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a(n) = floor(6*n*sqrt(n)/Pi^2).
1

%I #20 May 03 2024 20:46:52

%S 0,0,1,3,4,6,8,11,13,16,19,22,25,28,31,35,38,42,46,50,54,58,62,67,71,

%T 75,80,85,90,94,99,104,110,115,120,125,131,136,142,148,153,159,165,

%U 171,177,183,189,195,202,208,214,221,227,234,241,247,254,261,268,275,282

%N a(n) = floor(6*n*sqrt(n)/Pi^2).

%C Conjecture: the sum of the divisors of n is less than a(n) for n exceeding 12. - _Robert G. Wilson v_, May 14 2014

%H Harry J. Smith, <a href="/A060903/b060903.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = A000203(n) + A058208(n).

%F a(n) = floor(6*n^(3/2)/Pi^2).

%t f[n_] := Floor[6 n^(3/2)/Pi^2]; Array[f, 61, 0] (* _Robert G. Wilson v_, May 14 2014 *)

%o (PARI) { default(realprecision, 100); t=Pi^2/6; for (n=0, 1000, write("b060903.txt", n, " ", n*sqrt(n)\t); ) } \\ _Harry J. Smith_, Jul 14 2009

%Y Cf. A000093, A000203, A000796, A058208.

%K nonn

%O 0,4

%A _Henry Bottomley_, May 05 2001