cp's OEIS Frontend

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

%I A060143 #70 Jan 19 2025 14:28:11
%S A060143 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,
%T A060143 17,17,18,19,19,20,21,21,22,22,23,24,24,25,25,26,27,27,28,29,29,30,30,
%U A060143 31,32,32,33,33,34,35,35,36,37,37,38,38,39,40,40,41,42,42,43,43,44,45,45
%N A060143 a(n) = floor(n/tau), where tau = (1 + sqrt(5))/2.
%C A060143 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]
%C A060143 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
%C A060143 a(10^n) gives the first few digits of phi=(sqrt(5)-1)/2.
%C A060143 Comment corrected, two alternative ways, by _Peter Munn_, Jan 10 2018: (Start)
%C A060143 (a(n) = a(n+1) or a(n) = a(n-1)) if and only if a(n) is in A066096.
%C A060143 a(n+1) = a(n+2) if and only if n is in A003622.
%C A060143 (End)
%C A060143 From _Wolfdieter Lang_, Jun 28 2011: (Start)
%C A060143 a(n+1) counts for n >= 1 the number of Wythoff A-numbers not exceeding n.
%C A060143 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.
%C A060143 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).
%C A060143 a(n+1) = A(n+1)-(n+1) (serving, together with the last equation, as definition for A(n+1), given the rabbit sequence).
%C A060143 a(n+1) = A005206(n), n >= 0.
%C A060143 (End)
%C A060143 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
%H A060143 William A. Tedeschi, <a href="/A060143/b060143.txt">Table of n, a(n) for n = 0..10000</a> [This replaces an earlier b-file computed by Harry J. Smith]
%H A060143 M. Celaya and F. Ruskey (Proposers), <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.121.06.549">Another property of only the golden ratio</a>, Problem 11651, Amer. Math. Monthly, 121 (2014), 550-551.
%H A060143 F. Michel Dekking, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Dekking/dekk4.html">Morphisms, Symbolic Sequences, and Their Standard Forms</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
%H A060143 Vincent Granville and Jean-Paul Rasson, <a href="http://dx.doi.org/10.1016/0022-314X(88)90020-0">A strange recursive relation</a>, J. Number Theory 30 (1988), no. 2, 238--241. MR0961919(89j:11014). - From _N. J. A. Sloane_, Jun 13 2014
%H A060143 Jeffrey Shallit, <a href="https://arxiv.org/abs/2501.08823">The Hurt-Sada Array and Zeckendorf Representations</a>, arXiv:2501.08823 [math.NT], 2025. See p. 2.
%F A060143 a(n) = floor(phi(n)), where phi=(sqrt(5)-1)/2. [corrected by _Casey Mongoven_, Jul 18 2008]
%F A060143 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]
%F A060143 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]
%F A060143 A006336(n) = A006336(n-1) + A006336(a(n)) for n>1. - _Reinhard Zumkeller_, Oct 24 2007
%F A060143 a(n) = floor(n*phi) - n, where phi = (1+sqrt(5))/2. - _William A. Tedeschi_, Mar 06 2008
%F A060143 Celaya and Ruskey give an interesting formula for a(n). - _N. J. A. Sloane_, Jun 13 2014
%e A060143 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
%e A060143 From _Wolfdieter Lang_, Jun 28 2011: (Start)
%e A060143 There are a(4) = 2 (positive) Wythoff A-numbers <= 3, namely 1 and 3.
%e A060143 There are a(4) = 2 (positive) Wythoff B-numbers < A(4) = 6, namely 2 and 5.
%e A060143 a(4) = 2 = A(4) - 4 = 6 - 4.
%e A060143 (End)
%t A060143 Floor[Range[0,80]/GoldenRatio] (* _Harvey P. Dale_, May 09 2013 *)
%o A060143 (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
%o A060143 (Magma) [Floor(2*n/(1+Sqrt(5))): n in [0..80]]; // _Vincenzo Librandi_, Mar 29 2015
%o A060143 (Python)
%o A060143 from math import isqrt
%o A060143 def A060143(n): return (n+isqrt(5*n**2)>>1)-n # _Chai Wah Wu_, Aug 10 2022
%Y A060143 Cf. A000045 (Fibonacci numbers), A003622, A022342, A035336.
%Y A060143 Terms that occur only once: A001950.
%Y A060143 Terms that occur twice: A066096 (a version of A000201).
%Y A060143 Numerator sequences for other values, as described in Robert A. Stump's 2002 comment: A074065 (sqrt(3)), A074840 (sqrt(2)).
%Y A060143 Apart from initial terms, same as A005206.
%Y A060143 First differences: A096270 (a version of A005614).
%Y A060143 Partial sums: A183136.
%K A060143 nonn,frac,easy
%O A060143 0,5
%A A060143 _Clark Kimberling_, Mar 05 2001
%E A060143 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
%E A060143 More terms from _William A. Tedeschi_, Mar 06 2008