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%I #19 Sep 08 2022 08:44:45
%S 2,6,17,45,112,276,671,1627,3934,9506,22957,55433,133836,323120,
%T 780091,1883319,4546746,10976830,26500425,63977701,154455848,
%U 372889420,900234711,2173358867,5246952470,12667263834
%N Sum of Floor[ 2*(1+sqrt(2))^(n-k) ] for k from 1 to infinity.
%H G. C. Greubel, <a href="/A020963/b020963.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.
%F From _Chai Wah Wu_, Jun 02 2016: (Start)
%F a(n) = 3*a(n-1) - 4*a(n-3) + a(n-4) + a(n-5) for n > 5 (conjectured).
%F G.f.: x*(x^4 - 2*x^3 + x^2 - 2)/((x - 1)^2*(x + 1)*(x^2 + 2*x - 1)) (conjectured). (End)
%t Table[t=0; k=0; While[k++; s=Floor[2*(1+Sqrt[2])^(n-k)]; s>0, t=t+s]; t, {n, 26}]
%t Table[Sum[Floor[2*(1 + Sqrt[2])^(n - k)], {k, 1, 1000}], {n,1,50}] (* _G. C. Greubel_, Sep 30 2018 *)
%o (PARI) for(n=1,50, print1(sum(k=1,2*n, floor(2*(1+sqrt(2))^(n-k))), ", ")) \\ _G. C. Greubel_, Sep 30 2018
%o (Magma) [(&+[Floor(2*(1+sqrt(2))^(n-k)): k in [1..2*n]]): n in [1..50]] // _G. C. Greubel_, Sep 30 2018
%K nonn
%O 1,1
%A _Clark Kimberling_
%E Revised Feb 03 1999. Revised Nov 30 2010.
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