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%I #25 Sep 08 2022 08:46:16
%S 0,729,970299,997002999,999700029999,999970000299999,
%T 999997000002999999,999999700000029999999,999999970000000299999999,
%U 999999997000000002999999999,999999999700000000029999999999,999999999970000000000299999999999,999999999997000000000002999999999999
%N a(n) = (10^n-1)^3.
%C The sum of the digits of a(n) is divisible by 18. For example, 9^3 = 729 and 7 + 2 + 9 = 18 * 1.
%C Number of 9 in a(n) is 2*n-1 for n > 0. - _Seiichi Manyama_, Sep 18 2018
%F a(n) = A002283(n)^3.
%F From _Ilya Gutkovskiy_, Apr 19 2016: (Start)
%F O.g.f.: 729*x*(1 + 220*x + 1000*x^2)/((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
%F E.g.f.: (-1 + 3*exp(9*x) - 3*exp(99*x) + exp(999*x))*exp(x). (End)
%e From _Seiichi Manyama_, Sep 18 2018: (Start)
%e n| a(n) can be divided into 3 parts for n > 1.
%e -+--------------------------------------------
%e 1| 72 9
%e 2| 9 702 99
%e 3| 99 7002 999
%e 4| 999 70002 9999
%e (End)
%p A272066:=n->(10^n-1)^3: seq(A272066(n), n=0..15); # _Wesley Ivan Hurt_, Apr 19 2016
%t (10^Range[0, 10] - 1)^3 (* _Wesley Ivan Hurt_, Apr 19 2016 *)
%o (Ruby)
%o (0..n).each{|i| p ('9' * i).to_i ** 3}
%o (PARI) a(n) = (10^n-1)^3; \\ _Michel Marcus_, Apr 19 2016
%o (Magma) [(10^n-1)^3 : n in [0..10]]; // _Wesley Ivan Hurt_, Apr 19 2016
%Y Cf. A002283, A059988, A272067, A272068, A319358.
%K nonn,easy
%O 0,2
%A _Seiichi Manyama_, Apr 19 2016
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