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A294662
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Least k > a(n-1) such that k^3 has no digit in common with a(n-1) and a(n+1), a(0)=0.
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1
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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
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0,3
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COMMENTS
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This is the sequence which corresponds to the original definition of A030290, before it was corrected to reproduce the data (and the intended meaning).
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LINKS
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EXAMPLE
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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.
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PROG
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(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))}
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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