cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A151548 When A160552 is regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ..., this is what the rows converge to.

Original entry on oeis.org

1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31, 5, 11, 17, 19, 21, 39, 49, 35, 21, 39, 53, 59, 81, 127, 129, 63, 5, 11, 17, 19, 21, 39, 49, 35, 21, 39, 53, 59, 81, 127, 129, 67, 21, 39, 53, 59, 81, 127, 133, 91, 81, 131, 165, 199, 289, 383, 321, 127, 5, 11, 17, 19, 21, 39
Offset: 0

Views

Author

David Applegate, May 18 2009

Keywords

Comments

When convolved with A151575: (1, 0, 2, -2, 6, -10, 22, -42, 86, -170, 342, ...) equals the toothpick sequence A139250: (1, 3, 7, 11, 15, 23, 35, 43, ...). - Gary W. Adamson, May 25 2009
Equals A160552: [1, 1, 3, 1, 3, 5, ...] convolved with [1, 2, 0, 0, 0, ...], equivalent to a(n) = 2*A160552(n) + A160552(n+1). - Gary W. Adamson, Jun 04 2009
Equals (1, 0, -2, 2, -2, 2, ...) convolved with the Toothpick sequence, A139250. - Gary W. Adamson, Mar 06 2012
It appears that the sums of two successive terms give A147646. - Omar E. Pol, Feb 18 2015

Examples

			From _Omar E. Pol_, Jul 24 2009: (Start)
When written as a triangle:
1;
3;
5,7;
5,11,17,15;
5,11,17,19,21,39,49,31;
5,11,17,19,21,39,49,35,21,39,53,59,81,127,129,63;
5,11,17,19,21,39,49,35,21,39,53,59,81,127,129,67,21,39,53,59,81,127,133,91,...
(End)
		

Crossrefs

Programs

  • Maple
    G := 1/(1+x) + 4*x*mul(1+x^(2^k-1)+2*x^(2^k),k=1..20); # N. J. A. Sloane, May 23 2009
    S2:=proc(n) option remember; local i,j;
    if n <= 1 then RETURN(2*n+1); fi;
    i:=floor(log(n)/log(2));
    j:=n-2^i;
    if j=0 then 5 elif j=2^i-1 then 2*n+1
    else 2*S2(j)+S2(j+1); fi;
    end; # - N. J. A. Sloane, May 22 2009
  • Mathematica
    terms = 70; CoefficientList[1/(1 + x) + 4*x*Product[1 + x^(2^k - 1) + 2*x^(2^k), {k, 1, Log[2, terms] // Ceiling}] + O[x]^terms, x] (* Jean-François Alcover, Nov 14 2017, after N. J. A. Sloane *)

Formula

a(2^k-1) = 2^(k+1)-1 for k >= 0; otherwise a(2^k) = 5 for k >= 1; otherwise a(2^i+j) = 2a(j)+a(j+1) for i >= 2, 1 <= j <= 2^i-2. - N. J. A. Sloane, May 22 2009
G.f.: 1/(1+x) + 4*x*mul(1+x^(2^k-1)+2*x^(2^k),k=1..oo). - N. J. A. Sloane, May 23 2009
a(n) = A147646(n) - a(n-1), n >= 1. - Omar E. Pol, Feb 19 2015

A255747 Partial sums of A160552.

Original entry on oeis.org

0, 1, 2, 5, 6, 9, 14, 21, 22, 25, 30, 37, 42, 53, 70, 85, 86, 89, 94, 101, 106, 117, 134, 149, 154, 165, 182, 201, 222, 261, 310, 341, 342, 345, 350, 357, 362, 373, 390, 405, 410, 421, 438, 457, 478, 517, 566, 597, 602, 613, 630, 649, 670, 709, 758, 793, 814, 853, 906, 965, 1046, 1173, 1302, 1365, 1366, 1369, 1374
Offset: 0

Views

Author

Omar E. Pol, Mar 05 2015

Keywords

Comments

It appears that the sums of two successive terms give the positive terms of the toothpick sequence A139250.
It appears that the odd terms (a bisection) give A162795.
It appears that a(n) is also the total number of ON cells at stage n+1 in one of the four wedges of two-dimensional cellular automaton "Rule 750" using the von Neumann neighborhood (see A169707). Therefore a(n) is also the total number of ON cells at stage n+1 in one of the four quadrants of the NW-NE-SE-SW version of that cellular automaton.
See also the formula section.
First differs from A169779 at a(11).

Examples

			Also, written as an irregular triangle in which the row lengths are the terms of A011782 (the number of compositions of n, n >= 0), the sequence begins:
0;
1;
2,   5;
6,   9, 14, 21;
22, 25, 30, 37, 42, 53, 70, 85;
86, 89, 94,101,106,117,134,149,154,165,182,201,222,261,310,341;
...
It appears that the first column gives 0 together with the terms of A047849, hence the right border gives A002450.
It appears that this triangle only shares with A151920 the positive elements of the columns 1, 2, 4, 8, 16, ... (the powers of 2).
		

Crossrefs

Programs

  • Mathematica
    Accumulate[Nest[Join[#, 2 # + Append[Rest@#, 1]] &, {0}, 6]] (* Ivan Neretin, Feb 09 2017 *)

Formula

It appears that a(n) + a(n-1) = A139250(n), n >= 1.
It appears that a(2n-1) = A162795(n), n >= 1.
It appears that a(n) = (A169707(n+1) - 1)/4.

A170903 a(n) = 2*A160552(n)-1.

Original entry on oeis.org

1, 1, 5, 1, 5, 9, 13, 1, 5, 9, 13, 9, 21, 33, 29, 1, 5, 9, 13, 9, 21, 33, 29, 9, 21, 33, 37, 41, 77, 97, 61, 1, 5, 9, 13, 9, 21, 33, 29, 9, 21, 33, 37, 41, 77, 97, 61, 9, 21, 33, 37, 41, 77, 97, 69, 41, 77, 105, 117, 161, 253, 257, 125, 1, 5, 9, 13, 9, 21, 33, 29, 9, 21, 33, 37, 41, 77
Offset: 1

Views

Author

Gary W. Adamson, Jan 21 2010

Keywords

Examples

			When written as a triangle:
1
1, 5;
1, 5, 9, 13;
1, 5, 9, 13, 9, 21, 33, 29;
...
Rows sums are A006516 (this is immediate from the definition).
From _Omar E. Pol_, Feb 17 2015: (Start)
Also, written as an irregular triangle in which the row lengths are the terms of A011782:
1;
1;
5,1;
5,9,13,1;
5,9,13,9,21,33,29,1;
5,9,13,9,21,33,29,9,21,33,37,41,77,97,61,1;
5,9,13,9,21,33,29,9,21,33,37,41,77,97,61,9,21,33,37,41,77,97,69,41,77,105,117,161,253,257,125,1;
Row sums give 1 together with the positive terms of A006516.
It appears that the right border (A000012) gives the smallest difference between A160164 and A169707 in every period.
(End)
		

Crossrefs

Formula

It appears that a(n) = A160164(n) - A169707(n). - Omar E. Pol, Feb 17 2015

A266529 Terms of A160552 repeated.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 3, 3, 1, 1, 3, 3, 5, 5, 7, 7, 1, 1, 3, 3, 5, 5, 7, 7, 5, 5, 11, 11, 17, 17, 15, 15, 1, 1, 3, 3, 5, 5, 7, 7, 5, 5, 11, 11, 17, 17, 15, 15, 5, 5, 11, 11, 17, 17, 19, 19, 21, 21, 39, 39, 49, 49, 31, 31, 1, 1, 3, 3, 5, 5, 7, 7, 5, 5, 11, 11, 17, 17, 15, 15, 5, 5, 11, 11, 17, 17, 19, 19, 21, 21
Offset: 1

Views

Author

Omar E. Pol, Jan 02 2016

Keywords

Comments

First differs from A266509 at a(55), with which it shares infinitely many terms.
First differs from A266539 at a(25), with which it shares infinitely many terms.

Examples

			Written as an irregular triangle in which the row lengths are twice the terms of A011782 the sequence begins:
0,0;
1,1;
1,1,3,3;
1,1,3,3,5,5,7,7;
1,1,3,3,5,5,7,7,5,5,11,11,17,17,15,15;
1,1,3,3,5,5,7,7,5,5,11,11,17,17,15,15,5,5,11,11,17,17,19,19,21,21,39,39,49,49,31,31;
...
Row sums give 0 together with A004171.
		

Crossrefs

Programs

  • Mathematica
    Riffle[#, #] &@ Table[SeriesCoefficient[x (1 + 2 x)/(1 + x) + (4 x^2/(1 + 2 x)) (Product[1 + x^(2^k - 1) + 2 x^(2^k), {k, 20}] - 1), {x, 0, n}], {n, 0, 44}] (* Michael De Vlieger, Jan 05 2016, based on Maple by N. J. A. Sloane at A160552 *)

A160570 Triangle read by rows, A160552 convolved with (1, 2, 2, 2, ...); row sums = A139250, the Toothpick sequence.

Original entry on oeis.org

1, 1, 2, 3, 2, 2, 1, 6, 2, 2, 3, 2, 6, 2, 2, 5, 6, 2, 6, 2, 2, 7, 10, 6, 2, 6, 2, 2, 1, 14, 10, 6, 2, 6, 2, 2, 3, 2, 14, 10, 6, 2, 6, 2, 2, 5, 6, 2, 14, 10, 6, 2, 6, 2, 2, 7, 10, 6, 2, 14, 10, 6, 2, 6, 2, 2, 5, 14, 10, 6, 2, 14, 10, 6, 2, 6, 2, 2, 11, 10, 14, 10, 6, 2, 14, 10, 6, 2, 6
Offset: 1

Views

Author

Gary W. Adamson, May 19 2009

Keywords

Examples

			First few rows of the triangle:
  1;
  1,  2;
  3,  2,  2;
  1,  6,  2,  2;
  3,  2,  6,  2,  2;
  5,  6,  2,  6,  2,  2;
  7, 10,  6,  2,  6,  2,  2;
  1, 14, 10,  6,  2,  6,  2,  2;
  3,  2, 14, 10,  6,  2,  6,  2,  2;
  5,  6,  2, 14, 10,  6,  2,  6,  2,  2;
  ...
Example: Row 4 = (1, 6, 2, 2) = (1, 3, 1, 1) dot (1, 2, 2, 2); where (1 + 6 + 2 + 2) = A139250(4), i.e., 4th term in the Toothpick sequence.
		

Crossrefs

Programs

  • Maple
    T:=proc(n,k)if(k=1)then return A160552(n):else return 2*A160552(n-k+1):fi:end:
    for n from 1 to 8 do for k from 1 to n do print(T(n,k));od:od: # Nathaniel Johnston, Apr 13 2011

Formula

Construct triangle M = an infinite lower triangular Toeplitz matrix with A160552: (1, 1, 3, 1, 3, 5, 7, ...) in every column. Let Q = an infinite lower triangular matrix with (1, 2, 2, 2, 2, ...) as the main diagonal and the rest zeros. A160570 = M * Q.

A255045 a(n) = (1 + A160552(n))/2.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 4, 1, 2, 3, 4, 3, 6, 9, 8, 1, 2, 3, 4, 3, 6, 9, 8, 3, 6, 9, 10, 11, 20, 25, 16, 1, 2, 3, 4, 3, 6, 9, 8, 3, 6, 9, 10, 11, 20, 25, 16, 3, 6, 9, 10, 11, 20, 25, 18, 11, 20, 27, 30, 41, 64, 65, 32, 1, 2, 3, 4, 3, 6, 9, 8, 3, 6, 9, 10, 11, 20, 25, 16, 3, 6, 9, 10, 11, 20, 25, 18, 11, 20, 27, 30, 41, 64, 65, 32, 3, 6, 9, 10, 11, 20, 25, 18
Offset: 1

Views

Author

Omar E. Pol, Feb 13 2015

Keywords

Comments

Triangle of numbers related to cellular automata.
It appears that this is also a triangle read by rows in which row n lists the first 2^(n-1) terms of A255046, with n >= 1.

Examples

			Written as an irregular triangle in which the length of row j is 2^j, j >= 0, the sequence begins:
1;
1,2;
1,2,3,4;
1,2,3,4,3,6,9,8;
1,2,3,4,3,6,9,8,3,6,9,10,11,20,25,16;
1,2,3,4,3,6,9,8,3,6,9,10,11,20,25,16,3,6,9,10,11,20,25,18,11,20,27,30,41,64,65,32;
...
It appears that the right border gives A000079.
It appears that the row sums give A007582.
It appears that rows converge to A255046.
		

Crossrefs

A255049 a(n) = 2*A160552(n).

Original entry on oeis.org

0, 2, 2, 6, 2, 6, 10, 14, 2, 6, 10, 14, 10, 22, 34, 30, 2, 6, 10, 14, 10, 22, 34, 30, 10, 22, 34, 38, 42, 78, 98, 62, 2, 6, 10, 14, 10, 22, 34, 30, 10, 22, 34, 38, 42, 78, 98, 62, 10, 22, 34, 38, 42, 78, 98, 70, 42, 78, 106, 118, 162, 254, 258, 126, 2, 6, 10, 14, 10
Offset: 0

Views

Author

Omar E. Pol, Feb 13 2015

Keywords

Examples

			Written as an irregular triangle in which row lengths are the terms of A011782 the sequence begins:
0;
2;
2,6;
2,6,10,14;
2,6,10,14,10,22,34,30;
2,6,10,14,10,22,34,30,10,22,34,38,42,78,98,62;
2,6,10,14,10,22,34,30,10,22,34,38,42,78,98,62,10,22,34,38,42,78,98,70,42,78,106,118,162,254,258,126;
It appears that row sums give 0 together with A004171, (see also A081294).
It appears that right border gives the nonnegative terms of A000918, (see also A095121).
		

Crossrefs

Formula

It appears that a(n) = A169708(n)/2, n >= 1.

Extensions

Edited by Omar E. Pol, Feb 18 2015

A139250 Toothpick sequence (see Comments lines for definition).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Apr 24 2008

Keywords

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

Examples

			a(10^10) = 52010594272060810683. - _David A. Corneth_, Mar 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), 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.

Crossrefs

Programs

  • 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 *)
  • 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

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

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

A139251 First differences of toothpicks numbers A139250.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Apr 24 2008

Keywords

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)

Examples

			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)
		

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").

Programs

  • 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 *)

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

Extensions

Partially edited by Omar E. Pol, Feb 28 2019

A169707 Total number of ON cells at stage n of two-dimensional cellular automaton defined by "Rule 750" using the von Neumann neighborhood.

Original entry on oeis.org

1, 5, 9, 21, 25, 37, 57, 85, 89, 101, 121, 149, 169, 213, 281, 341, 345, 357, 377, 405, 425, 469, 537, 597, 617, 661, 729, 805, 889, 1045, 1241, 1365, 1369, 1381, 1401, 1429, 1449, 1493, 1561, 1621, 1641, 1685, 1753, 1829, 1913, 2069, 2265, 2389, 2409, 2453, 2521
Offset: 1

Views

Author

N. J. A. Sloane, Apr 17 2010

Keywords

Comments

Square grid, 4 neighbors per cell (N, E, S, W cells), turn ON iff exactly 1 or 3 neighbors are ON; once ON, cells stay ON.
The terms agree with those of A246335 for n <= 11, although the configurations are different starting at n = 7. - N. J. A. Sloane, Sep 21 2014
Offset 1 is best for giving a formula for a(n), although the Maple and Mathematica programs index the states starting at state 0.
It appears that this shares infinitely many terms with both A162795 and A147562, see Formula section and Example section. - Omar E. Pol, Feb 19 2015

Examples

			Divides naturally into blocks of sizes 1,2,4,8,16,...:
1,
5, 9,
21, 25, 37, 57,
85, 89, 101, 121, 149, 169, 213, 281, <- terms 8 through 15
341, 345, 357, 377, 405, 425, 469, 537, 597, 617, 661, 729, 805, 889, 1045, 1241,
1365, 1369, 1381, 1401, 1429, 1449, 1493, 1561, 1621, 1641, 1685, 1753, 1829, 1913, 2069, 2265, 2389, 2409, 2453, 2521, ...
From _Omar E. Pol_, Feb 18 2015: (Start)
Also, written as an irregular triangle T(j,k), k>=1, in which the row lengths are the terms of A011782:
1;
5;
9,   21;
25,  37,   57,  85;
89,  101, 121, 149, 169, 213, 281, 341;
345, 357, 377, 405, 425, 469, 537, 597, 617, 661, 729, 805, 889, 1045, 1241, 1365;
The right border gives the positive terms of A002450.
It appears that T(j,k) = A162795(j,k) = A147562(j,k), if k is a power of 2, for example: it appears that the three mentioned triangles only share the elements from the columns 1, 2, 4, 8, 16, ...
(End)
		

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 928.

Crossrefs

Cf. A169708 (first differences), A147562, A147582, A169648, A169649, A169709, A169710, A246333, A246334, A246335, A246336, A253098 (partial sums).
See A253088 for the analogous CA using Rule 750 and a 9-celled neighborhood.

Programs

  • Maple
    (Maple program that uses the actual definition of the automaton, rather than the (conjectured) formula, from N. J. A. Sloane, Feb 15 2015):
    # Count terms in a polynomial:
    C := f->`if`(type(f, `+`), nops(f), 1);
    # Replace all nonzero coeffts by 1:
    bool := proc(f) local ix, iy, f2, i, t1, t2, A;
    f2:=expand(f);
    if whattype(f) = `+` then
    t1:=nops(f2); A:=0;
    for i from 1 to t1 do t2:=op(i, f2); ix:=degree(t2, x); iy:=degree(t2, y);
    A:=A+x^ix*y^iy; od: A;
    else ix:=degree(f2, x); iy:=degree(f2, y); x^ix*y^iy;
    fi;
    end;
    # a loop that produces M steps of A169707 and A169708:
    M:=20;
    F:=x*y+x/y+1/x*y+1/x/y mod 2;
    GG[0]:=1;
    for n from 1 to M do dd[n]:=expand(F*GG[n-1]) mod 2;
    GG[n]:=bool(GG[n-1]+dd[n]);
    lprint(n,C(GG[n]), C(GG[n]-GG[n-1])); od:
  • Mathematica
    Map[Function[Apply[Plus,Flatten[ #1]]], CellularAutomaton[{ 750, {2,{{0,2,0},{2,1,2},{0,2,0}}},{1,1}},{{{1}},0},100]]
    ArrayPlot /@ CellularAutomaton[{750, {2, {{0, 2, 0}, {2, 1, 2}, {0, 2, 0}}}, {1, 1}}, {{{1}}, 0}, 23]
    (* The next two lines deal with the equivalent CA based on neighbors NW, NE, SE, SW. This is to facilitate the comparison with A246333 and A246335 *)
    Map[Function[Apply[Plus, Flatten[ #1]]], CellularAutomaton[{ 750, {2, {{2, 0, 2}, {0, 1, 0}, {2, 0, 2}}}, {1, 1}}, {{{1}}, 0}, 100]]
    ArrayPlot /@ CellularAutomaton[{750, {2, {{2, 0, 2}, {0, 1, 0}, {2, 0, 2}}}, {1, 1}}, {{{1}}, 0}, 23]

Formula

a(2^k + i) = (4^(k+1)-1)/3 + 4*A246336(i), for k >= 0, 0 <= i < 2^k. For example, if n = 15 = 2^3 + 7, so k=3, i=7, we have a(15) = (4^4-1)/3 + 4*A246336(7) = 85 + 4*49 = 281.
a(n) = 1 + 2*(A139250(n) - A160552(n)) = A160164(n) - A170903(n) = A187220(n) + 2*(A160552(n-1)). - Omar E. Pol, Feb 18 2015
It appears that a(n) = A162795(n) = A147562(n), if n is a member of A048645, otherwise a(n) > A162795(n) > A147562(n). - Omar E. Pol, Feb 19 2015
It appears that a(n) = 1 + 4*A255747(n-1). - Omar E. Pol, Mar 05 2015
It appears that a(n) = 1 + 4*(A139250(n-1) - (a(n-1) - 1)/4), n > 1. - Omar E. Pol, Jul 24 2015
It appears that a(2n) = 1 + 4*A162795(n). - Omar E. Pol, Jul 04 2017

Extensions

Edited (added formula, illustration, etc.) by N. J. A. Sloane, Aug 30 2014
Offset changed to 1 by N. J. A. Sloane, Feb 09 2015
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