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a(n) = [Sum_{k=1..n} (3k/8)], where [ ]=floor, ( )=fractional part.
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%I #21 Dec 12 2024 11:29:18

%S 0,1,1,1,2,2,3,3,3,4,4,5,6,6,7,7,7,8,8,8,9,9,10,10,10,11,11,12,13,13,

%T 14,14,14,15,15,15,16,16,17,17,17,18,18,19,20,20,21,21,21,22,22,22,23,

%U 23,24,24,24,25,25,26,27,27,28,28,28,29,29,29,30,30,31,31,31,32

%N a(n) = [Sum_{k=1..n} (3k/8)], where [ ]=floor, ( )=fractional part.

%H G. C. Greubel, <a href="/A194237/b194237.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Rec#order_17">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1).

%F From _Chai Wah Wu_, Jun 10 2020: (Start)

%F a(n) = a(n-1) + a(n-16) - a(n-17) for n > 17.

%F G.f.: x*(x^14 + x^12 + x^11 + x^9 + x^6 + x^4 + x)/(x^17 - x^16 - x + 1). (End)

%t r = 3/8;

%t a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]

%t Table[a[n], {n, 1, 90}] (* A194237 *)

%t s[n_] := Sum[a[k], {k, 1, n}]

%t Table[s[n], {n, 1, 100}] (* A194238 *)

%Y Cf. A194238.

%K nonn

%O 1,5

%A _Clark Kimberling_, Aug 20 2011