (3*2+1)^3 = 343.
a(6) = (3*6+1)^3 = 6859.
		

References

• S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.3.
• 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).

Crossrefs

Cf. A016791, A054966, A226735.

Programs

Magma

[(3*n+1)^3: n in [0..30]]; // Vincenzo Librandi, May 09 2011

Mathematica

Table[(3n+1)^3,{n,0,100}] (* Mohammad K. Azarian, Jun 15 2016 *)
LinearRecurrence[{4,-6,4,-1},{1,64,343,1000},40] (* Harvey P. Dale, Oct 31 2016 *)

PARI

a(n)=(3*n+1)^3 \\ Charles R Greathouse IV, Jan 02 2012

Formula

Sum_{n>=0} 1/a(n) = 2*Pi^3 / (81*sqrt(3)) + 13*zeta(3)/27.

O.g.f.: (1 + 60*x + 93*x^2 + 8*x^3)/(1 - x)^4. - R. J. Mathar, May 07 2008

E.g.f.: (1 + 63*x + 108*x^2 + 27*x^3)*exp(x). - Ilya Gutkovskiy, Jun 16 2016

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, Oct 02 2020

Sum_{n>=1} (-1)^n/a(n) = A226735. - R. J. Mathar, Feb 07 2024