

A294662


Least k > a(n1) such that k^3 has no digit in common with a(n1) and a(n+1), a(0)=0.


1



0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 29, 55, 88, 90, 111, 200, 211, 400, 518, 654, 888, 889, 1111, 2825, 3131, 4244, 11111, 28222, 31535, 42449, 53355, 90000, 111181, 590000, 618181, 900000, 1111115, 9000000, 11111115, 60660090, 114144155
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OFFSET

0,3


COMMENTS

This is the sequence which corresponds to the original definition of A030290, before it was corrected to reproduce the data (and the intended meaning).


LINKS



EXAMPLE

a(3) cannot be 3 because 3^3 = 27 would have the digit '2' in common with a(2) = 2, therefore a(3) = 4, which does not violate this condition.
After a(9) = 10, none of the numbers { 11, ..., 19 } can follow, because they have the digit '1' in common with a(9)^3 = 1000. Numbers { 20, ..., 28 } are excluded because their cube would have a digit '0' or '1' in common with a(9). Therefore, a(10) = 29 which hasn't a digit in common with a(9)^3, nor has 29^3 = 24389 a digit in common with a(9).
a(38) = 11111115 with 11111115^3 = 1371743552812575445875 using all digits except for 0, 6 and 9. So a(39) = 60660090 is possible, with a(39)^3 = 223207688999086038729000 having all digits except for 1, 4 and 5.


PROG

(PARI) nxt(a, L=oo, D(a)=Set(digits(a)), S=D(a), T=D(a^3))={for(k=a+1, L, #setintersect(D(k), T)#setintersect(D(k^3), S)return(k))}


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



