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a(n) = A001235(n) - floor(A001235(n)^(1/3))^3.
1

%I #13 Jun 10 2016 01:23:34

%S 1,8,8,1000,64,8,729,27,232,1728,64,216,1728,512,8000,4913,729,27,125,

%T 512,64,5832,13331,216,13580,125,4913,1000,1856,3375,13824,7073,343,

%U 2547,8,1331,12167,512,1728,8000,13824,13768,24389,9736,16496,216,12167,13824,19683,1

%N a(n) = A001235(n) - floor(A001235(n)^(1/3))^3.

%C Noncube terms of this sequence are 232, 13331, 13580, 1856, 7073, 2547, ...

%C How is the distribution of noncube terms in this sequence? See also A273592 that is related with this question.

%F a(n) = A055400(A001235(n)). - _Michel Marcus_, May 25 2016

%o (PARI) T = thueinit(x^3+1, 1);

%o isA001235(n) = my(v=thue(T, n)); sum(i=1, #v, v[i][1]>=0 && v[i][2]>=v[i][1])>1;

%o lista(nn) = for(n=1, nn, if(isA001235(n), print1(n-sqrtnint(n,3)^3, ", ")));

%Y Cf. A001235, A055400, A273592.

%K nonn

%O 1,2

%A _Altug Alkan_, May 25 2016