

A139250


Toothpick sequence (see Comments lines for definition).


404



0, 1, 3, 7, 11, 15, 23, 35, 43, 47, 55, 67, 79, 95, 123, 155, 171, 175, 183, 195, 207, 223, 251, 283, 303, 319, 347, 383, 423, 483, 571, 651, 683, 687, 695, 707, 719, 735, 763, 795, 815, 831, 859, 895, 935, 995, 1083, 1163, 1199, 1215, 1243, 1279, 1319, 1379
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OFFSET

0,3


COMMENTS

A toothpick is a copy of the closed interval [1,1]. (In the paper, we take it to be a copy of the unit interval [1/2, 1/2].)
We start at stage 0 with no toothpicks.
At stage 1 we place a toothpick in the vertical direction, anywhere in the plane.
In general, given a configuration of toothpicks in the plane, at the next stage we add as many toothpicks as possible, subject to certain conditions:
 Each new toothpick must lie in the horizontal or vertical directions.
 Two toothpicks may never cross.
 Each new toothpick must have its midpoint touching the endpoint of exactly one existing toothpick.
The sequence gives the number of toothpicks after n stages. A139251 (the first differences) gives the number added at the nth stage.
Call the endpoint of a toothpick "exposed" if it does not touch any other toothpick. The growth rule may be expressed as follows: at each stage, new toothpicks are placed so their midpoints touch every exposed endpoint.
This is equivalent to a twodimensional cellular automaton. The animations show the fractallike behavior.
After 2^k  1 steps, there are 2^k exposed endpoints, all located on two lines perpendicular to the initial toothpick. At the next step, 2^k toothpicks are placed on these lines, leaving only 4 exposed endpoints, located at the corners of a square with side length 2^(k1) times the length of a toothpick.  M. F. Hasler, Apr 14 2009 and others. For proof, see the ApplegatePolSloane paper.
If the third condition in the definition is changed to " Each new toothpick must have at exactly one of its endpoints touching the midpoint of an existing toothpick" then the same sequence is obtained. The configurations of toothpicks are of course different from those in the present sequence. But if we start with the configurations of the present sequence, rotate each toothpick a quarterturn, and then rotate the whole configuration a quarterturn, we obtain the other configuration.
If the third condition in the definition is changed to " Each new toothpick must have at least one of its endpoints touching the midpoint of an existing toothpick" then the sequence n^2  n + 1 is obtained, because there are no holes left in the grid.
A "toothpick" of length 2 can be regarded as a polyedge with 2 components, both on the same line. At stage n, the toothpick structure is a polyedge with 2*a(n) components.
Conjecture: Consider the rectangles in the sieve (including the squares). The area of each rectangle (A = b*c) and the edges (b and c) are powers of 2, but at least one of the edges (b or c) is <= 2.
In the toothpick structure, if n >> 1, we can see some patterns that look like "canals" and "diffraction patterns". For example, see the Applegate link "A139250: the movie version", then enter n=1008 and click "Update". See also "Tsquare (fractal)" in the Links section.  Omar E. Pol, May 19 2009, Oct 01 2011
From Benoit Jubin, May 20 2009: The web page "Gallery" of Chris Moore (see link) has some nice pictures that are somewhat similar to the pictures of the present sequence. What sequences do they correspond to?
For a connection to SierpiĆski triangle and Gould's sequence A001316, see the leftist toothpick triangle A151566.
Eric Rowland comments on Mar 15 2010 that this toothpick structure can be represented as a 5state CA on the square grid. On Mar 18 2010, David Applegate showed that three states are enough. See links.
Equals row sums of triangle A160570 starting with offset 1; equivalent to convolving A160552: (1, 1, 3, 1, 3, 5, 7,...) with (1, 2, 2, 2,...). Equals A160762: (1, 0, 2, 2, 2, 2, 2, 6,...) convolved with 2*n  1: (1, 3, 5, 7,...). Starting with offset 1 equals A151548: [1, 3, 5, 7, 5, 11, 17, 15,...] convolved with A078008 signed (A151575): [1, 0, 2, 2, 6, 10, 22, 42, 86, 170, 342,...].  Gary W. Adamson, May 19 2009, May 25 2009
For a threedimensional version of the toothpick structure, see A160160.  Omar E. Pol, Dec 06 2009
From Omar E. Pol, May 20 2010: (Start)
Observation about the arrangement of rectangles:
It appears there is a nice pattern formed by distinct modular substructures: a central cross surrounded by asymmetrical crosses (or "hidden crosses") of distinct sizes and also by "nuclei" of crosses.
Conjectures: after 2^k stages, for k >= 2, and for m = 1 to k  1, there are 4^(m1) substructures of size s = k  m, where every substructure has 4*s rectangles. The total number of substructures is equal to (4^(k1)1)/3 = A002450(k1). For example: If k = 5 (after 32 stages) we can see that:
a) There is a central cross, of size 4, with 16 rectangles.
b) There are four hidden crosses, of size 3, where every cross has 12 rectangles.
c) There are 16 hidden crosses, of size 2, where every cross has 8 rectangles.
d) There are 64 nuclei of crosses, of size 1, where every nucleus has 4 rectangles.
Hence the total number of substructures after 32 stages is equal to 85. Note that in every arm of every substructure, in the potential growth direction, the length of the rectangles are the powers of 2. (See illustrations in the links. See also A160124.) (End)
It appears that the number of grid points that are covered after nth stage of the toothpick structure, assuming the toothpicks have length 2*k, is equal to (2*k2)*a(n) + A147614(n), k > 0. See the formulas of A160420 and A160422.  Omar E. Pol, Nov 13 2010
Version "Gullwing": on the semiinfinite square grid, at stage 1, we place a horizontal "gull" with its vertices at [(1, 2), (0, 1), (1, 2)]. At stage 2, we place two vertical gulls. At stage 3, we place four horizontal gulls. a(n) is also the number of gulls after nth stage. For more information about the growth of gulls see A187220.  Omar E. Pol, Mar 10 2011
From Omar E. Pol, Mar 12 2011: (Start)
Version "Itoothpick": we define an "Itoothpick" to consist of two connected toothpicks, as a bar of length 2. An Itoothpick with length 2 is formed by two toothpicks with length 1. The midpoint of an Itoothpick is touched by its two toothpicks. a(n) is also the number of Itoothpicks after nth stage in the Itoothpick structure. The Itoothpick structure is essentially the original toothpick structure in which every toothpick is replaced by an Itoothpick. Note that in the physical model of the original toothpick structure the midpoint of a wooden toothpick of the new generation is superimposed on the endpoint of a wooden toothpick of the old generation. However, in the physical model of the Itoothpick structure the wooden toothpicks are not overlapping because all wooden toothpicks are connected by their endpoints. For the number of toothpicks in the Itoothpick structure see A160164 which also gives the number of gullwing in a gullwing structure because the gullwing structure of A160164 is equivalent to the Itoothpick structure. It also appears that the gullwing sequence A187220 is a supersequence of the original toothpick sequence A139250 (this sequence).
For the connection with the UlamWarburton cellular automaton see the ApplegatePolSloane paper and see also A160164 and A187220.
(End)
A version in which the toothpicks are connected by their endpoints: on the semiinfinite square grid, at stage 1, we place a vertical toothpick of length 1 from (0, 0). At stage 2, we place two horizontal toothpicks from (0,1), and so on. The arrangement looks like half of the Itoothpick structure. a(n) is also the number of toothpicks after the nth.  Omar E. Pol, Mar 13 2011
Version "Quartercircle" (or Qtoothpick): a(n) is also the number of Qtoothpicks after the nth stage in a Qtoothpick structure in the first quadrant. We start from (0,1) with the first Qtoothpick centered at (1, 1). The structure is asymmetric. For a similar structure but starting from (0, 0) see A187212. See A187210 and A187220 for more information.  Omar E. Pol, Mar 22 2011
Version "Tree": It appears that a(n) is also the number of toothpicks after the nth stage in a toothpick structure constructed following a special rule: the toothpicks of the new generation have length 4 when they are placed on the infinite square grid (note that every toothpick has four components of length 1), but after every stage, one (or two) of the four components of every toothpick of the new generation is removed, if such component contains an endpoint of the toothpick and if such endpoint is touching the midpoint or the endpoint of another toothpick. The truncated endpoints of the toothpicks remain exposed forever. Note that there are three sizes of toothpicks in the structure: toothpicks of lengths 4, 3 and 2. A159795 gives the total number of components in the structure after the nth stage. A153006 (the corner sequence of the original version) gives 1/4 of the total of components in the structure after the nth stage.  Omar E. Pol, Oct 24 2011
From Omar E. Pol, Sep 16 2012: (Start)
It appears that a(n)/A147614(n) converges to 3/4.
It appears that a(n)/A160124(n) converges to 3/2.
It appears that a(n)/A139252(n) converges to 3.
Also:
It appears that A147614(n)/A160124(n) converges to 2.
It appears that A160124(n)/A139252(n) converges to 2.
It appears that A147614(n)/A139252(n) converges to 4.
(End)
It appears that a(n) is also the total number of ON cells after nth stage in a quadrant of the structure of the cellular automaton described in A169707 plus the total number of ON cells after n+1 stages in a quadrant of the mentioned structure, without its central cell. See the illustration of the NWNESESW version in A169707. See also the connection between A160164 and A169707.  Omar E. Pol, Jul 26 2015


REFERENCES

D. Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157191
L. D. Pryor, The Inheritance of Inflorescence Characters in Eucalyptus, Proceedings of the Linnean Society of New South Wales, V. 79, (1954), p. 81, 83.
Richard P. Stanley, Enumerative Combinatorics, volume 1, second edition, chapter 1, exercise 95, figure 1.28, Cambridge University Press (2012), p. 120, 166.


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 0..65535
David Applegate, The movie version
David Applegate, Animation of first 32 stages
David Applegate, Animation of first 64 stages
David Applegate, Animation of first 128 stages
David Applegate, Animation of first 256 stages
David Applegate, C++ program to generate these animations  creates postscript for a specific n
David Applegate, Generates many postscripts, converts them to gifs, and glues the gifs together into an animation
David Applegate, Generates bfiles for A139250, A139251, A147614
David Applegate, The bfiles for A139250, A139251, A147614 sidebyside
David Applegate, A threestate CA for the toothpick structure
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, which is also available at arXiv:1004.3036v2
Barry Cipra, What comes next?, Science (AAAS) 327: 943.
Steven Finch, Toothpicks and live cells
Mats Granvik, Additional illustration: Number blocks where each number tells how many times a point on the square grid is crossed or connected to by a toothpick, Jun 21 2009.
Gordon Hamilton, Three integer sequences from recreational mathematics, Video (2013?).
M. F. Hasler, Illustration of initial terms
M. F. Hasler, Illustrations (Three slides)
Brian Hayes, Joshua Trees and Toothpicks
Brian Hayes, Idealized Joshua tree, a figure from "Joshua Trees and Toothpicks" (see preceding link)
Brian Hayes, The Toothpick Sequence  BitPlayer
Benoit Jubin, Illustration of initial terms
Chris Moore, Gallery, see the section on David Griffeath's Cellular Automata.
Omar E. Pol, Illustration of initial terms
Omar E. Pol, Illustration of the toothpick structure (after 23 steps)
Omar E. Pol, Illustration of patterns in the toothpick structure (after 32 steps)
Omar E. Pol, Illustration of patterns in the toothpick structure (after 32 steps) [Cached copy, with permission]
Omar E. Pol, Illustration of initial terms of A139250, A160120, A147562 (Overlapping figures)
Omar E. Pol, Illustration of initial terms of A160120, A161206, A161328, A161330 (triangular grid and toothpick structure)
Omar E. Pol, Illustration of the substructures in the first quadrant (As pieces of a puzzle), after 32 stages
Omar E. Pol, Illustration of the potential growth direction of the arms of the substructures, after 32 stages
L. D. Pryor, Illustration of initial terms (Fig. 2a)
L. D. Pryor, The Inheritance of Inflorescence Characters in Eucalyptus, Proceedings of the Linnean Society of New South Wales, V. 79, (1954), p. 7989.
E. Rowland, Toothpick sequence from cellular automaton on square grid
E. Rowland, Initial stages of toothpick sequence from cellular automaton on square grid (includes Mathematica code)
K. Ryde, ToothpickTree
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Wikipedia, H tree
Wikipedia, Toothpick sequence
Wikipedia, Tsquare (fractal)
Index entries for sequences related to toothpick sequences
Index entries for sequences related to cellular automata


FORMULA

a(2^k) = A007583(k), if k >= 0.
a(2^k1) = A006095(k+1), if k >= 1.
a(A000225(k))  a((A000225(k)1)/2) = A006516(k), if k >= 1.
a(A000668(k))  a((A000668(k)1)/2) = A000396(k), if k >= 1.
G.f.: (x/((1x)*(1+2*x))) * (1 + 2*x*Product(1+x^(2^k1)+2*x^(2^k),k=0..oo)).  N. J. A. Sloane, May 20 2009, Jun 05 2009
One can show that lim sup a(n)/n^2 = 2/3, and it appears that lim inf a(n)/n^2 is 0.451...  Benoit Jubin, Apr 15 2009 and Jan 29 2010, N. J. A. Sloane, Jan 29 2010
Observation: a(n) mod 4 == 3 for n>=2.  Jaume Oliver Lafont, Feb 05 2009
a(2^k1) = A000969(2^k2), if k >= 1.  Omar E. Pol, Feb 13 2010
It appears that a(n) = (A187220(n+1)  1)/2.  Omar E. Pol, Mar 08 2011
a(n) = 4*A153000(n2) + 3, if n >= 2.  Omar E. Pol, Oct 01 2011
It appears that a(n) = A160552(n) + (A169707(n)  1)/2, n >= 1.  Omar E. Pol, Feb 15 2015
It appears that a(n) = A255747(n) + A255747(n1), n >= 1.  Omar E. Pol, Mar 16 2015
Let n = msb(n) + j where msb(n) = A053644(n) and let a(0) = 0. Then a(n) = (2 * msb(n)^2 + 1) / 3 + 2 * a(j) + a(j + 1)  1.  David A. Corneth, Mar 26 2015
It appears that a(n) = (A169707(n)  1)/4 + (A169707(n+1)  1)/4, n >= 1.  Omar E. Pol, Jul 24 2015


EXAMPLE

a(10^10) = 52010594272060810683  David A. Corneth, Mar 26 2015


MAPLE

G := (x/((1x)*(1+2*x))) * (1 + 2*x*mul(1+x^(2^k1)+2*x^(2^k), k=0..20)); # N. J. A. Sloane, May 20 2009, Jun 05 2009
# From N. J. A. Sloane, Dec 25 2009: A139250 is T, A139251 is a.
a:=[0, 1, 2, 4]; T:=[0, 1, 3, 7]; M:=10;
for k from 1 to M do
a:=[op(a), 2^(k+1)];
T:=[op(T), T[nops(T)]+a[nops(a)]];
for j from 1 to 2^(k+1)1 do
a:=[op(a), 2*a[j+1]+a[j+2]];
T:=[op(T), T[nops(T)]+a[nops(a)]];
od: od: a; T;


MATHEMATICA

CoefficientList[ Series[ (x/((1  x)*(1 + 2x))) (1 + 2x*Product[1 + x^(2^k  1) + 2*x^(2^k), {k, 0, 20}]), {x, 0, 53}], x] (* Robert G. Wilson v, Dec 06 2010 *)


PROG

(PARI) A139250(n, print_all=0)={my(p=[] /* set of "used" points. Points are written as complex numbers, c=x+iy. Toothpicks are of length 2 */,
ee=[[0, 1]] /* list of (exposed) endpoints. Exposed endpoints are listed as [c, d] where c=x+iy is the position of the endpoint, and d (unimodular) is the direction */
c, d, ne, cnt=1); print_all & print1("0, 1"); n<2 & return(n);
for(i=2, n, p=setunion(p, Set(Mat(ee~)[, 1])); /* add endpoints (discard directions) from last move to "used" points */
ne=[]; /* new (exposed) endpoints */
for( k=1, #ee, /* add endpoints of new toothpicks if not among the used points */
setsearch(p, c=ee[k][1]+d=ee[k][2]*I)  ne=setunion(ne, Set([[c, d]])); \\
setsearch(p, c2*d)  ne=setunion(ne, Set([[c2*d, d]]));
); /* using Set() we have the points sorted, so it's easy to remove those which finally are not exposed because they touch a new toothpick */
forstep( k=#ee=eval(ne), 2, 1, ee[k][1]==ee[k1][1] & k & ee=vecextract(ee, Str("^"k"..", k+1)))\
cnt+=#ee; /* each exposed endpoint will give a new toothpick */ print_all & print1(", "cnt)); cnt} /* M. F. Hasler, Apr 14 2009 */
(PARI)
\\works for n > 0
a(n) = {my(k = (2*msb(n)^2 + 1) / 3); if(n==msb(n), k , k + 2*a(nmsb(n)) + a(n  msb(n) + 1)  1)}
msb(n)=my(t=0); while(n>>t>0, t++); 2^(t1)\\ David A. Corneth, Mar 26 2015


CROSSREFS

Cf. A000079, A139251, A139252, A139253, A147614.
Cf. A139560, A152968, A152978, A152980, A152998, A153000, A153001, A153003, A153004, A153006, A153007.
Cf. A000217, A007583, A007683, A000396, A000225, A000668, A006516, A006095, A019988, A160570, A160552.
Cf. A000969, A001316, A151566, A160406, A160408, A160702, A078008, A151548, A001045, A147562, A160120.
Cf. A160160, A160170, A160172, A161206, A161328, A161330.
Cf. A002450, A160124.  Omar E. Pol, May 20 2010
Sequence in context: A169626 A160808 A151567 * A256265 A182634 A173530
Adjacent sequences: A139247 A139248 A139249 * A139251 A139252 A139253


KEYWORD

nonn,look


AUTHOR

Omar E. Pol, Apr 24 2008, May 08 2008, Dec 20 2008


EXTENSIONS

Verified and extended, a(49)a(53), using the given PARI code by M. F. Hasler, Apr 14 2009
Edited by N. J. A. Sloane, Apr 29 2009, incorporating comments from Omar E. Pol, M. F. Hasler, Rob Pratt, Jaume Oliver Lafont, Franklin T. AdamsWatters, R. J. Mathar, David W. Wilson, David Applegate, Benoit Jubin and others.
Further edited by N. J. A. Sloane, Jan 28 2010


STATUS

approved



