OFFSET
1,1
COMMENTS
Suppose that S = (x(1),..,x(h)) is a vector of nonnegative integers. Let m = max(S) and F(S) = (f(0),..,f(m)), where f(i) is the number of occurrences of i in S. Define F(0) = S and F(q) = F(F(q-1)) for q>=1. By Theorem 1 at A225660, the vector F(q) is eventually periodic with period 6.
Theorem 2. If S is not one of the ten vectors listed below, then F(q) = (2, 2) for some q, and the concatenation of F(q), F(q+1),... comprises the periodic sequence A225869. Seven of the exceptional vectors are indicated by (0) -> (1) -> (0,1) -> (1,1) -> (1,0,1) -> (1,2) -> (0,1,1) -> (1,2) -> ..., and the remaining three, by (2) -> (0,0,1) -> (2,1) -> (0,1,1). (The second appearances of (1,2) and (0,1,1) are not counted.)
A proof of Theorem 2 consists of easy (omitted) examinations of cases. Note, in particular, that A225869 is the limiting sequence for every S having more than 3 components.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
FORMULA
The sequence is periodic with fundamental period 2, 2, 0, 0, 2, 2, 0, 1, 1, 1, 1, 0, 3, 1, 0, 0, 1 .
EXAMPLE
S = (6,1,0,5,5,3) -> (1,1,0,1,0,2,1) -> (2,4,1) -> (0,1,1,0,1) -> (2,3) -> (0,0,1,1) -> (2,2)* -> (0,0,2) -> (2,0,1) -> (1,1,1) -> (0,3) -> (1,0,0,1) -> (2,2). The asterisk (*) shows where the limiting sequence A225869 begins. The sequence is the concatenation of the repeating vectors starting with (2,2).
MATHEMATICA
t[n_] := t[n] = Table[Count[t[n - 1], k], {k, 0, Max[t[n - 1]]}];
t[0] = {2, 2}; (* t[0] is the vector S*)
u = Table[t[n], {n, 0, 36}] (* list of vectors F(q) *)
lst = Flatten[u] (* A225869 as a sequence *)
PadRight[{}, 100, {2, 2, 0, 0, 2, 2, 0, 1, 1, 1, 1, 0, 3, 1, 0, 0, 1}] (* Harvey P. Dale, Sep 10 2016 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 18 2013
STATUS
approved