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a(n) = Sum_{k>=0} floor(n/(5*k + 2)).
3

%I #19 Dec 23 2022 09:11:34

%S 0,0,1,1,2,2,3,4,5,5,6,6,8,8,10,10,11,12,13,13,14,15,17,17,19,19,20,

%T 21,23,23,24,24,26,26,28,29,31,32,33,33,34,34,37,37,39,39,40,41,43,44,

%U 45,46,48,48,50,50,52,53,54,54,56,56,58,59,61,61,63,64,66,66,68,68,71,71,73,73,74,76,77,77,78

%N a(n) = Sum_{k>=0} floor(n/(5*k + 2)).

%C Partial sums of A001877.

%H Seiichi Manyama, <a href="/A218445/b218445.txt">Table of n, a(n) for n = 0..10000</a>

%t Table[Sum[Floor[n/(5k+2)],{k,0,n}],{n,0,80}] (* _Harvey P. Dale_, Dec 08 2022 *)

%o (PARI) a(n)=sum(k=0,n\5,(n\(5*k+2)))

%o (Maxima) A218445[n]:=sum(floor(n/(5*k+2)),k,0,n)$

%o makelist(A218445[n],n,0,80); /* _Martin Ettl_, Oct 29 2012 */

%Y Cf. A218444, A218446, A218447.

%K nonn

%O 0,5

%A _Benoit Cloitre_, Oct 28 2012