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A054347 Partial sums of A000201. 5
0, 1, 4, 8, 14, 22, 31, 42, 54, 68, 84, 101, 120, 141, 163, 187, 212, 239, 268, 298, 330, 363, 398, 435, 473, 513, 555, 598, 643, 689, 737, 787, 838, 891, 946, 1002, 1060, 1119, 1180, 1243, 1307, 1373, 1440, 1509, 1580, 1652, 1726, 1802, 1879 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

From Michel Dekking, Aug 19 2019: (Start)

As n-> oo,  a(n)/(n*(n+1)) -> phi/2.

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

CROSSREFS

Sequence in context: A004797 A053459 A024398 * A194149 A003682 A011897

Adjacent sequences:  A054344 A054345 A054346 * A054348 A054349 A054350

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane, May 06 2000

STATUS

approved

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Last modified October 22 15:57 EDT 2019. Contains 328318 sequences. (Running on oeis4.)