A078970 Cycle of the inventory sequence (as in A063850) starting with n consists of prime numbers.

39, 93, 349, 394, 439, 493, 934, 943, 999, 1139, 1193, 1319, 1391, 1913, 1931, 1999, 3139, 3193, 3319, 3339, 3391, 3393, 3913, 3931, 3933, 9111, 9139, 9193, 9319, 9391, 9399, 9913, 9931, 9939, 9993, 11129, 11192, 11219, 11291, 11912, 11921, 12119, 12191, 12239

Offset: 1

Author

Joseph L. Pe, Jan 14 2003

Keywords

• base
• nice
• nonn

Comments

It can be proved that any inventory sequence ends in a cycle all of whose terms are <= 10^20.

Examples

			The inventory sequence starting with 39 is: 39, 1319, 211319, 12311319, 41122319, 1431221319, 4114232219, 2431321319, 2214333119, 2231143319, 2233311419, 2233311419, .... The cycle is 2233311419, 2233311419, .... and 2233311419 is prime, so 39 is in the sequence.
		

Links

• Carlos Rivera, Puzzle 207. The Inventory Sequences and Self-Inventoried Numbers, The Prime Puzzles & Problems Connection.

Crossrefs

Cf. A063850, A078786.

Programs

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]];
pr[n_] := Module[{r, t, p1, p, a}, r = {}; t = g[n]; a = True; While[ ! MemberQ[r, t], r = Append[r, t]; t = g[t]]; r = Append[r, t]; p1 = Flatten[Position[r, t]]; p = PrimeQ[Drop[r, p1[[1]]]]; If[MemberQ[p, False], a = False]; a];
l = {}; For[k = 1, k <= 10^4, k++, If[pr[k], l = Append[l, k]]]; l

Extensions

Missing terms inserted by Sean A. Irvine, Jul 25 2025