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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

%I

%S 1,2,3,4,5,6,7,8,9,100,121,132,143,154,105,176,187,108,0,200,231,242,

%T 253,264,225,286,297,0,0,300,341,352,363,374,315,396,0,0,0,400,451,

%U 462,473,484,405,0,0,0,0,500,561,572,583,594,0,0,0,0,0,600,671,682,693,0,0

%N 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.

%C 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.

%C 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

%Y Cf. A080436, A110735.

%K base,easy,nonn

%O 1,2

%A _Amarnath Murthy_, Aug 09 2005

%E Edited, corrected and extended by _Franklin T. Adams-Watters_, Sep 25 2006

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