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a(n) = [1/r]+[2/r]+...+[n/r], where r=sqrt(2) and []=floor.
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%I #10 Sep 08 2022 08:45:55

%S 0,1,3,5,8,12,16,21,27,34,41,49,58,67,77,88,100,112,125,139,153,168,

%T 184,200,217,235,254,273,293,314,335,357,380,404,428,453,479,505,532,

%U 560,588,617,647,678,709,741,774,807,841,876

%N a(n) = [1/r]+[2/r]+...+[n/r], where r=sqrt(2) and []=floor.

%C A183139 + A183140 = A000217 (the triangular numbers).

%H G. C. Greubel, <a href="/A183139/b183139.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = [1/r]+[2/r]+...+[n/r], where r=sqrt(2) and []=floor.

%t Accumulate[Floor[Range[100]/(Sqrt[2])]] (* _G. C. Greubel_, Apr 10 2018 *)

%o (PARI) for(n=1, 100, print1(sum(k=1,n, floor(k/sqrt(2))), ", ")) \\ _G. C. Greubel_, Apr 10 2018

%o (Magma) [(&+[Floor(k/Sqrt(2)):k in [1..n]]): n in [1..100]]; // _G. C. Greubel_, Apr 10 2018

%Y Cf. A183139, A183140, A000217.

%K nonn

%O 1,3

%A _Clark Kimberling_, Dec 26 2010