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%I #41 Jul 16 2016 16:11:59
%S 0,2,5,35,296,2600,25317,251416,2504474,25010000,250044723,2500100000,
%T 25000316228,250002000003,2500004472137,25000010000000,
%U 250000044721361,2500000141421358,25000000316227767
%N Smallest number m such that A272677(m) = n.
%C Given n, this is the smallest number m with the property that the smallest square beginning with m has n more digits than n.
%C a(n) >= 25*10^(n-3). Conjecture: a(n)/(25*10^(n-3)) -> 1 as n -> oo. - _Chai Wah Wu_, May 21 2016
%C For odd n > 2, it seems that a(n) is about 25 * 10^(n-3) + 10^(floor((n-1)/2)), although a(13) breaks that pattern. - _David A. Corneth_, May 22 2016
%C Except for n = 1 and 13, a(n) appears to be approximately equal to either 25*10^(n-3)+sqrt(10^(n-1)) (for n = 0, 2, 3, 5, 6, 9, 11, 12, 15, 18, ... ) or 25*10^(n-3)+sqrt(2*10^(n-1)) (for n = 4, 7, 8, 14, 16, 17, ...). For n = 1, a(n) is approximately 25*10^(n-3)+sqrt(3*10^(n-1)) and for n = 13, a(n) is about equal to 25*10^(n-3)+sqrt(4*10^(n-1)). Conjecture: a(n) is always approximately to 25*10^(n-3)+sqrt(k*10^(n-1)) for some small integer k > 0. - _Chai Wah Wu_, May 22 2016
%C Using the above conjecture as a guide, upper bounds for a(n) can be computed (see file in links) which coincide with a(n) for n <= 19. - _Chai Wah Wu_, May 23 2016
%H Chai Wah Wu, <a href="/A272678/a272678.txt">Upper bounds for a(n), n = 0..1000</a>
%e The smallest square beginning with 5 is 529, which has two more digits than 5, and corresponds to a(2) = 5.
%Y Cf. A018851, A018796, A272677.
%K nonn,more,base
%O 0,2
%A _Nathan Fox_, _Brooke Logan_, and _N. J. A. Sloane_, May 21 2016
%E a(6)-a(8) from _Chai Wah Wu_, May 21 2016
%E a(9)-a(10), a(15)-a(18) and corrected a(12) from _Chai Wah Wu_, May 22 2016
%E a(11)-a(14) from _David A. Corneth_, May 22 2016
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