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%I #12 Nov 01 2019 15:53:27
%S 3,11,31,93,281,841,2525,7575,22727,68183,204551,613655,1840965,
%T 5522897,16568691,49706071,149118215,447354647,1342063941,4026191825,
%U 12078575475,36235726425,108707179277,326121537829,978364613487
%N Difference between two consecutive squares enclosing 3^(2n+1).
%F a(n) = 2*Floor(3^((2n + 1)/2)) + 1.
%e a(2)=31 because 3^(2*2+1)=3^5=243, 15^2<243<16^2 and 16^2-15^2=256-225=31.
%t f[n_] := 2*Floor[3^((2n + 1)/2)] + 1; Table[f[n], {n, 0, 25}] (* _Ray Chandler_, Jun 09 2006 *)
%K nonn
%O 0,1
%A _Zak Seidov_, May 27 2006
%E Extended by _Ray Chandler_, Jun 09 2006