A054347 Partial sums of A000201.

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

Offset: 0

Author

N. J. A. Sloane, May 06 2000

Keywords

• nonn
• easy

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.

Crossrefs

Cf. A000201.

Programs

Mathematica

Accumulate[Table[Floor[GoldenRatio n], {n, 0, 30}]] (* Birkas Gyorgy, May 06 2011 *)

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))
A054347_list = list(islice(A054347_gen(),30)) # Chai Wah Wu, Aug 29 2022

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