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

0,3

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

Crossrefs

Cf. A000201.

Sequence in context: A004797 A053459 A024398 * A194149 A351362 A003682

Adjacent sequences: A054344 A054345 A054346 * A054348 A054349 A054350

Keyword

nonn,easy

Author

N. J. A. Sloane, May 06 2000

Status

approved