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A030290 a(n) is the smallest k > a(n-1) such that k^3 has no digit in common with a(n-1)^3.

%I A030290 #17 Nov 13 2017 21:59:36
%S A030290 0,1,2,3,4,5,7,8,16,18,40,45,67,98,150,204,237,44216,46443,78742,
%T A030290 79930,130714,173000,185604,1000000,1304963,10000000,13049563,
%U A030290 100000000,130495593,1000000000,1304955895,10000000000,13049558812,100000000000,130495588186,1000000000000,1304955880707
%N A030290 a(n) is the smallest k > a(n-1) such that k^3 has no digit in common with a(n-1)^3.
%C A030290 From a(24) = 10^6 on, we have a(2k) = 10^(k-6) and a(2k+1) ~ c*a(2k) with c = (20/9)^(1/3) = 1.30495588... Indeed, a(2k)^3 = 1000^(k-6) has then only digits 0 and 1, and the next term must have a cube >= 2.2222...*1000^(k-6), so a(2k+1) will be the cube root of the next larger cube with no digit 0 and 1. - _M. F. Hasler_, Nov 12 2017
%H A030290 David W. Wilson, <a href="/A030290/b030290.txt">Table of n, a(n) for n = 0..90</a>
%F A030290 a(n) = A030289(n)^(1/3). - _David W. Wilson_, Nov 08 2017
%F A030290 For k >= 12, a(2k) = 10^(k-6), and a(2k+1) > c*a(2k) with approximate equality, where c = (20/9)^(1/3) = 1.30495588... - _M. F. Hasler_, Nov 12 2017
%e A030290 a(5) = 5 and 5^3 = 125 has no digit in common with the cube of a(4) = 4, 4^3 = 64.
%e A030290 But a(6) cannot be equal to 6, because 6^3 = 216 has digits '1' and '2' in common with 5^3 = 125.
%o A030290 (PARI) next_A030290(n, S=Set(digits(n^3)))={if(n>18e4,S[1]&&return(10^logint(n<<3,10));n\=sqrtn(.45,3));for(k=n+1,oo, #setintersect(Set(digits(k^3)), S)||return(k))} \\ _M. F. Hasler_, Nov 12 2017
%o A030290 print1(a=0); for(i=1, 99, print1(", "a=next_A030290(a))) \\ _M. F. Hasler_, Nov 08 2017
%Y A030290 Cf. A030289.
%K A030290 nonn,base
%O A030290 0,3
%A A030290 _Patrick De Geest_