Let r(n) be the smallest value for which r(n)^2 starts with n. Let n >= 1. Let k(n) = floor(log10(n)) + 2 = (number of digits in n) + 1. Then, if m(n) = ceil(sqrt(n * 10^k(n))), m(n)^2 starts with n. If r(n) >= 1 is the smallest positive integer with such that r(n)^2 starts with n, then r(n) = ceil(sqrt(n * 10^k)) for some k <= k(n). r(n)^2 <= 40*n is a tight bound. This bound is approached by numbers slightly larger than 25*10^j, for example
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Neil Sloane Sent: Saturday, May 21, 2016 8:45 PM To: fun Subject: [math-fun] squares beginning with n
Given n, A018851 and A018796 give the smallest square that begins with n
Question: does such a square always exist, and if so how big can the smallest example be? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun