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Partial sums of A153001.
+20
0
4, 11, 20, 30, 45, 67, 88, 102, 117, 140, 168, 203, 255, 319, 368, 390, 405, 428, 456, 491, 543, 608, 664, 707, 760, 834, 925, 1047, 1215, 1391, 1504, 1542, 1557, 1580, 1608, 1643, 1695, 1760, 1816, 1859, 1912, 1986, 2077, 2199, 2367, 2544, 2664, 2723, 2776, 2850
OFFSET
1,1
LINKS
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
CROSSREFS
KEYWORD
nonn
AUTHOR
Omar E. Pol, May 18 2009
EXTENSIONS
More terms from Jinyuan Wang, Mar 15 2020
STATUS
approved
Toothpick sequence (see Comments lines for definition).
+10
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
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 n-th 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 two-dimensional cellular automaton. The animations show the fractal-like 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^(k-1) times the length of a toothpick. - M. F. Hasler, Apr 14 2009 and others. For proof, see the Applegate-Pol-Sloane 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 quarter-turn, and then rotate the whole configuration a quarter-turn, 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 "T-square (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 5-state 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 three-dimensional 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^(m-1) substructures of size s = k - m, where every substructure has 4*s rectangles. The total number of substructures is equal to (4^(k-1)-1)/3 = A002450(k-1). 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 n-th stage of the toothpick structure, assuming the toothpicks have length 2*k, is equal to (2*k-2)*a(n) + A147614(n), k > 0. See the formulas of A160420 and A160422. - Omar E. Pol, Nov 13 2010
Version "Gullwing": on the semi-infinite 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 n-th 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 "I-toothpick": we define an "I-toothpick" to consist of two connected toothpicks, as a bar of length 2. An I-toothpick with length 2 is formed by two toothpicks with length 1. The midpoint of an I-toothpick is touched by its two toothpicks. a(n) is also the number of I-toothpicks after n-th stage in the I-toothpick structure. The I-toothpick structure is essentially the original toothpick structure in which every toothpick is replaced by an I-toothpick. 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 I-toothpick structure the wooden toothpicks are not overlapping because all wooden toothpicks are connected by their endpoints. For the number of toothpicks in the I-toothpick structure see A160164 which also gives the number of gullwing in a gullwing structure because the gullwing structure of A160164 is equivalent to the I-toothpick 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 Ulam-Warburton cellular automaton see the Applegate-Pol-Sloane paper and see also A160164 and A187220.
(End)
A version in which the toothpicks are connected by their endpoints: on the semi-infinite 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 I-toothpick structure. a(n) is also the number of toothpicks after the n-th. - Omar E. Pol, Mar 13 2011
Version "Quarter-circle" (or Q-toothpick): a(n) is also the number of Q-toothpicks after the n-th stage in a Q-toothpick structure in the first quadrant. We start from (0,1) with the first Q-toothpick 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 n-th 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 n-th stage. A153006 (the corner sequence of the original version) gives 1/4 of the total of components in the structure after the n-th 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 n-th 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 NW-NE-SE-SW 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 non-adjacent 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 n-th stage. A160164(n) gives the number of ON cells in the structure after n-th stage. - Omar E. Pol, Mar 29 2018
From Omar E. Pol, Mar 06 2019: (Start)
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 n-th 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 Ulam-Warburton 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 n-th 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 "one-step bishop" of the Ulam-Warburton cellular automaton of A147562. (End)
From Omar E. Pol, Nov 27 2020: (Start)
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^(k-1) 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^(k-1) - 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^(k-1) - 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 connection to Mersenne primes (A000668) and perfect numbers (A000396) see A153006.
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), 157-191
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
David Applegate, The movie version
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.], which is also available at arXiv:1004.3036v2
Ayliean, Toothpick fractal and breaking rules, Youtube video (2021)
Joe Champion, Ultimate toothpick pattern, Photo 1, Photo 2, Photo 3, Photo 4, Boise Math Circles, Boise State University. [Links updated by P. Michael Hutchins, Mar 03 2018]
Barry Cipra, What comes next?, Science (AAAS) 327: 943.
Steven R. Finch, Toothpicks and Live Cells, July 21, 2015. [Cached copy, with permission of the author]
Ulrich Gehmann, Martin Reiche, World mountain machine, Berlin, (2014), first edition, p. 205, 238, 253.
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.
J. K. Hamilton, I. R. Hooper, and C. R. Lawrence, Exploring microwave absorption by non-periodic metasurfaces, Advanced Electromagnetics, 10(3), 1-6 (2021).
Brian Hayes, Idealized Joshua tree, a figure from "Joshua Trees and Toothpicks" (see preceding link)
Mathemaesthetics, 2'796'203 Toothpicks, 2'048 Generations, Youtube video (2021).
Chris Moore, Gallery, see the section on David Griffeath's Cellular Automata.
Olbaid Fractalium, Toothpick sequence Part 1, (Watch from minute 0:00 until 3:13), Youtube video (2023).
Programing Puzzles & Code Golf Stack Exchange, Generate toothpick sequence
L. D. Pryor, The Inheritance of Inflorescence Characters in Eucalyptus, Proceedings of the Linnean Society of New South Wales, V. 79, (1954), p. 79-89.
K. Ryde, ToothpickTree.
Daniel Shiffman, Coding Challenge #126: Toothpicks, The Coding Train video (2018)
N. J. A. Sloane and Brady Haran, Terrific Toothpick Patterns, Numberphile video (2018)
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).
Wikipedia, H tree
FORMULA
a(2^k) = A007583(k), if k >= 0.
a(2^k-1) = 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/((1-x)*(1+2*x))) * (1 + 2*x*Product_{k>=0} (1 + x^(2^k-1) + 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
Observation: a(n) == 3 (mod 4) for n >= 2. - Jaume Oliver Lafont, Feb 05 2009
a(2^k-1) = A000969(2^k-2), 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(n-2) + 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(n-1), 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/((1-x)*(1+2*x))) * (1 + 2*x*mul(1+x^(2^k-1)+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 *)
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}] (* Jean-François 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, c-2*d) || ne=setunion(ne, Set([[c-2*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[k-1][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(n-msb(n)) + a(n - msb(n) + 1) - 1)}
msb(n)=my(t=0); while(n>>t>0, t++); 2^(t-1)\\ David A. Corneth, Mar 26 2015
(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
[a(n) for n in range(101)] # Indranil Ghosh, Jul 01 2017, after David A. Corneth's PARI script
KEYWORD
nonn,look,nice
AUTHOR
Omar E. Pol, Apr 24 2008
EXTENSIONS
Verified and extended, a(49)-a(53), using the given PARI code by M. F. Hasler, Apr 14 2009
Further edited by N. J. A. Sloane, Jan 28 2010
STATUS
approved
First differences of toothpicks numbers A139250.
+10
236
0, 1, 2, 4, 4, 4, 8, 12, 8, 4, 8, 12, 12, 16, 28, 32, 16, 4, 8, 12, 12, 16, 28, 32, 20, 16, 28, 36, 40, 60, 88, 80, 32, 4, 8, 12, 12, 16, 28, 32, 20, 16, 28, 36, 40, 60, 88, 80, 36, 16, 28, 36, 40, 60, 88, 84, 56, 60, 92, 112, 140, 208, 256, 192, 64, 4, 8, 12, 12, 16, 28, 32, 20, 16, 28
OFFSET
0,3
COMMENTS
Number of toothpicks added to the toothpick structure at the n-th step (see A139250).
It appears that if n is equal to 1 plus a power of 2 with positive exponent then a(n) = 4. (For proof see the second Applegate link.)
It appears that there is a relation between this sequence, even superperfect numbers, Mersenne primes and even perfect numbers. Conjecture: The sum of the toothpicks added to the toothpick structure between the stage A061652(k) and the stage A000668(k) is equal to the k-th even perfect number, for k >= 1. For example: A000396(1) = 2+4 = 6. A000396(2) = 4+4+8+12 = 28. A000396(3) = 16+4+8+12+12+16+28+32+20+16+28+36+40+60+88+80 = 496. - Omar E. Pol, May 04 2009
Concerning this conjecture, see David Applegate's comments on the conjectures in A153006. - N. J. A. Sloane, May 14 2009
In the triangle (See example lines), the sum of row k is equal to A006516(k), for k >= 1. - Omar E. Pol, May 15 2009
Equals (1, 2, 2, 2, ...) convolved with A160762: (1, 0, 2, -2, 2, 2, 2, -6, ...). - Gary W. Adamson, May 25 2009
Convolved with the Jacobsthal sequence A001045 = A160704: (1, 3, 9, 19, 41, ...). - Gary W. Adamson, May 24 2009
It appears that the sums of two successive terms of A160552 give the positive terms of this sequence. - Omar E. Pol, Feb 19 2015
From Omar E. Pol, Feb 28 2019: (Start)
The study of the toothpick automaton on triangular grid (A296510), and other C.A. of the same family, reveals that some cellular automata that have recurrent periods can be represented in general by irregular triangles (of first differences) whose row lengths are the terms of A011782 multiplied by k, where k >= 1, is the length of an internal cycle. This internal cycle is called "word" of a cellular automaton. For example: A160121 has word "a", so k = 1. This sequence has word "ab", so k = 2. A296511 has word "abc", so k = 3. A299477 has word "abcb" so k = 4. A299479 has word "abcbc", so k = 5.
The structure of this triangle (with word "ab" and k = 2) for the nonzero terms is as follows:
a,b;
a,b;
a,b,a,b;
a,b,a,b,a,b,a,b;
a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b;
...
The row lengths are the terms of A011782 multiplied by 2, equaling the column 2 of the square array A296612: 2, 2, 4, 8, 16, ...
This arrangement has the property that the odd-indexed columns (a) contain numbers of the toothpicks that are parallel to initial toothpick, and the even-indexed columns (b) contain numbers of the toothpicks that are orthogonal to the initial toothpick (see the third triangle in the Example section).
An associated sound to the animation could be (tick, tock), (tick, tock), ..., the same as the ticking clock sound.
For further information about the "word" of a cellular automaton see A296612. (End)
LINKS
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.], Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
David Applegate, The movie version
FORMULA
Recurrence from N. J. A. Sloane, Jul 20 2009: a(0) = 0; a(2^i)=2^i for all i; otherwise write n=2^i+j, 0 < j < 2^i, then a(n) = 2a(j)+a(j+1). Proof: This is a simplification of the following recurrence of David Applegate. QED
Recurrence from David Applegate, Apr 29 2009: (Start)
Write n=2^(i+1)+j, where 0 <= j < 2^(i+1). Then, for n > 3:
for j=0, a(n) = 2*a(n-2^i) (= n = 2^(i+1))
for 1 <= j <= 2^i - 1, a(n) = a(n-2^i)
for j=2^i, a(n) = a(n-2^i)+4 (= 2^(i+1)+4)
for 2^i+1 <= j <= 2^(i+1)-2, a(n) = 2*a(n-2^i) + a(n-2^i+1)
for j=2^(i+1)-1, a(n) = 2*a(n-2^i) + a(n-2^i+1)-4
and a(n) = 2^(n-1) for n=1,2,3. (End)
G.f.: (x/(1+2*x)) * (1 + 2*x*Product_{k>=0} (1 + x^(2^k-1) + 2*x^(2^k))). - N. J. A. Sloane, May 20 2009, Jun 05 2009
With offset 0 (which would be more natural, but offset 1 is now entrenched): a(0) = 1, a(1) = 2; for i >= 1, a(2^i) = 4; otherwise write n = 2^i +j, 0 < j < 2^i, then a(n) = 2 * Sum_{ k >= 0 } 2^(wt(j+k)-k)*binomial(wt(j+k),k). - N. J. A. Sloane, Jun 03 2009
It appears that a(n) = A187221(n+1)/2. - Omar E. Pol, Mar 08 2011
It appears that a(n) = A160552(n-1) + A160552(n), n >= 1. - Omar E. Pol, Feb 18 2015
EXAMPLE
From Omar E. Pol, Dec 16 2008: (Start)
Triangle begins:
1;
2;
4,4;
4,8,12,8;
4,8,12,12,16,28,32,16;
4,8,12,12,16,28,32,20,16,28,36,40,60,88,20,32;
(End)
From David Applegate, Apr 29 2009: (Start)
The layout of the triangle was adjusted to reveal that the columns become constant as shown below:
. 0;
. 1;
. 2,4;
. 4,4,8,12;
. 8,4,8,12,12,16,28,32;
.16,4,8,12,12,16,28,32,20,16,28,36,40,60,88,80;
.32,4,8,12,12,16,28,32,20,16,28,36,40,60,88,80,36,16,28,36,40,60,88,84,56,...
...
The row sums give A006516.
(End)
From Omar E. Pol, Feb 28 2018: (Start)
Also the nonzero terms can write as an irregular triangle in which the row lengths are the terms of A011782 multiplied by 2 as shown below:
1,2;
4,4;
4,8,12,8;
4,8,12,12,16,28,32,16;
4,8,12,12,16,28,32,20,16,28,36,40,60,88,20,32;
...
(End)
MAPLE
G := (x/(1+2*x)) * (1 + 2*x*mul(1+x^(2^k-1)+2*x^(2^k), k=0..20)); # N. J. A. Sloane, May 20 2009, Jun 05 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;
# N. J. A. Sloane, Dec 25 2009
MATHEMATICA
CoefficientList[Series[((x - x^2)/((1 - x) (1 + 2 x))) (1 + 2 x Product[1 + x^(2^k - 1) + 2 x^(2^k), {k, 0, 20}]), {x, 0, 60}], x] (* Vincenzo Librandi, Aug 22 2014 *)
CROSSREFS
Equals 2*A152968 and 4*A152978 (if we ignore the first couple of terms).
See A147646 for the limiting behavior of the rows. See also A006516.
Row lengths in A011782.
Cf. A160121 (word "a"), A296511 (word "abc"), A299477 (word "abcb"), A299479 (word "abcbc").
KEYWORD
nonn,tabf,look
AUTHOR
Omar E. Pol, Apr 24 2008
EXTENSIONS
Partially edited by Omar E. Pol, Feb 28 2019
STATUS
approved
Toothpick sequence starting at the outside corner of an infinite square from which protrudes a half toothpick.
+10
54
0, 1, 3, 6, 9, 13, 20, 28, 33, 37, 44, 53, 63, 78, 100, 120, 129, 133, 140, 149, 159, 174, 196, 217, 231, 246, 269, 297, 332, 384, 448, 496, 513, 517, 524, 533, 543, 558, 580, 601, 615, 630, 653, 681, 716, 768, 832, 881, 903, 918, 941
OFFSET
0,3
COMMENTS
a(n) is the total number of integer toothpicks after n steps.
It appears that this sequence is related to triangular numbers, Mersenne primes and even perfect numbers. Conjecture: a(A000668(n))=A000217(A000668(n)). Conjecture: a(A000668(n))=A000396(n), assuming there are no odd perfect numbers.
The main entry for this sequence is A139250. See also A152980 (the first differences) and A147646.
The Mersenne prime conjectures are true, but aren't really about Mersenne primes. a(2^i-1) = 2^i (2^i-1)/2 for all i (whether or not i or 2^i-1 is prime). This follows from the formulas for A139250(2^i-1) and A139250(2^i). - David Applegate, May 11 2009
Then we can write a(A000225(k)) = A006516(k), for k > 0. - Omar E. Pol, May 23 2009
Equals A151550 convolved with [1, 2, 2, 2, ...]. (This is equivalent to the observation that the g.f. is x((1+x)/(1-x)) * Product_{n >= 1} (1 + x^(2^n-1) + 2*x^(2^n)).) Equivalently, equals A151555 convolved with A151575. - Gary W. Adamson, May 25 2009
It appears that a(n) is also 1/4 of the total path length of a toothpick structure as A139250 after n-th stage which is constructed following a special rule: toothpicks of the new generation have length 4 when are placed on the square grid (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 a 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 length 4, 3 and 2. a(n) is also 1/4 of the number of grid points that are covered after n-th stage, except the central point of the structure. A159795 gives the total path length and also the total number of components in the structure after n-th stage. - Omar E. Pol, Oct 24 2011
LINKS
David Applegate, The movie version
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
FORMULA
G.f.: x*((1 + x)/(1 - x)) * Product_{n >= 1} (1 + x^(2^n-1) + 2*x^(2^n)). - N. J. A. Sloane, May 20 2009
MAPLE
G:=x*((1 + x)/(1 - x)) * mul( (1 + x^(2^n-1) + 2*x^(2^n)), n=1..20); # N. J. A. Sloane, May 20 2009
KEYWORD
nonn
AUTHOR
Omar E. Pol, Dec 17 2008, Dec 19 2008, Apr 28 2009
EXTENSIONS
Edited by N. J. A. Sloane, Dec 19 2008
STATUS
approved
First differences of toothpick corner sequence A153006.
+10
50
1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 9, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23, 28, 35, 52, 64, 48, 17, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23, 28, 35, 52, 64, 49, 22, 15, 23, 28, 35, 52, 65, 56, 43, 53, 74, 91, 122, 168, 176, 112, 33, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23, 28, 35, 52
OFFSET
0,2
COMMENTS
Rows of A152978 when written as a triangle converge to this sequence. - Omar E. Pol, Jul 19 2009
LINKS
David Applegate, The movie version
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
FORMULA
G.f.: (1 + x) * Prod_{ n >= 1} (1 + x^(2^n-1) + 2*x^(2^n)). - N. J. A. Sloane, May 20 2009, corrected May 21 2009
For formula see A147646 (or, better, see the Maple code below).
EXAMPLE
Triangle begins:
.1;
.2;
.3,3;
.4,7,8,5;
.4,7,9,10,15,22,20,9;
.4,7,9,10,15,22,21,14,15,23,28,35,52,64,48,17;
....
Rows converge to A153001. - N. J. A. Sloane, Jun 07 2009
MAPLE
Maple code from N. J. A. Sloane, May 18 2009. First define old version with offset 1:
S:=proc(n) option remember; local i, j;
if n <= 0 then RETURN(0); fi;
if n <= 2 then RETURN(2^(n-1)); fi;
i:=floor(log(n)/log(2));
j:=n-2^i;
if j=0 then RETURN(n/2+1); fi;
if j<2^i-1 then RETURN(2*S(j)+S(j+1)); fi;
if j=2^i-1 then RETURN(2*S(j)+S(j+1)-1); fi;
-1;
end;
# Now change the offset:
T:=n->S(n+1);
G := (1 + x) * mul(1 + x^(2^k-1) + 2*x^(2^k), k=1..20);
MATHEMATICA
nmax = 78;
G = x*((1 + x)/(1 - x)) * Product[ (1 + x^(2^n - 1) + 2*x^(2^n)), {n, 1, Log2[nmax] // Ceiling}];
CoefficientList[G + O[x]^nmax, x] // Differences (* Jean-François Alcover, Jul 21 2022 *)
CROSSREFS
Equals A151688 divided by 2. - N. J. A. Sloane, Jun 03 2009
For generating functions of the form Product_{k>=c} (1+a*x^(2^k-1)+b*x^2^k) for the following values of (a,b,c) see: (1,1,0) A160573, (1,1,1) A151552, (1,1,2) A151692, (2,1,0) A151685, (2,1,1) A151691, (1,2,0) A151688 and A152980, (1,2,1) A151550, (2,2,0) A151693, (2,2,1) A151694.
Equals A147646/4. - N. J. A. Sloane, May 01 2009
KEYWORD
nonn,look
AUTHOR
Omar E. Pol, Dec 16 2008, Dec 19 2008, Jan 02 2009
EXTENSIONS
More terms (based on A147646) from N. J. A. Sloane, May 01 2009
Offset changed by N. J. A. Sloane, May 18 2009
STATUS
approved
a(n) = A139251(n+2)/4 = A152968(n+1)/2.
+10
23
1, 1, 1, 2, 3, 2, 1, 2, 3, 3, 4, 7, 8, 4, 1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 8, 1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 9, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23, 28, 35, 52, 64, 48, 16, 1, 2, 3, 3, 4, 7, 8, 5, 4, 7, 9, 10, 15, 22, 20, 9, 4, 7, 9, 10, 15, 22, 21, 14, 15, 23
OFFSET
1,4
COMMENTS
Also, first differences of toothpick numbers A153000.
LINKS
David Applegate, The movie version
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
FORMULA
G.f.: (1+x)*(Prod(1+x^(2^k-1)+2*x^(2^k),k=1..oo)-1)/(1+2*x). - N. J. A. Sloane, May 20 2009
EXAMPLE
If written as a triangle, begins:
.1,1;
.1,2,3,2;
.1,2,3,3,4,7,8,4;
.1,2,3,3,4,7,8,5,4,7,9,10,15,22,20,8;
....
Rows converge to A152980.
It appears that row sums give A004171. [From Omar E. Pol, May 25 2010]
CROSSREFS
Cf. toothpick sequence A139250.
Cf. A004171. [From Omar E. Pol, May 25 2010]
KEYWORD
nonn,easy
AUTHOR
Omar E. Pol, Dec 16 2008, Dec 20 2008
EXTENSIONS
More terms from Omar E. Pol, Jul 26 2009
STATUS
approved
Sequence to which the rows of the triangle of A160121 converge.
+10
2
9, 21, 15, 21, 27, 51, 33, 21, 27, 51, 51, 57, 69, 117, 87, 33, 27, 51, 51, 57, 75, 129, 117, 75, 69, 117, 135
OFFSET
1,1
CROSSREFS
KEYWORD
more,nonn
AUTHOR
Omar E. Pol, Jun 18 2009
STATUS
approved

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