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Count down from 2*k to 1, then from 2*(k+1) to 1 and so on.
3

%I #17 Jan 15 2016 11:44:33

%S 2,1,4,3,2,1,6,5,4,3,2,1,8,7,6,5,4,3,2,1,10,9,8,7,6,5,4,3,2,1,12,11,

%T 10,9,8,7,6,5,4,3,2,1,14,13,12,11,10,9,8,7,6,5,4,3,2,1,16,15,14,13,12,

%U 11,10,9,8,7,6,5,4,3,2,1,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1

%N Count down from 2*k to 1, then from 2*(k+1) to 1 and so on.

%F a(n) = 1 - n + floor(sqrt(n)+3/2)*floor(sqrt(n)+1/2).

%F a(n) = A130829(n) - A074294(n).

%t a253515[n_] := Block[{f}, f[x_] := 1 - x + Floor[Sqrt[x] + 3/2] * Floor[Sqrt[x] + 1/2]; Array[f, n]]; a253515[100] (* _Michael De Vlieger_, Jan 03 2015 *)

%t Flatten[Table[Range[2n,1,-1],{n,10}]] (* _Harvey P. Dale_, Jan 15 2016 *)

%Y Cf. A074294.

%K nonn

%O 1,1

%A _Mikael Aaltonen_, Jan 03 2015