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A118514
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Define sequence S_n by: initial term = n, reverse digits and add 2 to get next term. It is conjectured that S_n always reaches a cycle. Sequence gives number of steps for S_n to reach a cycle.
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10
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1, 3, 0, 2, 0, 1, 0, 0, 0, 0, 0, 9, 0, 7, 10, 0, 9, 4, 0, 3, 8, 8, 8, 7, 15, 5, 5, 3, 12, 1, 11, 16, 0, 7, 0, 5, 8, 0, 7, 2, 0, 6, 6, 6, 6, 5, 13, 3, 3, 1, 10, 6, 9, 14, 0, 5, 0, 3, 6, 0, 5, 4, 0, 4, 4, 4, 4, 3, 11, 1, 1, 13, 8, 4, 7, 12, 0, 3, 0, 1, 4, 2, 3, 2, 0, 2, 2, 2, 2, 1, 9, 12, 0
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OFFSET
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1,2
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COMMENTS
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Initial cycles have length 81 or 90.
There is one cycle of length 81 (least component is 3, all components have at most three digits, cf. A117521), 22 cycles of length 90 with 4-digit components (least components are 1013 + 2*k for k = 0, ..., 21, cf. A120214) and one cycle of length 45 with 4-digit components (least component is 1057, cf. A120215). Furthermore there are 22 cycles of length 1890 (least components are 100013 + 2*k for k = 0, ..., 21, cf. A120216), one cycle of length 945 (least component is 100057, cf. A120217) and 225 cycles of length 900 (least components are 100103 + 2*k for k = 0, ..., 224, cf. A120218), all having 6-digit components. It is conjectured that there are also cycles of increasing length with 8-, 10-, 12-, ... digit components. - Klaus Brockhaus, Jun 10 2006
There are 22 cycles of length 19890 (least components are 10000013 + 2*k for k = 0, ..., 21), one cycle of length 9945 (least component 10000057), 225 cycles of length 18900 (least components are 10000103 + 2*k for k = 0, ..., 224) and 2250 cycles of length 9000 (least components are 10001003 + 2*k for k = 0, ..., 2249), all having 8-digit components.
These patterns continue. Specifically, there is one cycle of length 10^(n/2) - 55 (least component 10^(n-1) + 57), and there are 22 cycles of length 2*(10^(n/2) - 55) (least components 10^(n-1) + 13 + 2*k for k = 0, ..., 21), each for n = 4, 6, 8, 10, 12, 14, 16. (End)
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LINKS
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PROG
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(Python) # see linked program
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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