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

%I A078786 #5 Jul 26 2025 16:06:40
%S A078786 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,
%T A078786 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,
%U A078786 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
%N A078786 Period of cycle of the inventory sequence (as in A063850) starting with n.
%C A078786 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.
%H A078786 Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_207.htm">Puzzle 207. The Inventory Sequences and Self-Inventoried Numbers</a>, The Prime Puzzles & Problems Connection.
%e A078786 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.
%t A078786 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]];
%t A078786 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}]
%Y A078786 Cf. A063850, A078970.
%K A078786 base,nice,nonn
%O A078786 1,1
%A A078786 _Joseph L. Pe_, Jan 14 2003