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A078786
Period of cycle of the inventory sequence (as in A063850) starting with n.
1
2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 4, 2, 1, 2, 1, 1, 4, 4, 4, 4, 3, 1, 1, 1, 1, 1, 4, 3, 3, 3, 2, 1, 1, 1, 4, 4, 1, 3, 2, 2, 2, 1, 1, 1, 4, 3, 3, 1, 2, 2, 2, 1, 1, 1, 4, 3, 2, 2, 1, 2, 2, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1
OFFSET
1,1
COMMENTS
It can be proved that any inventory sequence ends in a cycle all of whose terms are <= 10^20. Conjecture: a(n) <= 4 for all n. It suffices to check this for all inventory sequences starting with n, where n <= 10^20.
EXAMPLE
The inventory sequence starting with 1 is: 1, 11, 21, 1211, 3112, 132112, 311322, 232122, 421311, 14123113, 41141223, 24312213, 32142321, 23322114, 32232114, 23322114, .... which ends in the cycle 32232114, 23322114 of period 2. Hence a(1) = 2.
MATHEMATICA
g[n_] := Module[{seen, r, d, l, i, t}, seen = {}; r = {}; d = IntegerDigits[n]; l = Length[d]; For[i = 1, i <= l, i++, t = d[[i]]; If[ ! MemberQ[seen, t], r = Join[r, IntegerDigits[Count[d, t]]]; r = Join[r, {t}]; seen = Append[seen, t]]]; FromDigits[r]];
per[n_] := Module[{r, t, p1, p}, r = {}; t = g[n]; While[ ! MemberQ[r, t], r = Append[r, t]; t = g[t]]; r = Append[r, t]; p1 = Flatten[Position[r, t]]; p = p1[[2]] - p1[[1]]; p]; Table[per[i], {i, 1, 100}]
CROSSREFS
Sequence in context: A279185 A161385 A152907 * A102677 A145884 A030368
KEYWORD
base,nice,nonn
AUTHOR
Joseph L. Pe, Jan 14 2003
STATUS
approved