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Search: id:A078786
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| A078786 |
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Period of cycle of the inventory sequence (as in A063850) starting with n. |
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2
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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.
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LINKS
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Carlos Rivera, The Inventory Sequences and Self-Inventoried Numbers
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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Cf. A063850, A078970.
Sequence in context: A049241 A101080 A130836 this_sequence A102677 A030368 A037805
Adjacent sequences: A078783 A078784 A078785 this_sequence A078787 A078788 A078789
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KEYWORD
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base,nice,nonn
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AUTHOR
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Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Jan 14 2003
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