%I #11 Oct 12 2017 16:33:16

%S 0,6534,65934,659934,6599934,65346534,65999934,653406534,659999934,

%T 6534006534,6593465934,6599999934,65340006534,65934065934,65999999934,

%U 653400006534,653465346534,659340065934,659934659934,659999999934,6534000006534,6534659346534,6593400065934,6599340659934,6599999999934

%N Numbers n such that reversal(n)=2n/3.

%C If n=0 or n>1 then 66*(10^n-1) is in the sequence (the first five terms of this sequence are of this form) so this sequence is infinite. Let g(s,t,r) be (s.(0)(t))(r).s where dot between numbers means concatenation and "(m)(n)" means number of m's is n, for example g(2005,1,2)=20050200502005. It is interesting that, if n is in the sequence then all numbers of the form g(n,t,r) for nonnegative integers t and r are in the sequence, for example since 6534 is in the sequence so g(6534,1,2)=(6534.(0)(1))(2).6534=65340653406534 is in the sequence.

%C It seems that all similar sequences (sequences with the definition "numbers n such that reversal(n) =r*n for a fixed rational number r" ) have the same property (see A101705 and A101706). All sequences of the form 10^s*A002113 are in this category.

%C There are Fibonacci(floor((n-2)/2)) terms with n digits, n>1 (this is essentially A103609). - _Ray Chandler_, Oct 12 2017

%H Ray Chandler, <a href="/A101704/b101704.txt">Table of n, a(n) for n = 1..10000</a>

%e g(65934,3,4)=6593400065934000659340006593400065934 is in the sequence

%e because reversal(6593400065934000659340006593400065934)

%e = 4395600043956000439560004395600043956

%e =2/3*6593400065934000659340006593400065934.

%t Do[If[FromDigits[Reverse[IntegerDigits[n]]] == 2/3*n, Print[n]], {n, 150000000}]

%Y Cf. A101703, A101705.

%K base,nonn

%O 1,2

%A _Farideh Firoozbakht_, Dec 31 2004

%E a(8)-a(25) from _Max Alekseyev_, Aug 18 2013