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%I #53 Aug 30 2022 02:12:30
%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 Limit_{n->oo} a(n)/(n*(n+1)) = phi/2.
%C Proof: Let {alpha} be the fractional part of a real number alpha and let [alpha] = floor(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
%o (Python)
%o from math import isqrt
%o from itertools import count, islice, accumulate
%o def A054347_gen(): # generator of terms
%o return accumulate(n+isqrt(5*n**2)>>1 for n in count(0))
%o A054347_list = list(islice(A054347_gen(),30)) # _Chai Wah Wu_, Aug 29 2022
%Y Cf. A000201.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, May 06 2000
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