A061903 - OEIS

%I #18 Oct 27 2023 22:00:46

%S 1,1,4,1,3,3,1,2,2,1,1,4,1,2,2,1,2,3,1,2,4,1,2,2,2,2,3,2,3,2,1,2,3,2,

%T 2,2,2,3,2,1,3,2,2,3,3,1,2,2,1,3,3,1,2,3,2,2,2,2,2,2,1,2,3,3,3,2,2,3,

%U 2,2,2,2,2,3,3,2,3,3,2,2,2,2,3,3,2,2,3,3,3,3,1,3,3,3,3,2,2,3,3,2,1,4,1,2,2

%N Number of distinct elements of the iterative cycle: n -> sum of digits of n^2.

%C It seems that any such iterative cycle can contain at most 4 distinct elements.

%C a(197483417) = 5 is the first counterexample: 136 -> 28 -> 19 -> 10 -> 1. In fact this sequence is unbounded, since you can extend any chain leftward with the number k999...999 for suitably chosen k. In particular this gives the (pessimistic) bound that there is some n < 10^21942602 with a(n) = 6. - _Charles R Greathouse IV_, May 30 2014

%H Robert Israel, <a href="/A061903/b061903.txt">Table of n, a(n) for n = 0..10000</a>

%e a(2) = 4 since 2 -> 4 -> 1+6 = 7 -> 4+9 = 13 -> 1+6+9 = 16 -> 2+5+6 = 13, thus {4,7,13,16} are the distinct elements of the iterative cycle of 2. a(6) = 1 since 6 -> 3+6 = 9 -> 8+1 = 9 thus 9 is the only element in the iterative cycle of 6.

%p A:= proc(n) local L,m,x;

%p   L:= {}; x:= n;

%p   do

%p     x:= convert(convert(x^2,base,10),`+`);

%p     if member(x,L) then return nops(L)  fi;

%p     L:= L union {x};

%p   od:

%p end proc:

%p seq(A(n), n=0..200); # _Robert Israel_, May 30 2014

%o (PARI) a(n)=my(v=List()); while(1, n=sumdigits(n^2); for(i=1, #v, if(n==v[i], return(#v))); listput(v,n)) \\ _Charles R Greathouse IV_, May 30 2014

%Y Cf. A007953, A004159, A061904-A061910.

%K nonn,base,easy

%O 0,3

%A _Asher Auel_, May 17 2001

%E Corrected a(0) and example, _Robert Israel_, May 30 2014