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A110734
Let n = a_1a_2...a_k, where the a_i are digits. a(n) = least multiple of n of the form a_1b_1a_2b_2...b_{k-1}a_k, obtained by inserting single digits b_i in the gaps; 0 if no such number exists.
2
1, 2, 3, 4, 5, 6, 7, 8, 9, 100, 121, 132, 143, 154, 105, 176, 187, 108, 0, 200, 231, 242, 253, 264, 225, 286, 297, 0, 0, 300, 341, 352, 363, 374, 315, 396, 0, 0, 0, 400, 451, 462, 473, 484, 405, 0, 0, 0, 0, 500, 561, 572, 583, 594, 0, 0, 0, 0, 0, 600, 671, 682, 693, 0, 0
OFFSET
1,2
COMMENTS
Conjecture: for large n, a(n) is nonzero. For a k-digit number there are k-1 gaps and 10^(k-1) candidates, so the chances that one of them is a multiple of n increases with k on the one hand though the probability decreases because n becomes large.
Pursuing the probability argument, the probability that a(n) is zero is (1-1/n)^{10^{k-1}}, which has an expected value of e^{-(10^{k-1})/n}. (10^{k-1})/n varies from 1/100 to 1/10, depending on the leading digits of n, so the probability a(n) is zero is between e^{-1/10} and e^{-1/100}; thus we would expect that only a small but nonzero fraction of all n have a(n) nonzero. Of course, it is not clear that the probability argument is accurate. Franklin T. Adams-Watters, Sep 25 2006
CROSSREFS
Sequence in context: A088472 A087019 A044911 * A080436 A135295 A122625
KEYWORD
base,easy,nonn
AUTHOR
Amarnath Murthy, Aug 09 2005
EXTENSIONS
Edited, corrected and extended by Franklin T. Adams-Watters, Sep 25 2006
STATUS
approved