A063850 -id:A063850 - cp's OEIS frontend

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

Joseph L. Pe, Jan 14 2003

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

			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.

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

Cf. A063850, 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]];
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

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

A112512 Say what you see in previous term, same as A063850, but starting with 2.

Original entry on oeis.org

2, 12, 1112, 3112, 132112, 311322, 232122, 421311, 14123113, 41141223, 24312213, 32142321, 23322114, 32232114, 23322114, 32232114, 23322114, 32232114

Offset: 1

Michele Dondi (blazar(AT)lcm.mi.infn.it), Sep 09 2005

Eventually periodic, eventually identical (to a shift of) A063850.

Cf. A005150, A005151, A063850, A112512-A112516.

#!/usr/bin/perl -l use strict; use warnings; die "Usage: $0  " unless @ARGV==1; $=shift; { print; my (%cnt, %saw); $cnt{$}++ for /./g; s/./ $saw{$&}++ ? '' : $cnt{$&} . $& /ge; <>, redo; } _END_

A112515 Say what you see in previous term, same as A063850 but starting with 5.

Original entry on oeis.org

5, 15, 1115, 3115, 132115, 31131215, 23411215, 2213143115, 2241231415, 3224311315, 3322143115, 3322311415, 3322311415, 3322311415, 3322311415, 3322311415, 3322311415

Offset: 1

Michele Dondi (blazar(AT)lcm.mi.infn.it), Sep 09 2005

Cf. A005150, A005151, A063850, A112512-A112515.

#!/usr/bin/perl -l use strict; use warnings; die "Usage: $0 " unless @ARGV==1; $=shift; { print; my (%cnt, %saw); $cnt{$}++ for /./g; s/./ $saw{$&}++ ? '' : $cnt{$&} . $& /ge; <>, redo; } _END_

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

Original entry on oeis.org

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

Joseph L. Pe, Jan 14 2003

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.

			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.

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

Cf. A063850, A078970.

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}]

A079466 Numbers k such that the "inventory" A063850 of k is a palindrome.

Original entry on oeis.org

1, 22, 112, 121, 211, 333, 1113, 1131, 1311, 3111, 4444, 11114, 11141, 11411, 14111, 22233, 22323, 22332, 23223, 23232, 23322, 32223, 32232, 32322, 33222, 41111, 55555

Offset: 1

Joseph L. Pe, Jan 14 2003

			The "inventory" of 112 is 2112 (two "1"s, one "2"), which is a palindrome. Hence 112 belongs to the sequence.

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

Cf. A063850. Different from A079676.

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]]; isPalin[n_] := (n == FromDigits[Reverse[IntegerDigits[n]]]); Select[Range[10^5], isPalin[g[ # ]] &]

A079464 Numbers k such that the "inventory" A063850 of k is prime.

Original entry on oeis.org

1, 3, 7, 9, 17, 23, 27, 33, 39, 51, 63, 69, 81, 93, 99, 111, 113, 127, 131, 133, 137, 193, 199, 203, 209, 223, 232, 233, 271, 299, 301, 311, 313, 331, 359, 361, 367, 371, 377, 414, 431, 433, 439, 441, 447, 451, 463, 469, 474, 477, 479, 481, 497, 499, 503, 523

Offset: 1

Joseph L. Pe, Jan 14 2003

			The "inventory" of 299 is 1229 (one "2", two "9"s), which is prime. Hence 299 belongs to the sequence.

Carlos Rivera, Puzzle 207. The Inventory Sequences and Self-Inventoried Numbers, The Prime Puzzles & Problems Connection. See Question 7.

Cf. A063850.

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]]; Select[Range[10^3], PrimeQ[g[ # ]] &]

A079465 Numbers k such that the "inventory" A063850 of k is a perfect square.

Original entry on oeis.org

6, 55, 116, 161, 255, 511, 666, 969, 996, 5311, 9666, 9999, 12255, 12525, 12552, 41199, 41919, 41991, 54246, 54264, 54426, 71177, 71717, 71771, 72255, 72525, 72552, 77117, 77171, 77711, 78055, 83399, 83939, 83993, 89999, 97117, 97171

Offset: 1

Joseph L. Pe, Jan 14 2003

			The "inventory" of 511 is 1521 (one "5", two "1"s) = 39^2. Hence 1521 belongs to the sequence.

Carlos Rivera, Puzzle 207. The Inventory Sequences and Self-Inventoried Numbers, The Prime Puzzles & Problems Connection. See Question 7.

Cf. A063850.

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]]; Select[Range[10^5], IntegerQ[Sqrt[g[ # ]]] &]

A079470 Primes with prime inventory number (as in A063850).

Original entry on oeis.org

3, 7, 17, 23, 113, 127, 131, 137, 193, 199, 223, 233, 271, 311, 313, 331, 359, 367, 431, 433, 439, 463, 479, 499, 503, 523, 587, 607, 641, 677, 691, 733, 773, 797, 809, 821, 823, 829, 853, 997, 1009, 1069, 1123, 1129, 1187, 1213, 1217, 1223, 1231, 1277, 1291

Offset: 1

Joseph L. Pe, Jan 15 2003

			The prime 127 has inventory number 111217 (one "1", one "2", one "7"), which is also prime. Hence 127 belongs to the sequence.

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

Cf. A063850.

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]]; s = {}; For[j = 1, j <= 10^3, j++, temp = Prime[j]; If[PrimeQ[g[temp]], s = Append[s, temp]]]; s

A112513 Same as A063850, Say what you see in previous term, but starting with 3.

Original entry on oeis.org

3, 13, 1113, 3113, 2321, 221311, 223113, 222321, 421311, 14123113, 41141223, 24312213, 32142321, 23322114, 32232114, 23322114, 32232114, 23322114, 32232114, 23322114

Offset: 1

Michele Dondi (blazar(AT)lcm.mi.infn.it), Sep 09 2005

Eventually periodic.

Cf. A005150, A005151, A063850, A112512-A112515.

#!/usr/bin/perl -l use strict; use warnings; die "Usage: $0  " unless @ARGV==1; $=shift; { print; my (%cnt, %saw); $cnt{$}++ for /./g; s/./ $saw{$&}++ ? '' : $cnt{$&} . $& /ge; <>, redo; } _END_

A112514 Say what you see in previous term, same as A063850 but starting with 4.

Original entry on oeis.org

4, 14, 1114, 3114, 132114, 31131214, 23411214, 22132431, 32212314, 23322114, 32232114, 23322114, 32232114, 23322114, 32232114, 23322114, 32232114, 23322114, 32232114

Offset: 1

Michele Dondi (blazar(AT)lcm.mi.infn.it), Sep 09 2005

Eventually periodic.

Cf. A005150, A005151, A063850, A112512-A112515.

#!/usr/bin/perl -l use strict; use warnings; die "Usage: $0 " unless @ARGV==1; $=shift; { print; my (%cnt, %saw); $cnt{$}++ for /./g; s/./ $saw{$&}++ ? '' : $cnt{$&} . $& /ge; <>, redo; } _END_

cp's OEIS Frontend

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

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A112512 Say what you see in previous term, same as A063850, but starting with 2.

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A112515 Say what you see in previous term, same as A063850 but starting with 5.

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

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A079466 Numbers k such that the "inventory" A063850 of k is a palindrome.

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A079464 Numbers k such that the "inventory" A063850 of k is prime.

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A079465 Numbers k such that the "inventory" A063850 of k is a perfect square.

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A079470 Primes with prime inventory number (as in A063850).

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A112513 Same as A063850, Say what you see in previous term, but starting with 3.

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