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a(n) = Sum_{k >=1} floor(n*sqrt(2)^(3-k)).
1

%I #16 Sep 08 2022 08:44:45

%S 3,9,16,22,28,36,40,49,55,62,66,76,82,87,94,103,111,116,121,130,134,

%T 141,150,157,163,170,176,182,190,196,200,212,217,227,231,238,245,250,

%U 256,266,270,277,283,291,295,307,312,320,327,333,341,347,352,360,364

%N a(n) = Sum_{k >=1} floor(n*sqrt(2)^(3-k)).

%H G. C. Greubel, <a href="/A020967/b020967.txt">Table of n, a(n) for n = 1..1000</a>

%H C. Kimberling, <a href="http://www.jstor.org/stable/2975195">Problem 10520</a>, Amer. Math. Mon. 103 (1996) p. 347.

%t Table[Sum[Floor[n*Sqrt[2]^(3-k)], {k, 1, 1000}] -1, {n,1,50}] (* _G. C. Greubel_, May 31 2018 *)

%o (PARI) j=[]; for(n=1,100,j=concat(j,suminf(k=1,floor(n*sqrt(2)^(3-k) )))); j

%o (Magma) [(&+[Floor(n*Sqrt(2)^(3-k)): k in [1..1000]]): n in [1..30]]; // _G. C. Greubel_, May 31 2018

%K nonn

%O 1,1

%A _Clark Kimberling_

%E More terms from _Jason Earls_, Jul 28 2001