OFFSET
0,2
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
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).
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
EXAMPLE
a(2) = (3*2+1)^3 = 343.
a(6) = (3*6+1)^3 = 6859.
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 *)
PROG
(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
(Magma) [(3*n+1)^3: n in [0..30]]; // Vincenzo Librandi, May 09 2011
(PARI) a(n)=(3*n+1)^3 \\ Charles R Greathouse IV, Jan 02 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved