A060143 a(n) = floor(n/tau), where tau = (1 + sqrt(5))/2.

0, 0, 1, 1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45

Offset: 0

Author

Clark Kimberling, Mar 05 2001

Keywords

• nonn
• frac
• easy

Comments

Fibonacci base shift right: for n >= 0, a(n+1) = Sum_{k in A_n} F_{k-1}, where n = Sum_{k in A_n} F_k (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >=2). - Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001 [corrected, and aligned with sequence offset by Peter Munn, Jan 10 2018]

Numerators a(n) of fractions slowly converging to phi, the golden ratio: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to (1 + sqrt(5))/2. For all n, a(n) / b(n) < (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002

a(10^n) gives the first few digits of phi=(sqrt(5)-1)/2.

Comment corrected, two alternative ways, by Peter Munn, Jan 10 2018: (Start)

(a(n) = a(n+1) or a(n) = a(n-1)) if and only if a(n) is in A066096.

a(n+1) = a(n+2) if and only if n is in A003622.

(End)

From Wolfdieter Lang, Jun 28 2011: (Start)

a(n+1) counts for n >= 1 the number of Wythoff A-numbers not exceeding n.

a(n+1) counts also the number of Wythoff B-numbers smaller than A(n+2), with the Wythoff A- and B-sequences A000201 and A001950, respectively.

a(n+1) = Sum_{j=1..n} A005614(j-1) for n >= 1 (no rounding problems like in the above definition, because the rabbit sequence A005614(n-1) for n >= 1, can be defined by a substitution rule).

a(n+1) = A(n+1)-(n+1) (serving, together with the last equation, as definition for A(n+1), given the rabbit sequence).

a(n+1) = A005206(n), n >= 0.

(End)

Let b(n) = floor((n+1)/phi). Then b(n) + b(b(n-1)) = n [Granville and Rasson]. - N. J. A. Sloane, Jun 13 2014

Examples

			a(6)= 3 so b(6) = 6 - 3 = 3. a(7) = 4 because (a(6) + 1) / b(6) = 4/3 which is < (1 + sqrt(5))/2. So b(7) = 7 - 4 = 3. a(8) = 4 because (a(7) + 1) / b(7) = 5/3 which is > (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002
From _Wolfdieter Lang_, Jun 28 2011: (Start)
There are a(4) = 2 (positive) Wythoff A-numbers <= 3, namely 1 and 3.
There are a(4) = 2 (positive) Wythoff B-numbers < A(4) = 6, namely 2 and 5.
a(4) = 2 = A(4) - 4 = 6 - 4.
(End)
		

Links

• William A. Tedeschi, Table of n, a(n) for n = 0..10000 [This replaces an earlier b-file computed by Harry J. Smith]
• M. Celaya and F. Ruskey (Proposers), Another property of only the golden ratio, Problem 11651, Amer. Math. Monthly, 121 (2014), 550-551.
• F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
• Vincent Granville and Jean-Paul Rasson, A strange recursive relation, J. Number Theory 30 (1988), no. 2, 238--241. MR0961919(89j:11014). - From _N. J. A. Sloane_, Jun 13 2014
• Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 2.

Crossrefs

Cf. A000045 (Fibonacci numbers), A003622, A022342, A035336.

Terms that occur only once: A001950.

Terms that occur twice: A066096 (a version of A000201).

Numerator sequences for other values, as described in Robert A. Stump's 2002 comment: A074065 (sqrt(3)), A074840 (sqrt(2)).

Apart from initial terms, same as A005206.

First differences: A096270 (a version of A005614).

Partial sums: A183136.

Programs

Magma

[Floor(2*n/(1+Sqrt(5))): n in [0..80]]; // Vincenzo Librandi, Mar 29 2015

Mathematica

Floor[Range[0,80]/GoldenRatio] (* Harvey P. Dale, May 09 2013 *)

PARI

{ default(realprecision, 10); p=(sqrt(5) - 1)/2; for (n=0, 1000, write("b060143.txt", n, " ", floor(n*p)); ) } \\ Harry J. Smith, Jul 02 2009

Python

from math import isqrt
def A060143(n): return (n+isqrt(5*n**2)>>1)-n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = floor(phi(n)), where phi=(sqrt(5)-1)/2. [corrected by Casey Mongoven, Jul 18 2008]

a(F_n + 1) = F_{n-1} if F_n is the n-th Fibonacci number. [aligned with sequence offset by Peter Munn, Jan 10 2018]

a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002 [corrected by Peter Munn, Jan 07 2018]

A006336(n) = A006336(n-1) + A006336(a(n)) for n>1. - Reinhard Zumkeller, Oct 24 2007

a(n) = floor(n*phi) - n, where phi = (1+sqrt(5))/2. - William A. Tedeschi, Mar 06 2008

Celaya and Ruskey give an interesting formula for a(n). - N. J. A. Sloane, Jun 13 2014

Extensions

I merged three identical sequences to create this entry. Some of the formulas may need their initial terms adjusting now. - N. J. A. Sloane, Mar 05 2003

More terms from William A. Tedeschi, Mar 06 2008