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A061903
Number of distinct elements of the iterative cycle: n -> sum of digits of n^2.
7
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
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
KEYWORD
nonn,base,easy
AUTHOR
Asher Auel, May 17 2001
EXTENSIONS
Corrected a(0) and example, Robert Israel, May 30 2014
STATUS
approved