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a(0)=2; a(n) is the smallest k > a(n-1) such that the fractional part of k^(1/11) starts with n.
2

%I #21 Dec 09 2017 04:01:51

%S 2,3,8,18,41,87,176,343,643,1165,3503,3691,3888,4095,4311,4538,4776,

%T 5025,5285,5558,5844,6143,6455,6782,7125,7482,7856,8247,8656,9083,

%U 9529,9994,10481,10988,11518,12072,12649,13251,13879,14534,15217,15929,16672

%N a(0)=2; a(n) is the smallest k > a(n-1) such that the fractional part of k^(1/11) starts with n.

%F For n > 0, a(n) = ceiling((d + n/10^d)^11) where d = 1 + floor(log_10(n)). - _Jon E. Schoenfield_, Nov 28 2017

%e a(4) = 41 -> 41^(1/11) = 1.{4}01576...;

%e a(5) = 87 -> 87^(1/11) = 1.{5}0079001... and a(4)=41 < a(5)=87.

%e From _Jon E. Schoenfield_, Nov 28 2017: (Start)

%e n a(n) a(n)^(1/11)

%e -- ------ ---------------

%e 0 2 1.{0}6504108...

%e 1 3 1.{1}1612317...

%e 2 8 1.{2}1481404...

%e 3 18 1.{3}0051594...

%e 4 41 1.{4}0157620...

%e 5 87 1.{5}0079001...

%e 6 176 1.{6}0006459...

%e 7 343 1.{7}0012668...

%e 8 643 1.{8}0008041...

%e 9 1165 1.{9}0001444...

%e 10 3503 2.{10}001226...

%e 11 3691 2.{11}001635...

%e 12 3888 2.{12}001410...

%e ...

%e 99 170759 2.{99}000033...

%e 100 254085 3.{100}00025...

%e 101 254988 3.{101}00020... (End)

%t fps[n_,i_]:=Module[{c=RealDigits[Surd[n,11],10,10]},FromDigits[ Take[ Drop[ c[[1]],c[[2]]],IntegerLength[i]]]]; nxt[{i_,n_}]:={i+1,Module[ {x=n+1}, While[fps[x,i+1]!=i+1,x++];x]}; Transpose[NestList[nxt,{0,2},50]][[2]] (* _Harvey P. Dale_, Nov 14 2013 *)

%Y Cf. A034066, A034086.

%K nonn,base

%O 0,1

%A _Patrick De Geest_, Sep 15 1998

%E Name edited by _Jon E. Schoenfield_, Nov 28 2017