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A194222 a(n) = floor(Sum_{k=1..n} frac(k/5)), where frac() = fractional part.

%I A194222 #35 Jun 06 2024 18:47:13
%S A194222 0,0,1,2,2,2,2,3,4,4,4,4,5,6,6,6,6,7,8,8,8,8,9,10,10,10,10,11,12,12,
%T A194222 12,12,13,14,14,14,14,15,16,16,16,16,17,18,18,18,18,19,20,20,20,20,21,
%U A194222 22,22,22,22,23,24,24,24,24,25,26,26,26,26,27,28,28,28,28,29,30
%N A194222 a(n) = floor(Sum_{k=1..n} frac(k/5)), where frac() = fractional part.
%H A194222 G. C. Greubel, <a href="/A194222/b194222.txt">Table of n, a(n) for n = 1..5000</a>
%H A194222 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F A194222 From _Chai Wah Wu_, Jun 10 2020: (Start)
%F A194222 a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6.
%F A194222 G.f.: x^3*(x + 1)/((x-1)^2*(1+x+x^2+x^3+x^4)). (End)
%F A194222 a(n) = floor((n+1)/5) + floor((n+2)/5). - _Ridouane Oudra_, Dec 14 2021
%F A194222 a(n) = A002266(n+1)+A002266(n+2). - _R. J. Mathar_, Nov 21 2023
%p A194222 seq(floor((n+1)/5)+floor((n+2)/5), n=1..80); # _Ridouane Oudra_, Dec 14 2021
%t A194222 r = 1/5;
%t A194222 a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]
%t A194222 Table[a[n], {n, 1, 90}]    (* A194222 *)
%t A194222 s[n_] := Sum[a[k], {k, 1, n}]
%t A194222 Table[s[n], {n, 1, 100}]   (* A118015 *)
%t A194222 LinearRecurrence[{1,0,0,0,1,-1},{0,0,1,2,2,2},80] (* _Harvey P. Dale_, Jun 06 2024 *)
%Y A194222 Cf. A118015.
%K A194222 nonn,easy
%O A194222 1,4
%A A194222 _Clark Kimberling_, Aug 19 2011