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a(n) is the smallest number not already in the sequence such that a(n)^3 begins with n.
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%I #10 Aug 01 2019 00:19:52

%S 1,3,7,16,8,4,9,2,21,10,48,5,11,52,25,55,12,57,27,59,6,61,62,29,63,64,

%T 14,66,31,67,68,32,15,70,33,154,72,73,34,74,161,35,76,164,77,36,78,

%U 169,17,37,80,174,81,38,82,178,83,18,39,182,85,184,86,40,87,188,189,19,191,89,193,90,194,42,91,197,92,199,43

%N a(n) is the smallest number not already in the sequence such that a(n)^3 begins with n.

%C Conjectured to be a permutation of the natural numbers.

%H Derek Orr, <a href="/A260465/b260465.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) >= n^(1/3) for all n > 0. If a(n) = n^(1/3), then n is a cube. Note the converse is false: a(27) = 14.

%o (PARI) v=[]; k=1; while(#v<100, d=digits(k^3); D=digits(#v+1); if(#D<=#d, c=1; for(i=1, #D, if(D[i]!=d[i], c=0; break)); if(c&&!vecsearch(vecsort(v), k), v=concat(v, k); k=0)); k++); v

%Y Cf. A018852.

%K nonn,base

%O 1,2

%A _Derek Orr_, Jul 26 2015