

A006340


An "etasequence": [ (n+1)*tau + 1/2 ]  [ n*tau + 1/2 ], tau = (1 + sqrt(5))/2.
(Formerly M0100)


4



2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1
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OFFSET

0,1


COMMENTS

Equals its own "derivative", which is formed by counting the strings of 1's that lie between 2's.
Conjecture: A006340 = continued fraction expansion of (2.729967741... = sup{f(n,1)}), where f(1,x) = x + 1 and thereafter f(n,x) = x + 1 if n is in the lower Wythoff sequence (A000201), else f(n,x) = 1/x. The first 12 values of f(n,1) are given in Example at A245216.  Clark Kimberling, Jul 14 2014
From Michel Dekking, Mar 05 2018: (Start)
The description of this sequence is not correct, since the derivative of a equals
a' = 1,2,2,1,2,2,1,2,1,2,2,1,2,1,2,2,...
The claim by Hofstadter in formula (4) in the 1977 letter to Sloane is also not correct, since the second derivative of a is equal to
a'' = 2,2,1,2,1,2,2,1,2,1,2,2,1,...
so a is not equal to its own second derivative.
Nevertheless, this sequence has a selfsimilarity property: if one replaces every chunk 212 with 1 and every chunk 21212 with 2, then one obtains back the original sequence. In other words, (a(n)) is the unique fixed point of the morphism sigma given by sigma: 1>212, 2>21212.
This can be proved following the ideas of Chapter 2 in Lothaire's book and Section 4 of my paper "Substitution invariant Sturmian words and binary trees".
To comply with these references change the alphabet to {0,1}. This changes sigma into the morphism 0>101, 1>10101.
The fractional part {tau} of tau is larger than 1/2; as it is convenient to have it smaller than 1/2 we change to beta = 1tau = (3sqrt(5))/2.
This changes the morphism 0>101, 1>10101 to its mirror image psi given by 0>01010, 1>010.
Let psi_1 and psi_2 be the elementary Sturmian morphisms given by
psi_1(0)=01 , psi_1(1)=1, psi_2(0)=10, psi_2(1)=0.
Then psi = psi_2^2 psi_1.
This already shows that psi generates a Sturmian sequence with certain parameters alpha and rho: s(alpha,rho) = ([(n+1)*alpha+rho][n*alpha+rho]).
Since psi is the composition psi_2^2psi_1, the parameters of s(alpha,rho) are given by the composition T:=T_2^2T_1 of the fractional linear maps
T_1(x,y) = ((1x)/(2x),(1y)/(2x)),
T_2(x,y) = ((1x)/(2x), (2xy)/(2x)).
Since one can verify that T(beta,1/2)=(beta,1/2), it follows that
alpha = beta, and rho = 1/2.
(End)


REFERENCES

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n = 0..1000
Michel Dekking, Substitution invariant Sturmian words and binary trees, arXiv:1705.08607 [math.CO], (2017).
Michel Dekking, Substitution invariant Sturmian words and binary trees, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.
D. R. Hofstadter, EtaLore [Cached copy, with permission]
D. R. Hofstadter, PiMu Sequences [Cached copy, with permission]
D. R. Hofstadter and N. J. A. Sloane, Correspondence, 1977 and 1991
M. Lothaire, Algebraic combinatorics on words, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.


MATHEMATICA

Differences[ Table[ Round[ GoldenRatio*n], {n, 0, 93}]] (* JeanFrançois Alcover, Aug 13 2012 *)


PROG

(PARI) rt(n) = my(tau=(1 + sqrt(5))/2); round(tau*n)
a(n) = rt(n+1)rt(n) \\ Felix Fröhlich, Aug 26 2018


CROSSREFS

Differs from A014675 in many places. Cf. A245216.
Sequence in context: A172155 A080573 A186440 * A270641 A076371 A175044
Adjacent sequences: A006337 A006338 A006339 * A006341 A006342 A006343


KEYWORD

nonn,easy,nice


AUTHOR

D. R. Hofstadter, Jul 15 1977


EXTENSIONS

Extended by N. J. A. Sloane, Nov 07 2001


STATUS

approved



