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a(n) = 10^(2n-1) minus largest square less than 10^(2n-1).
2

%I #16 Mar 31 2012 14:40:14

%S 1,39,144,1756,49116,484471,4175271,38053824,10649244,1064924400,

%T 43246886799,529955487744,2399106211776,50173961567511,

%U 590207432515431,2099745368512359,20237877241133151,126421128012287511

%N a(n) = 10^(2n-1) minus largest square less than 10^(2n-1).

%C For even indices a(2k) = 2*10^k-1, hence only odd powers of 10 are considered in this sequence.

%H Hugo Pfoertner, <a href="/A128826/b128826.txt">Table of n, a(n) for n = 1..500</a>

%F a(n)=10^(2n-1)-A061433(2n-1).

%F a(n) = 10^(2*n-1) - floor(sqrt(10^(2*n-1)))^2.

%e a(1) = 1 because 10 - 3^2 = 1.

%e a(2) = 39 because 1000 - 31^2 = 39.

%e a(3) = 144 because 100000 - 316^2 = 144.

%t Table[10^n-Floor[(10^n-1)^(1/2)]^2,{n,1,40,2}]

%Y Cf. A061433, A051221.

%K nonn

%O 1,2

%A _Zak Seidov_, Apr 12 2007