|
%I #16 Sep 08 2022 08:44:45
%S 4,11,28,70,171,417,1010,2444,5905,14263,34440,83154,200759,484685,
%T 1170142,2824984,6820125,16465251,39750644,95966558,231683779,
%U 559334137,1350352074,3260038308,7870428713,19000895759
%N Sum of Floor[ 3*(1+sqrt(2))^(n-k) ] for k from 1 to infinity.
%H G. C. Greubel, <a href="/A020964/b020964.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[t=0; k=0; While[k++; s=Floor[3*(1+Sqrt[2])^(n-k)]; s>0, t= t+s]; t, {n, 26}]
%t Table[Sum[Floor[3*(1+Sqrt[2])^(n-k)], {k, 1, 1000}], {n,0,50}] (* _G. C. Greubel_, Sep 30 2018 *)
%o (PARI) for(n=1,50, print1(sum(k=1,2*n, floor(3*(1+sqrt(2))^(n-k))), ", ")) \\ _G. C. Greubel_, Sep 30 2018
%o (Magma) [(&+[Floor(3*(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.
|
