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

0,3

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

Links

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

Example

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.

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

Prog

(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

Crossrefs

Cf. A007953, A004159, A061904-A061910.

Sequence in context: A067277 A177951 A222130 * A309603 A328611 A084118

Adjacent sequences: A061900 A061901 A061902 * A061904 A061905 A061906

Keyword

nonn,base,easy

Author

Asher Auel, May 17 2001

Extensions

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

Status

approved