LINKS
Eric Angelini, <a href="http://list.seqfan.eu/pipermailoldermail/seqfan/2012-September/010124.html">Strings resurrection</a>, SeqFan mailing list, Sep 08 2012
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Revision History for A216557
(Bold, blue-underlined text is an addition; faded, red-underlined text is aShowing entries 1-10 | older changes
Number of iterations of A216556 until the initial value n appears as a substring of the iterate; 0 if this will never happen.
(history; published version)
(history; published version)
STATUS
editing
approved
COMMENTS
The nonzero a(n) has take only 18 different values: (9, 10, 19, 28, 29, 37, 39, 46, 49, 55, 59, 64, 69, 73, 79, 82, 89, 90). For n < 10^12 the corresponding counts are (108, 75, 829, 388, 306, 326, 302, 289, 291, 277, 303, 265, 315, 254, 327, 245, 339, 2). Specifically a(19) = a(210) = 90.
EXAMPLE
STATUS
proposed
editing
STATUS
editing
proposed
COMMENTS
A216587(0) = -1 but a(101) = 19 because 101 -> 212 -> 323 -> 434 -> 545 -> 656 -> 767 -> 878 -> 989 -> 10910 -> 211021 -> 322132 -> 433243 -> 544354 -> 655465 -> 766576 -> 877687 -> 988798 -> 10998109 -> 21(101)092110 and A216587(11) = -1 but a(2110)=9 because 2110 -> 3221 -> 4332 -> 5443 -> 6554 -> 7665 -> 8776 -> 9887 -> 10998 -> (2110)109 disprove the conjecture.
STATUS
proposed
editing
STATUS
editing
proposed
EXAMPLE
a(211) = 9 since under the action of A216556, 211 -> 322 -> 433 -> 544 -> 655 -> 766 -> 877 -> 988 -> 1099 -> 211010, which contains the substring 211.
a(111) = 0 since if some number has "111" as its substring, then its preimage for A216556 (cf. A216587) contains at least the substring "00" (e.g., A216587(21110) = 1009), and has in turn no preimage under A216556. Therefore, 111 cannot occur as a substring in the orbit of any number under A216556.
STATUS
proposed
editing
STATUS
editing
proposed
COMMENTS
Can someone prove (and maybe strengthen) the following conjecture? : a(n) = 0 whenever A216587(m) = -1 for all m obtained by concatenating any digit to the left and any digit to the right of n.
A216587(0) = -1 but a(101) = 19 because 101 -> 212 -> 323 -> 434 -> 545 -> 656 -> 767 -> 878 -> 989 -> 10910 -> 211021 -> 322132 -> 433243 -> 544354 -> 655465 -> 766576 -> 877687 -> 988798 -> 10998109 -> 21(101)092110 and A216587(11) = -1 but a(2110)=9 because 2110 -> 3221 -> 4332 -> 5443 -> 6554 -> 7665 -> 8776 -> 9887 -> 10998 -> (2110)109 disprove the conjecture.
Nonzero terms are becoming increasingly sparse. For k = 1..12 the number of nonzero a(n) for n < 10^k is (10, 92, 247, 489, 797, 1194, 1678, 2236, 2860, 3565, 4359, 5421). (End)
STATUS
reviewed
editing
STATUS
proposed
reviewed