OFFSET
0,3
COMMENTS
From Michel Dekking, Aug 19 2019: (Start)
Limit_{n->oo} a(n)/(n*(n+1)) = phi/2.
Proof: Let {alpha} be the fractional part of a real number alpha and let [alpha] = floor(alpha).
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}].
When we divide by n*(n+1) this tends to phi/2, since the second term is bounded by n.
(End)
LINKS
T. D. Noe, Table of n, a(n) for n = 0..10000
M. Griffiths, The Golden String, Zeckendorf Representations, and the Sum of a Series, Amer. Math. Monthly, 118 (2011), 497-507.
FORMULA
a(n) = floor(n*(n+1)/2*phi - n/2) + 0 or +1. - Benoit Cloitre, Oct 03 2003
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
MATHEMATICA
Accumulate[Table[Floor[GoldenRatio n], {n, 0, 30}]] (* Birkas Gyorgy, May 06 2011 *)
PROG
(PARI) for(n=0, 50, print1(sum(k=0, n, floor(k*(1+sqrt(5))/2)), ", ")) \\ G. C. Greubel, Oct 06 2017
(Python)
from math import isqrt
from itertools import count, islice, accumulate
def A054347_gen(): # generator of terms
return accumulate(n+isqrt(5*n**2)>>1 for n in count(0))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, May 06 2000
STATUS
approved