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%I #22 Apr 08 2016 03:29:38
%S 1,3,9,81,243,729,6561,19683,59049,1594323,4782969,14348907,129140163,
%T 387420489,1162261467,10460353203,31381059609,94143178827,
%U 2541865828329,7625597484987,22876792454961,205891132094649,617673396283947
%N Consider the power series (x+1)^(1/3)=1+x/3-x^2/9+5x^3/81+...; sequence gives denominators of coefficients.
%C All terms are powers of 3.
%F a(n) = 3^A004128(n).
%F a(n) = 3^n*a(floor(n/3)). - _Vladeta Jovovic_, Mar 01 2004
%F a(n) = denominator(binomial(1/3, n)). - _Peter Luschny_, Apr 07 2016
%p A067623 := n -> denom(binomial(1/3,n)):
%p seq(A067623(n), n=0..21); # _Peter Luschny_, Apr 07 2016
%t Table[Denominator@ Binomial[1/3, n], {n, 0, 22}] (* _Michael De Vlieger_, Apr 07 2016 *)
%Y Cf. A004128, A046161, A067622 (numerators), A123854.
%K nonn,frac
%O 0,2
%A _Benoit Cloitre_, Feb 02 2002
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