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a(n) = Sum_{k>=0} floor(n/(5*k + 3)).
4

%I #18 Dec 23 2022 09:11:29

%S 0,0,0,1,1,1,2,2,3,4,4,4,5,6,6,7,8,8,10,10,10,11,11,12,14,14,15,16,17,

%T 17,18,18,19,21,21,21,23,23,24,26,27,27,28,29,29,30,31,31,34,34,34,35,

%U 36,37,39,39,41,42,43,43,44,44,44,46,47,48,50,50,51,53,53,53,56,57,57,58,59,59,62,62,63

%N a(n) = Sum_{k>=0} floor(n/(5*k + 3)).

%C Partial sums of A001878.

%H Seiichi Manyama, <a href="/A218446/b218446.txt">Table of n, a(n) for n = 0..10000</a>

%t Accumulate[Table[Count[Divisors[n],_?(Mod[#,5]==3&)],{n,0,90}]] (* _Harvey P. Dale_, Nov 08 2012 *)

%o (PARI) a(n)=sum(k=0,n,(n\(5*k+3)))

%o (Maxima) A218446[n]:=sum(floor(n/(5*k+3)),k,0,n)$

%o makelist(A218446[n],n,0,80); /* _Martin Ettl_, Oct 29 2012 */

%Y Cf. A218444, A218445, A218447.

%K nonn

%O 0,7

%A _Benoit Cloitre_, Oct 28 2012