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a(n) = floor(n^(3/2))*floor(3+n^(3/2))/2.
1

%I #10 Jul 08 2017 12:48:39

%S 2,5,20,44,77,119,189,275,405,527,702,902,1127,1430,1769,2144,2555,

%T 3002,3485,4094,4752,5459,6215,7020,8000,8910,10010,11174,12402,13694,

%U 15050,16652,18144,19899,21735,23652,25650,27729,29889,32130,34715,37400,39902,42777,45752,48827,52325,55610,59339,62834,66794,70499,74690,79002,83435,88409,93095

%N a(n) = floor(n^(3/2))*floor(3+n^(3/2))/2.

%C Subsequence of A000096.

%H G. C. Greubel, <a href="/A185593/b185593.txt">Table of n, a(n) for n = 1..5000</a>

%t Table[Floor[n^(3/2)]*Floor[n^(3/2)+3]/2, {n,1,80}]

%o (PARI) for(n=1,50, print1(floor(n^(3/2))*floor(3+n^(3/2))/2, ", ")) \\ _G. C. Greubel_, Jul 08 2017

%Y Cf. A000096.

%K nonn

%O 1,1

%A _Clark Kimberling_, Feb 01 2011