A078786 Period of cycle of the inventory sequence (as in A063850) starting with n.

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.

Links

Table of n, a(n) for n=1..100.

Carlos Rivera, The Inventory Sequences and Self-Inventoried Numbers

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

Cf. A063850, A078970.

Sequence in context: A279185 A161385 A152907 * A102677 A145884 A030368

Adjacent sequences: A078783 A078784 A078785 * A078787 A078788 A078789

Keyword

base,nice,nonn

Author

Joseph L. Pe, Jan 14 2003

Status

approved