A061903 Number of distinct elements of the iterative cycle: n -> sum of digits of n^2.

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, 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, 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

Offset: 0

Author

Asher Auel, May 17 2001

Keywords

• nonn
• base
• easy

Comments

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

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

Examples

			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.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A007953, A004159, A061904-A061910.

Programs

Maple

A:= proc(n) local L,m,x;
  L:= {}; x:= n;
  do
    x:= convert(convert(x^2,base,10),`+`);
    if member(x,L) then return nops(L)  fi;
    L:= L union {x};
  od:
end proc:
seq(A(n), n=0..200); # Robert Israel, May 30 2014

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

Extensions

Corrected a(0) and example, Robert Israel, May 30 2014