

A000201


Lower Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi), where phi = (1+sqrt(5))/2 = A001622.
(Formerly M2322 N0917)


310



1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 46, 48, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 101, 103, 105, 106, 108, 110
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OFFSET

1,2


COMMENTS

This is the unique sequence a satisfying a'(n)=a(a(n))+1 for all n in the set N of natural numbers, where a' denotes the ordered complement (in N) of a.  Clark Kimberling, Feb 17 2003
This sequence and A001950 may be defined as follows. Consider the maps a > ab, b > a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0).  Philippe Deléham, Feb 20 2004
These are the numbers whose lazy Fibonacci representation (see A095791) includes 1; the complementary sequence (the upper Wythoff sequence, A001950) are the numbers whose lazy Fibonacci representation includes 2 but not 1.
a(n) is the unique monotonic sequence satisfying a(1)=1 and the condition "if n is in the sequence then n+(rank of n) is not in the sequence" (e.g. a(4)=6 so 6+4=10 and 10 is not in the sequence)  Benoit Cloitre, Mar 31 2006
Write A for A000201 and B for A001950 (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB,...,BBB,... appear in many complementary equations having solution A000201 (or equivalently, A001950). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864).  Clark Kimberling, Nov 14 2007
The lower Wythoff sequence also can be constructed by playing the socalled Mancalagame: n piles of total d(n) chips are standing in a row. The piles are numbered from left to right by 1, 2, 3, ... . The number of chips in a pile at the beginning of the game is equal to the number of the pile. One step of the game is described as follows: Distribute the pile on the very left one by one to the piles right of it. If chips are remaining, build piles out of one chip subsequently to the right. After f(n) steps the game ends in a constant row of piles. The lower Wythoff sequence is also given by n > f(n).  Roland Schroeder (florola(AT)gmx.de), Jun 19 2010
With the exception of the first term, a(n) gives the number of iterations required to reverse the list {1,2,3,...,n} when using the mapping defined as follows: remove the first term of the list, z(1), and add 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list. See A183110 where this mapping is used and other references given. This appears to be essentially the Mancalatype game interpretation given by R. Schroeder above.  John W. Layman, Feb 03 2011
Numbers k such that {k*phi} > phi^(2), where {} denotes the fractional part.
Proof: Write m = floor(k*phi).
If {k*phi} > phi^(2), take s = mk+1. From m < k*phi < m+1 we have k < (mk+1)*phi < k + phi, so floor(s*phi) = k or k+1. If floor(s*phi) = k+1, then (see A003622) floor((k+1)*phi) = floor(floor(s*phi)*phi) = floor(s*phi^2)1 = s+floor(s*phi)1 = m+1, but actually we have (k+1)*phi > m+phi+phi^(2) = m+2, a contradiction. Hence floor(s*phi) = k.
If floor(s*phi) = k, suppose otherwise that k*phi  m <= phi^(2), then m < (k+1)*phi <= m+2, so floor((k+1)*phi) = m+1. Suppose that A035513(p,q) = k for p,q >= 1, then A035513(p,q+1) = floor((k+1)*phi)  1 = m = A035513(s,1). But it is impossible for one number (m) to occur twice in A035513. (End)
The formula from Jianing Song above is a direct consequence of an old result by Carlitz et al. (1972). Their Theorem 11 states that (a(n)) consists of the numbers k such that {k*phi^(2)} < phi^(1). One has {k*phi^(2)} = {k*(2phi)} = {k*phi}. Using that 1phi^(1) = phi^(2), the Jianing Song formula follows.  Michel Dekking, Oct 14 2023


REFERENCES

Eric Friedman, Scott M. Garrabrant, Ilona K. PhippsMorgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, date?
M. Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
I. M. Yaglom, Two games with matchsticks, pp. 17 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.


LINKS

L. Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Fibonacci representations, Fib. Quart., Vol. 10, No. 1 (1972), pp. 128.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, 43 pages, no date, unpublished.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, unpublished, no date [Cached copy, with permission]
A. S. Fraenkel, Ratwyt, December 28 2011.
Clark Kimberling, Problem Proposals, The Fibonacci Quarterly, vol. 52 #5, 2015, p514.
Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319337.[See A317208 for a link.]
D. J. Newman, Problem 3117, Amer. Math. Monthly, 34 (1927), 158159.
D. J. Newman, Problem 5252, Amer. Math. Monthly, 72 (1965), 11441145.


FORMULA

Zeckendorf expansion of n (cf. A035517) ends with an even number of 0's.
Other properties: a(1)=1; for n>1, a(n) is taken to be the smallest integer greater than a(n1) which is consistent with the condition "n is in the sequence if and only if a(n)+1 is not in the sequence".
a(1) = 1; for n>0, a(n+1) = a(n)+1 if n is not in the sequence, a(n+1) = a(n)+2 if n is in the sequence.
a(a(n)) = floor(n*phi^2)  1 = A003622(n).
{a(k)} union {a(k)+1} = {1, 2, 3, 4, ...}. Hence a(1) = 1; for n>1, a(a(n)) = a(a(n)1)+2, a(a(n)+1) = a(a(n))+1.  Benoit Cloitre, Mar 08 2003
{a(n)} is a solution to the recurrence a(a(n)+n) = 2*a(n)+n, a(1)=1 (see Barbeau et al.).
a(Fibonacci(r1)+j) = Fibonacci(r)+a(j) for 0 < j <= Fibonacci(r2); 2 < r.  Paul Weisenhorn, Aug 18 2012


EXAMPLE

From Roland Schroeder (florola(AT)gmx.de), Jul 13 2010: (Start)
Example for n = 5; a(5) = 8;
(Start: [1,2,3,4,5]; 8 steps until [5,4,3,2,1]):
[1,2,3,4,5]; [3,3,4,5]; [4,5,6]; [6,7,1,1]; [8,2,2,1,1,1]: [3,3,2,2,2,1,1,1]; [4,3,3,2,1,1,1]; [4,4,3,2,1,1]; [5,4,3,2,1]. (End)


MAPLE

Digits := 100; t := evalf((1+sqrt(5))/2); A000201 := n>floor(t*n);


MATHEMATICA

Table[Floor[N[n*(1+Sqrt[5])/2]], {n, 1, 75}]


PROG

(PARI) a(n)=floor(n*(sqrt(5)+1)/2)
(Maxima) makelist(floor(n*(1+sqrt(5))/2), n, 1, 60); /* Martin Ettl, Oct 17 2012 */
(Haskell)
a000201 n = a000201_list !! (n1)
a000201_list = f [1..] [1..] where
f (x:xs) (y:ys) = y : f xs (delete (x + y) ys)
(Python)
def aupton(terms):
alst, aset = [None, 1], {1}
for n in range(1, terms):
an = alst[n] + (1 if n not in aset else 2)
alst.append(an); aset.add(an)
return alst[1:]
(Python)
from math import isqrt


CROSSREFS

A185615 gives values n such that n divides A000201(n)^m for some integer m>0.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841.  N. J. A. Sloane, Mar 11 2021


KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



