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.

Showing 1-10 of 14 results. Next

A152999 Primes in toothpick sequence A152998.

Original entry on oeis.org

3, 5, 7, 11, 17, 23, 47, 61, 97, 103, 151, 173, 191, 211, 241, 347, 353, 359, 367, 397, 467, 541, 599, 607, 659, 733, 1109, 1237, 1367, 1439, 1453, 1471, 1663, 2029, 2357, 2399, 2671, 2797, 3373, 3607, 3719, 3911, 4241, 5479, 5501, 5527, 5701, 5741, 5779, 5923
Offset: 1

Views

Author

Omar E. Pol, Dec 23 2008

Keywords

Crossrefs

Extensions

More terms from Jinyuan Wang, Mar 15 2020

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

A153000 Toothpick sequence in the first quadrant.

Original entry on oeis.org

0, 1, 2, 3, 5, 8, 10, 11, 13, 16, 19, 23, 30, 38, 42, 43, 45, 48, 51, 55, 62, 70, 75, 79, 86, 95, 105, 120, 142, 162, 170, 171, 173, 176, 179, 183, 190, 198, 203, 207, 214, 223, 233, 248, 270, 290, 299, 303, 310, 319, 329, 344, 366, 387
Offset: 0

Views

Author

Omar E. Pol, Dec 16 2008, Dec 20 2008, Jan 02 2009

Keywords

Comments

From Omar E. Pol, Nov 29 2009: (Start)
At stage 0, we start from a horizontal half toothpick at [(0,1),(1,1)]. This half toothpick represents one of the two components of the second toothpick placed in the toothpick structure of A139250. Consider only the toothpicks of length 2, so a(0) = 0.
At stage 1 we place an orthogonal toothpick of length 2 centered at the end, so a(1) = 1.
In each subsequent stage, for every exposed toothpick end, place an orthogonal toothpick centered at that end.
The sequence gives the number of toothpicks after n stages. Note that this sequence contains even numbers and odd numbers, the same as A152978 (the first differences) which gives the number of toothpicks added at n-th stage. For more information see A139250. (End)
A079559 gives the parity of this sequence, if n >= 1. - Omar E. Pol, Aug 13 2013

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

Crossrefs

Programs

  • Maple
    G := (1+x)*(mul(1+x^(2^k-1)+2*x^(2^k),k=1..20)-1)/((1-x)*(1+2*x)); # N. J. A. Sloane, May 20 2009
  • Python
    def msb(n):
        t=0
        while n>>t>0: t+=1
        return 2**(t - 1)
    def a139250(n):
        k=(2*msb(n)**2 + 1)//3
        return 0 if n==0 else k if n==msb(n) else k + 2*a139250(n - msb(n)) + a139250(n - msb(n) + 1) - 1
    def a(n): return 0 if n==0 else (a139250(n + 2) - 3)//4
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 01 2017

Formula

a(n) = (A139250(n+2)-3)/4 = (A152998(n+1)-1)/2.
G.f.: (1+x)*(Product_{k>=1} (1+x^(2^k-1)+2*x^(2^k))-1)/((1-x)*(1+2*x)). - N. J. A. Sloane, May 20 2009
Contribution from Omar E. Pol, Oct 01 2011: (Start)
a(n) = A152998(n+1) + A153003(n+1) - A139250(n+2) + 1.
a(n) = A139250(n+2) - A153003(n+1) - 2.
a(n) = A153003(n+1) - A152998(n+1).
(End)
a(n) = (A187220(n+3) - 7)/8. - Omar E. Pol, Feb 16 2013

A152978 a(n) = A139251(n+2)/4 = A152968(n+1)/2.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Dec 16 2008, Dec 20 2008

Keywords

Comments

Also, first differences of toothpick numbers A153000.

Examples

			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]

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

Extensions

More terms from Omar E. Pol, Jul 26 2009

A152968 a(n) = A139251(n+1)/2.

Original entry on oeis.org

1, 2, 2, 2, 4, 6, 4, 2, 4, 6, 6, 8, 14, 16, 8, 2, 4, 6, 6, 8, 14, 16, 10, 8, 14, 18, 20, 30, 44, 40, 16, 2, 4, 6, 6, 8, 14, 16, 10, 8, 14, 18, 20, 30, 44, 40, 18, 8, 14, 18, 20, 30, 44, 42, 28, 30, 46, 56, 70, 104, 128, 96, 32, 2
Offset: 1

Views

Author

Omar E. Pol, Dec 16 2008, Dec 20 2008

Keywords

Comments

Also, first differences of toothpicks numbers A152998. [From Omar E. Pol, Jan 02 2009]

Examples

			Triangle begins:
.1;
.2,2;
.2,4,6,4;
.2,4,6,6,8,14,16,8;
.2,4,6,6,8,14,16,10,8,14,18,20,30,44,40,16;
....
Rows approach A151688. - _N. J. A. Sloane_, Jun 03 2009
		

Crossrefs

Formula

Write n = 2^i +j, 0 <= j < 2^i; then a(n) = Sum_k 2^(wt(j+k)-k)*binomial(wt(j+k),k). except that a(2^r-1) = 2^(r-1). - N. J. A. Sloane, Jun 03 2009, Jul 16 2009
G.f.: x*(Prod(1+x^(2^k-1)+2*x^(2^k),k=0..oo)-1)/(1+2*x). - N. J. A. Sloane, Jun 05 2009

A151920 a(n) = (Sum_{i=1..n+1} 3^wt(i))/3, where wt() = A000120().

Original entry on oeis.org

1, 2, 5, 6, 9, 12, 21, 22, 25, 28, 37, 40, 49, 58, 85, 86, 89, 92, 101, 104, 113, 122, 149, 152, 161, 170, 197, 206, 233, 260, 341, 342, 345, 348, 357, 360, 369, 378, 405, 408, 417, 426, 453, 462, 489, 516, 597, 600, 609, 618, 645, 654, 681, 708, 789, 798, 825, 852, 933, 960
Offset: 0

Views

Author

N. J. A. Sloane, Aug 05 2009, Aug 06 2009

Keywords

Comments

Partial sums of A147610 (but with offset changed to 0).
It appears that the first bisection gives the positive terms of A147562. - Omar E. Pol, Mar 07 2015

Examples

			n=3: (3^1+3^1+3^2+3^1)/3 = 18/3 = 6.
n=18: the binary expansion of 18+1 is 10011, i.e., 19 = 2^4 + 2^1 + 2^0.
The exponents of these powers of 2 (4, 1 and 0) reoccur as exponents in the powers of 4: a(19) = 3^0 * [(4^4 - 1) / 3 + 1] + 3^1 * [(4^1 - 1) / 3 + 1] + 3^2 * [(4^0 - 1)/3 + 1] = 1 * 86 + 3 * 2 + 9 * 1 = 101. - _David A. Corneth_, Mar 21 2015
		

Crossrefs

Programs

  • Mathematica
    t = Nest[Join[#, # + 1] &, {0}, 14]; Table[Sum[3^t[[i + 1]], {i, 1, n}]/3, {n, 60}] (* Michael De Vlieger, Mar 21 2015 *)
  • PARI
    a(n) = sum(i=1, n+1, 3^hammingweight(i))/3; \\ Michel Marcus, Mar 07 2015
    
  • PARI
    a(n)={b=binary(n+1);t=#b;e=-1;sum(i=1,#b,e+=(b[i]==1);(b[i]==1)*3^e*((4^(#b-i)-1)/3+1))} \\ David A. Corneth, Mar 21 2015

Formula

a(n) = (A147562(n+2) - 1)/4 = (A151917(n+2) - 1)/2. - Omar E. Pol, Mar 13 2011
a(n) = (A130665(n+1) - 1)/3. - Omar E. Pol, Mar 07 2015
a(n) = a(n-1) + 3^A000120(n+1)/3. - David A. Corneth, Mar 21 2015

A153003 Toothpick sequence in the first three quadrants.

Original entry on oeis.org

0, 1, 4, 7, 10, 16, 25, 31, 34, 40, 49, 58, 70, 91, 115, 127, 130, 136, 145, 154, 166, 187, 211, 226, 238, 259, 286, 316, 361, 427, 487, 511, 514, 520, 529, 538, 550, 571, 595, 610, 622, 643, 670, 700, 745, 811, 871, 898, 910, 931
Offset: 0

Views

Author

Omar E. Pol, Jan 02 2009

Keywords

Comments

From Omar E. Pol, Oct 01 2011: (Start)
On the infinite square grid, consider only the first three quadrants and count only the toothpicks of length 2.
At stage 0, we start from a vertical half toothpick at [(0,0),(0,1)]. This half toothpick represents one of the two components of the first toothpick placed in the toothpick structure of A139250, so a(0) = 0.
At stage 1, we place an orthogonal toothpick of length 2 centered at the end, so a(1) = 1. Also we place half toothpick at [(0,-1),(1,-1)]. This last half toothpick represents one of the two components of the third toothpick placed in the toothpick structure of A139250.
At stage 2, we place three toothpicks, so a(2) = 1+3 = 4.
In each subsequent stage, for every exposed toothpick end, place an orthogonal toothpick centered at that end.
The sequence gives the number of toothpicks after n stages. A153004 (the first differences) gives the number of toothpicks added to the structure at n-th stage.
Note that this sequence is different from the toothpick "corner" sequence A153006. For more information see A139250. (End)

Crossrefs

Programs

  • Mathematica
    A139250[n_] := A139250[n] = Module[{m, k}, If[n == 0, Return[0]]; m = 2^(Length[IntegerDigits[n, 2]] - 1); k = (2 m^2 + 1)/3; If[n == m, k, k + 2 A139250[n - m] + A139250[n - m + 1] - 1]];
    a[n_] := If[n == 0, 0, (3/4)(A139250[n + 1] - 3) + 1];
    a /@ Range[0, 49] (* Jean-François Alcover, Apr 06 2020 *)
  • Python
    def msb(n):
        t=0
        while n>>t>0: t+=1
        return 2**(t - 1)
    def a139250(n):
        k=(2*msb(n)**2 + 1)/3
        return 0 if n==0 else k if n==msb(n) else k + 2*a139250(n - msb(n)) + a139250(n - msb(n) + 1) - 1
    def a(n): return 0 if n==0 else (a139250(n + 1) - 3)*3/4 + 1
    [a(n) for n in range(51)] # Indranil Ghosh, Jul 01 2017

Formula

a(n) = (A139250(n+1)-3)*3/4 + 1, if n >= 1.
From Omar E. Pol, Oct 01 2011: (Start)
a(n) = A139250(n+1) - A152998(n) + A153000(n-1) - 1, if n >= 1.
a(n) = A139250(n+1) - A153000(n-1) - 2, if n >= 1.
a(n) = A152998(n) + A153000(n-1), if n >= 1.
(End)

A153004 First differences of toothpick numbers A153003.

Original entry on oeis.org

1, 3, 3, 3, 6, 9, 6, 3, 6, 9, 9, 12, 21, 24, 12, 3, 6, 9, 9, 12, 21, 24, 15, 12, 21, 27, 30, 45, 66, 60, 24, 3, 6, 9, 9, 12, 21, 24, 15, 12, 21, 27, 30, 45, 66, 60, 27, 12, 21, 27, 30, 45, 66, 63, 42, 45, 69, 84, 105, 156, 192, 144, 48, 3
Offset: 1

Views

Author

Omar E. Pol, Jan 02 2009

Keywords

Examples

			Triangle begins:
1;
3,3;
3,6,9,6;
3,6,9,9,12,21,24,12;
3,6,9,9,12,21,24,15,12,21,27,30,45,66,60,24;
		

Crossrefs

Toothpick sequence: A139250.

Programs

  • Mathematica
    A139250[n_] := A139250[n] = Module[{m, k}, If[n == 0, Return[0]]; m = 2^(Length[IntegerDigits[n, 2]] - 1); k = (2 m^2 + 1)/3; If[n == m, k, k + 2 A139250[n - m] + A139250[n - m + 1] - 1]];
    b[n_] := If[n == 0, 0, (3/4) (A139250[n + 1] - 3) + 1];
    b /@ Range[0, 64] // Differences (* Jean-François Alcover, Apr 06 2020 *)

A159788 a(n) = A159786(n+1)/2.

Original entry on oeis.org

0, 0, 2, 4, 4, 6, 16, 24, 24, 26, 32, 36, 38, 52, 88, 112, 112, 114, 120, 124, 126, 140, 168, 184, 186, 196, 212, 222, 240, 304, 432, 480, 480, 482, 488, 492, 494, 508, 536, 552, 554, 564, 580, 590, 608, 672, 768, 816, 818, 828
Offset: 0

Views

Author

Omar E. Pol, Apr 28 2009, May 02 2009

Keywords

Crossrefs

Toothpick sequence: A139250.

Extensions

a(11)-a(49) from Robert Price, May 10 2019

A222176 Total number of ON states after n generations of cellular automaton based on pentagons starting near of the vertex of an infinite 72-degree wedge.

Original entry on oeis.org

0, 1, 3, 5, 7, 11, 17, 21, 23, 27, 35, 43, 49, 59, 73, 81, 83, 87, 95, 107, 123
Offset: 0

Views

Author

Omar E. Pol, Mar 21 2013

Keywords

Comments

The structure is one of the five wedges of A222180.
The sequence gives the number of pentagons in the structure after n-th stage. A222177 (the first differences) gives the number of pentagons added at n-th stage.
Also P-toothpick sequence starting near of the vertex of an infinite 72-degree wedge, since each pentagon can be replaced by a P-toothpick.
For more information see A222180.
First differs from A152998 at a(10).

Crossrefs

Formula

a(n) = (A222180(n+1) - 1)/5.

Extensions

Name improved by Omar E. Pol, Nov 24 2013
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