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A272066 a(n) = (10^n-1)^3.

%I A272066 #28 Mar 29 2025 23:57:26
%S A272066 0,729,970299,997002999,999700029999,999970000299999,
%T A272066 999997000002999999,999999700000029999999,999999970000000299999999,
%U A272066 999999997000000002999999999,999999999700000000029999999999,999999999970000000000299999999999,999999999997000000000002999999999999
%N A272066 a(n) = (10^n-1)^3.
%C A272066 The sum of the digits of a(n) is divisible by 18. For example, 9^3 = 729 and 7 + 2 + 9 = 18 * 1.
%C A272066 Number of 9 in a(n) is 2*n-1 for n > 0. - _Seiichi Manyama_, Sep 18 2018
%H A272066 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1111,-112110,1111000,-1000000).
%F A272066 a(n) = A002283(n)^3.
%F A272066 From _Ilya Gutkovskiy_, Apr 19 2016: (Start)
%F A272066 O.g.f.: 729*x*(1 + 220*x + 1000*x^2)/((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
%F A272066 E.g.f.: (-1 + 3*exp(9*x) - 3*exp(99*x) + exp(999*x))*exp(x). (End)
%e A272066 From _Seiichi Manyama_, Sep 18 2018: (Start)
%e A272066 n| a(n) can be divided into 3 parts for n > 1.
%e A272066 -+--------------------------------------------
%e A272066 1|        72    9
%e A272066 2|   9   702   99
%e A272066 3|  99  7002  999
%e A272066 4| 999 70002 9999
%e A272066 (End)
%p A272066 A272066:=n->(10^n-1)^3: seq(A272066(n), n=0..15); # _Wesley Ivan Hurt_, Apr 19 2016
%t A272066 (10^Range[0, 10] - 1)^3 (* _Wesley Ivan Hurt_, Apr 19 2016 *)
%o A272066 (Ruby)
%o A272066 (0..n).each{|i| p ('9' * i).to_i ** 3}
%o A272066 (PARI) a(n) = (10^n-1)^3; \\ _Michel Marcus_, Apr 19 2016
%o A272066 (Magma) [(10^n-1)^3 : n in [0..10]]; // _Wesley Ivan Hurt_, Apr 19 2016
%Y A272066 Cf. A002283, A059988, A272067, A272068, A319358.
%K A272066 nonn,easy
%O A272066 0,2
%A A272066 _Seiichi Manyama_, Apr 19 2016