A118740 - OEIS

%I #12 Dec 26 2017 03:22:31

%S 1,6,1,4,3,3,16,3,1,2,1,2,1,1,2,1,1,2,1,24,6,13,21,6,35,9,2,25,1,6,1,

%T 5,1,1,1,2,22,3,1,52,5,1,16,21,2,19,10,11,18,32,9,12,1,2,1,3,2,3,55,9,

%U 4,18,2,3,2,2,1,3,8,58,1,2,3,3,3,2,2,3,81,35,2,3,2,2,13,2,2,3,4,2,3,3,2,19,2,2

%N Let L_n be the infinite sequence formed by starting with 1 and repeatedly placing the first digit at the end of the number and adding n to get the next term. Sequence gives number of steps for L_n to enter a cycle, or -1 if no cycle is ever reached.

%C It is conjectured that L_n always reaches a cycle.

%C From _Robert Israel_, Dec 25 2017: (Start)

%C a(10^k-1) = 1 for k >= 1.

%C a(19*10^k-1) = 2.

%C Empirical:

%C   a(10^k) = k+1 for all k.

%C   a(2*10^k) = 9*k+15 for k >= 1.

%C   a(2*10^k-1) = 1 for k >= 1.

%C   a(10^k+1) = k for k >= 1.

%C   a(2*10^k+1) = 3*k+3 for k >= 1. (End)

%H Robert Israel, <a href="/A118740/b118740.txt">Table of n, a(n) for n = 1..10000</a>

%e L_2 = [1,3,5,7,9,11,13,33,35,55,57,77,79,99,101,13,...] enters a cycle of length 9 after 6 steps.

%p f:= proc(n) local t, k, S,d;

%p     t:= 1; S[t]:= 0;

%p     for k from 1 do

%p       d:= 10^ilog10(t);

%p       t:= 10*(t mod d)+ floor(t/d) + n;

%p       if assigned(S[t]) then return S[t] fi;

%p       S[t]:= k;

%p     od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Dec 25 2017

%Y Cf. A118739.

%K base,nonn

%O 1,2

%A Luc Stevens (lms022(AT)yahoo.com), May 24 2006

%E Corrected by _Robert Israel_, Dec 25 2017