%I #11 May 26 2019 19:31:16
%S 5,35,282,2600,25317,251000,2503163,25010000,250031623,2500100000,
%T 25000316228,250001000000,2500003162278,25000010000000,
%U 250000031622777,2500000100000000,25000000316227767,250000001000000000,2500000003162277661,25000000010000000000
%N a(n) is the smallest positive integer k such that there does not exist an m such that floor(m^2/10^n) = k.
%C For n >= 2, note that when k < 5*10^(n-1) we have (k + 1)^2 - k^2 = 2*k + 1 < 10^n, so there exists m such that floor(m^2/10^n) = 0, 1, 2, ..., 25*10^(n-2). For 0 <= t < sqrt(10^n), floor((5*10^(n-1) + t)^2/10^n) = 25*10^(n-2) + t; for t = ceiling(sqrt(10^n)), floor((5*10^(n-1) + t)^2/10^n) = 25*10^(n-2) + t + 1. So the number 25*10^(n-2) + ceiling(10^(n/2)) is skipped over. Take n = 3 as an example. When k < 500, (k + 1)^2 - k^2 < 1000, so there exists m such that floor(m^2/1000) = 0, 1, 2, ..., 250. Since 31^2 = 961 < 1000, 32^2 = 1024 > 1000, (500 + t)^2 is successively 251001, 252004, ..., 281961, 283024, so a(3) = 282.
%C The sum of digits for Sum_{i=1..n} a(2*n) is 8*n.
%F a(n) = 25*10^(n-2) + ceiling(10^(n/2)) for n >= 2.
%e floor(m^2/10) = 0, 0, 0, 0, 1, 2, 3, 4, 6 for m = 0..8, so a(1) = 5.
%e floor(m^2/100) = 0, ..., 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 36 for m = 0..60, so a(2) = 35.
%o (PARI) a(n) = if(n==1, 5, 25*10^(n-2) + ceil(10^(n/2)))
%Y Cf. A017936, A317774, A322667.
%K nonn
%O 1,1
%A _Jianing Song_, Dec 22 2018