A294662 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.
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
Offset: 0
Examples
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.
Comments