OFFSET
0,2
COMMENTS
The sum of the digits of a(n) is divisible by 18. For example, 9^3 = 729 and 7 + 2 + 9 = 18 * 1.
Number of 9 in a(n) is 2*n-1 for n > 0. - Seiichi Manyama, Sep 18 2018
FORMULA
a(n) = A002283(n)^3.
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 729*x*(1 + 220*x + 1000*x^2)/((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
E.g.f.: (-1 + 3*exp(9*x) - 3*exp(99*x) + exp(999*x))*exp(x). (End)
EXAMPLE
From Seiichi Manyama, Sep 18 2018: (Start)
n| a(n) can be divided into 3 parts for n > 1.
-+--------------------------------------------
1| 72 9
2| 9 702 99
3| 99 7002 999
4| 999 70002 9999
(End)
MAPLE
MATHEMATICA
(10^Range[0, 10] - 1)^3 (* Wesley Ivan Hurt, Apr 19 2016 *)
PROG
(Ruby)
(0..n).each{|i| p ('9' * i).to_i ** 3}
(PARI) a(n) = (10^n-1)^3; \\ Michel Marcus, Apr 19 2016
(Magma) [(10^n-1)^3 : n in [0..10]]; // Wesley Ivan Hurt, Apr 19 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Seiichi Manyama, Apr 19 2016
STATUS
approved