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A054347 Partial sums of A000201. 5

%I

%S 0,1,4,8,14,22,31,42,54,68,84,101,120,141,163,187,212,239,268,298,330,

%T 363,398,435,473,513,555,598,643,689,737,787,838,891,946,1002,1060,

%U 1119,1180,1243,1307,1373,1440,1509,1580,1652,1726,1802,1879

%N Partial sums of A000201.

%C From _Michel Dekking_, Aug 19 2019: (Start)

%C As n-> oo, a(n)/(n*(n+1)) -> phi/2.

%C Proof: Let {alpha} be the fractional part of a real number alpha.

%C a(n) = [phi]+[2*phi]+...+[n*phi] = phi+{phi} + 2*phi + {2*phi} +...+ n*phi + {n*phi} = n*(n+1)*phi/2 + [{phi}+{2*phi}+...+{n*phi}].

%C When we divide by n*(n+1) this tends to phi/2, since the second term is bounded by n.

%C (End)

%H T. D. Noe, <a href="/A054347/b054347.txt">Table of n, a(n) for n = 0..10000</a>

%H M. Griffiths, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.118.06.497">The Golden String, Zeckendorf Representations, and the Sum of a Series</a>, Amer. Math. Monthly, 118 (2011), 497-507.

%F a(n) = floor( n*(n+1)/2*phi -n/2) + 0 or +1 - _Benoit Cloitre_, Oct 03 2003

%F a(n) = floor( n*(n+1)/2*phi -n/2) + 0, +1, or -1 (n=7920, 18762, 18851,...), or +2 (n=12815, 15841, 30358, 30382,...) if n<2000000. - _Birkas Gyorgy_, May 06 2011

%t Accumulate[Table[Floor[GoldenRatio n], {n, 0, 30}]] (* _Birkas Gyorgy_, May 06 2011 *)

%o (PARI) for(n=0,50, print1(sum(k=0,n, floor(k*(1+sqrt(5))/2)), ", ")) \\ _G. C. Greubel_, Oct 06 2017

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, May 06 2000

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Last modified May 15 01:50 EDT 2021. Contains 343909 sequences. (Running on oeis4.)