

A001950


Upper Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi = (1+sqrt(5))/2.
(Formerly M1332 N0509)


160



2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 109, 112, 115, 117, 120, 123, 125, 128, 130, 133, 136, 138, 141, 143, 146, 149, 151, 154, 157
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OFFSET

1,1


COMMENTS

Indices at which blocks (1;0) occur in infinite Fibonacci word; i.e., n such that A005614(n2) = 0 and A005614(n1) = 1.  Benoit Cloitre, Nov 15 2003
A000201 and this sequence 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
a(n) = nth integer which is not equal to the floor of any multiple of phi, where phi = (1+sqrt(5))/2 = golden number.  Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), May 09 2007
a(n) = Min(m: A134409(m) = A006336(n)).  Reinhard Zumkeller, Oct 24 2007
Write A for A000201 and B for the present sequence (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, the present sequence). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864).  Clark Kimberling, Nov 14 2007
Apart from the initial 0 in A090909, is this the same as that sequence?  Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
If we define a basephi integer as a positive number whose representation in the golden ratio base consists only of nonnegative powers of phi, and if these basephi integers are ordered in increasing order (beginning 1, phi, ...), then it appears that the difference between the nth and (n1)th basephi integer is phi1 if and only if n belongs to this sequence, and the difference is 1 otherwise. Further, if each basephi integer is written in linear form as a + b*phi (for example, phi^2 is written as 1 + phi), then it appears that there are exactly two basephi integers with b=n if and only if n belongs to this sequence, and exactly three basephi integers with b=n otherwise.  Geoffrey Caveney, Apr 17 2014


REFERENCES

C. Berge, Graphs and Hypergraphs, NorthHolland, 1973; p. 324, Problem 2.
M. Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
Michel Rigo, Invariant games and nonhomogeneous Beatty sequences, Slides of a talk, Journée de Mathématiques Discrètes, 2015; http://orbi.ulg.be/bitstream/2268/177711/1/Rigo.pdf
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

T. D. Noe, Table of n, a(n) for n=1..1000
L. Carlitz, R. Scoville and T. Vaughan, Some arithmetic functions related to Fibonacci numbers, Fib. Quart., 11 (1973), 337386.
I. G. Connell, Some properties of Beatty sequences I, Canad. Math. Bull., 2 (1959), 190197.
N. Fox, On Aperiodic Subtraction Games with Bounded Nim Sequence, arXiv preprint arXiv:1407.2823, 2014
A. S. Fraenkel, How to beat your Wythoff games' opponent on three fronts, Amer. Math. Monthly, 89 (1982), 353361 (the case a=1).
A. S. Fraenkel, Ratwyt, December 28 2011.
Aviezri S. Fraenkel, Complementary iterated floor words and the Flora game, SIAM J. Discrete Math. 24 (2010), no. 2, 570588.  N. J. A. Sloane, May 06 2011
M. Griffiths, The Golden String, Zeckendorf Representations, and the Sum of a Series, Amer. Math. Monthly, 118 (2011), 497507.
C. Kimberling, A SelfGenerating Set and the Golden Mean, J. Integer Sequences, 3 (2000), #00.2.8.
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
C. Kimberling, Complementary equations and Wythoff Sequences, JIS 11 (2008) 08.3.3
D. J. Newman, Problem 5252, Amer. Math. Monthly, 72 (1965), 11441145.
Vincent Russo and Loren Schwiebert, Beatty Sequences, Fibonacci Numbers, and the Golden Ratio, The Fibonacci Quarterly, Vol 49, Number 2, May 2011.
X. Sun, Wythoff's sequence and NHeap Wythoff's conjectures, Discr. Math., 300 (2005), 180195.
J. C. Turner, The alpha and the omega of the Wythoff pairs, Fib. Q., 27 (1989), 7686.
Eric Weisstein's World of Mathematics, Beatty Sequence, MathWorld.
Eric Weisstein's World of Mathematics, Golden ratio, MathWorld.
Eric Weisstein's World of Mathematics, Wythoff's Game, MathWorld.
Eric Weisstein's World of Mathematics, Wythoff Array
Index entries for sequences related to Beatty sequences


FORMULA

a(n) = n + floor(2*n*phi). In general, floor(n*phi^m) = Fibonacci(m1)*n + floor(Fibonacci(m)*n*phi).  Benoit Cloitre, Mar 18 2003
a(n) = n + floor(n*phi) = n + A000201(n).  Paul Weisenhorn and Philippe Deléham
Append a 0 to the Zeckendorf expansion (cf. A035517) of nth term of A000201.
a(n) = A003622(n) + 1.  Philippe Deléham, Apr 30 2004
If a'=A000201 is the ordered complement (in N) of {a(n)}, then a(Fib(r2)+j) = Fib(r)+a(j) for 0<j<=Fib(r2), 3<r; and a'(Fib(r1)+j) = Fib(r)+a'(j) for 0<j<=Fib(r2), 2<r.  Paul Weisenhorn, Aug 18 2012
with a(1)=2, a(2)=5, a'(1)=1, a'(2)=3 and 1<k and a(k1)<n<=a(k) one gets a(n)=3*nk, a'(n)=2*nk.  Paul Weisenhorn, Aug 21 2012


EXAMPLE

From Paul Weisenhorn, Aug 18 2012 and Aug 21 2012: (Start)
a(14)=floor(14*phi^2)=36; a'(14)=floor(14*phi)=22;
with r=9 and j=1: a(13+1)=34+2=36;
with r=8 and j=1: a'(13+1)=21+1=22.
k=6 and a(5)=13 < n <= a(6)=15
a(14)=3*146=36; a'(14)=2*146=22;
a(15)=3*156=39; a'(15)=2*156=24. (End)


MATHEMATICA

Table[Floor[N[n*(1+Sqrt[5])^2/4]], {n, 1, 75}]
Array[ Floor[ #*GoldenRatio^2] &, 60] (* Robert G. Wilson v, Apr 17 2010 *)


PROG

(PARI) a(n)=floor(n*(sqrt(5)+3)/2)
(PARI) A001950(n)=(sqrtint(n^2*5)+n*3)\2 \\ M. F. Hasler, Sep 17 2014
(Haskell)
a001950 n = a000201 n + n  Reinhard Zumkeller, Mar 10 2013


CROSSREFS

a(n) = greatest k such that s(k) = n, where s = A026242. Complement of A000201.
A002251 maps between A000201 and A001950, in that A002251(A000201(n)) = A001950(n), A002251(A001950(n)) = A000201(n).
Cf. A026352, A004976, A004919.
Sequence in context: A018717 A188036 A090909 * A022841 A047480 A038127
Adjacent sequences: A001947 A001948 A001949 * A001951 A001952 A001953


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane, Apr 30 1991


EXTENSIONS

Corrected by Michael Somos, Jun 07 2000


STATUS

approved



