%I #18 Jul 12 2014 03:03:18

%S 2,2,3,5,2,3,5,8,6,2,5,9,2,5,9,14,2,6,5,11,6,5,11,18,8,2,9,11,2,9,11,

%T 20,6,2,11,21,2,5,15,20,2,6,5,17,12,11,17,30,14,8,9,11,2,9,17,26,6,2,

%U 11,27,2,11,21,32,2,6,5,17,12,11,23,36,2,2,21,17,8,15,17,38

%N 3*sum(k=1..n, floor(k^2/n)) - n^2.

%C According to the reference, a(p*q) = a(p) + a(q) - 2 whenever p and q are distinct primes with p congruent to q modulo 4.

%C The sequences of indices n where a(n)=2 is {1, 2, 5, 10, 13, 17, 26, ...}, which appears to be A020893 (squarefree sums of two squares). This has been confirmed for the first 500 terms. [_John W. Layman_, May 16 2011]

%H Walter Blumberg, <a href="http://www.jstor.org/stable/10.4169/000298910X515820">Problem 11529</a>, Amer. Math. Monthly, 117 (2010), 742.

%F a(n) = 3*A014817(n) - A000290(n). - _Wesley Ivan Hurt_, Jul 10 2014

%p A175908:=n->3*add(floor(k^2/n),k=1..n)-n^2: seq(A175908(n), n=1..60); # _Wesley Ivan Hurt_, Jul 10 2014

%t Table[3 Sum[Floor[k^2/n], {k, n}] - n^2, {n, 60}] (* _Wesley Ivan Hurt_, Jul 10 2014 *)

%o (PARI) a(n) = 3*sum(k=1, n, k^2\n) - n^2; \\ _Michel Marcus_, Jul 09 2014

%Y Cf. A020893.

%K nonn

%O 1,1

%A _John W. Layman_, Oct 14 2010