OFFSET
0,2
COMMENTS
Given n, this is the smallest number m with the property that the smallest square beginning with m has n more digits than n.
a(n) >= 25*10^(n-3). Conjecture: a(n)/(25*10^(n-3)) -> 1 as n -> oo. - Chai Wah Wu, May 21 2016
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
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
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
LINKS
EXAMPLE
The smallest square beginning with 5 is 529, which has two more digits than 5, and corresponds to a(2) = 5.
CROSSREFS
KEYWORD
nonn,more,base
AUTHOR
EXTENSIONS
a(6)-a(8) from Chai Wah Wu, May 21 2016
a(9)-a(10), a(15)-a(18) and corrected a(12) from Chai Wah Wu, May 22 2016
a(11)-a(14) from David A. Corneth, May 22 2016
STATUS
approved