

A139250


Toothpick sequence (see Comments lines for definition).


480



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 Sierpiński 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
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
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
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:
(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
On the infinite Cairo pentagonal tiling consider the symmetric figure formed by two nonadjacent pentagons connected by a line segment joining two trivalent nodes. At stage 1 we start with one of these figures turned ON. The rule for the next stages is that the concave part of the figures of the new generation must be adjacent to the complementary convex part of the figures of the old generation. a(n) gives the number of figures that are ON in the structure after nth stage. A160164(n) gives the number of ON cells in the structure after nth stage.  Omar E. Pol, Mar 29 2018
The "word" of this sequence is "ab". For further information about the word of cellular automata see A296612.
Version "triangular grid": a(n) is also the total number of toothpicks of length 2 after nth stage in the toothpick structure on the infinite triangular grid, if we use only two of the three axes. Otherwise, if we use the three axes, so we have the sequence A296510 which has word "abc".
The normal toothpick structure can be considered a superstructure of the UlamWarburton celular automaton since A147562(n) equals here the total number of "hidden crosses" after 4*n stages, including the central cross (beginning to count the crosses when their "nuclei" are totally formed with 4 quadrilaterals). Note that every quadrilateral in the structure belongs to a "hidden cross".
Also, the number of "hidden crosses" after n stages equals the total number of "flowers with six petals" after nth stage in the structure of A323650, which appears to be a "missing link" between this sequence and A147562.
Note that the location of the "nuclei of the hidden crosses" is very similar (essentially the same) to the location of the "flowers with six petals" in the structure of A323650 and to the location of the "ON" cells in the version "onestep bishop" of the UlamWarburton cellular automaton of A147562. (End)
The simplest substructures are the arms of the hidden crosses. Each closed region (square or rectangle) of the structure belongs to one of these arms. The narrow arms have regions of area 1, 2, 4, 8, ... The broad arms have regions of area 2, 4, 8, 16 , ... Note that after 2^k stages, with k >= 3, the narrow arms of the main hidden crosses in each quadrant frame the size of the toothpick structure after 2^(k1) stages.
Another kind of substructure could be called "bar chart" or "bar graph". This substructure is formed by the rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure after 2^k stages, with k >= 2. The height of these successive regions gives the first 2^(k1)  1 terms from A006519. For example: if k = 5 the respective heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1]. The area of these successive regions gives the first 2^(k1)  1 terms of A171977. For example: if k = 5 the respective areas are [2, 4, 2, 8, 2, 4, 2, 16, 2, 4, 2, 8, 2, 4, 2].
For a representation of the Wagstaff primes (A000979) using the toothpick structure see A194810.
For a connection to stained glass windows and a hidden curve see A336532. (End)
It appears that the graph of a(n) bears a striking resemblance to the cumulative distribution function F(x) for X the random variable taking values in [0,1], where the binary expansion of X is given by a sequence of independent coin tosses with probability 3/4 of being 1 at each bit. It appears that F(n/2^k)*(2^(2k+1)+1)/3 approaches a(n) for k large.  James Coe, Jan 10 2022


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

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.
Chris Moore, Gallery, see the section on David Griffeath's Cellular Automata.
Alex van den Brandhof and Paul Levrie, Tandenstokerrij, Pythagoras, Wiskundetijdschrift voor Jongeren, 55ste Jaargang, Nummer 6, Juni 2016, (see the cover, pages 1, 18, 19 and the back cover).


FORMULA

a(2^k1) = A006095(k+1), if k >= 1.
G.f.: (x/((1x)*(1+2*x))) * (1 + 2*x*Product_{k>=0} (1 + x^(2^k1) + 2*x^(2^k))).  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
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


EXAMPLE



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
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 *)
a[0] = 0; a[n_] := a[n] = Module[{m, k}, m = 2^(Length[IntegerDigits[n, 2]]  1); k = (2m^2+1)/3; If[n == m, k, k + 2 a[n  m] + a[n  m + 1]  1]]; Table[a[n], {n, 0, 100}] (* JeanFrançois Alcover, Oct 06 2018, after David A. Corneth *)


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)}
(Python)
def msb(n):
t = 0
while n>>t > 0:
t += 1
return 2**(t  1)
def a(n):
k = (2 * msb(n)**2 + 1) / 3
return 0 if n == 0 else k if n == msb(n) else k + 2*a(n  msb(n)) + a(n  msb(n) + 1)  1


CROSSREFS

Cf. A000079, A002450, A006519, A139251, A139252, A139253, A147614, A139560, A152968, A152978, A152980, A152998, A153000, A153001, A153003, A153004, A153006, A153007, A000217, A007583, A007683, A000396, A000225, A000668, A006516, A006095, A019988, A160570, A160552, A000969, A001316, A151566, A160406, A160408, A160702, A078008, A151548, A001045, A147562, A160124, A160120, A160160, A160170, A160172, A161206, A161328, A161330, A171977, A194810, A296510, A296612, A299476, A299478, A323650, A336532.


KEYWORD



AUTHOR



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.


STATUS

approved



