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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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

A078008 Expansion of (1 - x)/((1 + x)*(1 - 2*x)).

Original entry on oeis.org

1, 0, 2, 2, 6, 10, 22, 42, 86, 170, 342, 682, 1366, 2730, 5462, 10922, 21846, 43690, 87382, 174762, 349526, 699050, 1398102, 2796202, 5592406, 11184810, 22369622, 44739242, 89478486, 178956970, 357913942, 715827882, 1431655766, 2863311530, 5726623062
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

Conjecture: a(n) = the number of fractions in the infinite Farey row of 2^n terms with even denominators. Compare the Salamin & Gosper item in the Beeler et al. link. - Gary W. Adamson, Oct 27 2003
Counts closed walks starting and ending at the same vertex of a triangle. 3*a(n) = P(C_n, 3) chromatic polynomial for 3 colors on cyclic graph C_n. A078008(n) + 2*A001045(n) = 2^n provides decomposition of Pascal's triangle. - Paul Barry, Nov 17 2003
Permutations with one fixed point avoiding 123 and 132.
General form: iterate k -> 2^n - k. See also A001045. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
The inverse g.f. generates sequence 1, 0, -2, -2, -2, -2, ...
a(n) gives the number of oriented (i.e., unreduced for symmetry) meanders on an (n+2) X 3 rectangular grid; see A201145. - Jon Wild, Nov 22 2011
Pisano period lengths: 1, 1, 6, 1, 4, 6, 6, 2, 18, 4, 10, 6, 12, 6, 12, 2, 8, 18, 18, 4, ... - R. J. Mathar, Aug 10 2012
a(n) is the number of length n binary words that end in an odd length run of 0's if we do not include the first letter of the word in our run length count. a(4) =6 because we have 0000, 0010, 0110, 1000, 1010, 1110. - Geoffrey Critzer, Dec 16 2013
a(n) is the top left entry of the n-th power of any of the six 3 X 3 matrices [0, 1, 1; 1, 1, 1; 1, 0, 0], [0, 1, 1; 1, 1, 0; 1, 1, 0], [0, 1, 1; 1, 0, 1; 1, 1, 0], [0, 1, 1; 1, 1, 0; 1, 0, 1], [0, 1, 1; 1, 0, 1; 1, 0, 1] or [0, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 04 2014
a(n) is the number of compositions of n into parts of two kinds without part 1. - Gregory L. Simay, Jun 04 2018
a(n) is the number of words of length n over a binary alphabet whose position in the lexicographic order is a multiple of three. a(3) = 2: aba, bab. - Alois P. Heinz, Apr 13 2022
a(n) is the number of words of length n over a ternary alphabet starting with a fixed letter (say, 'a') and ending in a different letter, such that no two adjacent letters are the same. a(4) = 6: abab, abac, abcb, acab, acac, acbc. - Ignat Soroko, Jul 19 2023

Examples

			G.f. = 1 + 2*x^2 + 2*x^3 + 6*x^4 + 10*x^5 + 22*x^6 + ... - _Michael Somos_, Mar 18 2022
		

References

  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983, ex. 1.1.10a.

Crossrefs

First differences of A001045.
See A151575 for a signed version.
Bisections: A047849, A020988.

Programs

Formula

Euler expands(1-x)/(1 - x - 2*x^2) into an infinite series and finds that the coefficient of the n-th term is (2^n + (-1)^n 2)/3. Section 226 shows that Euler could have easily found the recursion relation: a(n) = a(n-1) + 2a(n-2) with a(0) = 1 and a(1) = 0. - V. Frederick Rickey (fred-rickey(AT)usma.edu), Feb 10 2006. [Typos corrected by Jaume Oliver Lafont, Jun 01 2009]
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n)+3*k) where f(n) = (0, 2, 1, 0, 2, 1, ...) = A080424(n). - Paul Barry, Feb 20 2003
E.g.f. (exp(2*x) + 2*exp(-x))/3. - Paul Barry, Apr 20 2003
a(n) = A001045(n) + (-1)^n = A000079(n) - 2*A001045(n). - Paul Barry, Feb 20 2003
a(n) = (2^n + 2*(-1)^n)/3. - Mario Catalani (mario.catalani(AT)unito.it), Aug 29 2003
a(n) = T(n, i/(2*sqrt(2)))*(-i*sqrt(2))^n - U(n-1, i/(2*sqrt(2)))*(-i*sqrt(2))^(n-1)/2. - Paul Barry, Nov 17 2003
From Paul Barry, Jul 30 2004: (Start)
a(n) = 2*a(n-1) + 2*(-1)^n for n > 0, with a(0)=1.
a(n) = Sum_{k=0..n} (-1)^k*(2^(n-k-1) + 0^(n-k)/2). (End)
a(n) = A014113(n-1) for n > 0; a(n) = A052953(n-1) - 2*(n mod 2) = sum of n-th row of the triangle in A108561. - Reinhard Zumkeller, Jun 10 2005
A137208(n+1) - 2*A137208(n) = a(n) signed. - Paul Curtz, Aug 03 2008
a(n) = A001045(n+1) - A001045(n) - Paul Curtz, Feb 09 2009
If p[1] =0, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
a(n) = 2*(a(n-2) + a(n-3) + a(n-4) ... + a(0)), that is, twice the sum of all the previous terms except the last; with a(0) = 1 and a(1) = 0. - Benoit Jubin, Nov 21 2011
a(n+1) = 2*A001045(n). - Benoit Jubin, Nov 22 2011
G.f.: 1 - x + x*Q(0), where Q(k) = 1 + 2*x^2 + (2*k+3)*x - x*(2*k+1 + 2*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 05 2013
G.f.: 1+ x^2*Q(0), where Q(k) = 1 + 1/(1 - x*(4*k+1+2*x)/(x*(4*k+3+2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 01 2014
a(n) = 3*a(n-2) + 2*a(n-3). - David Neil McGrath, Sep 10 2014
a(n) = (-1)^n * A151575(n). - G. C. Greubel, Jun 28 2019
a(n)+a(n+1) = 2^n. - R. J. Mathar, Feb 24 2021
a(n) = -a(2-n) * (-2)^(n-1) = (3/2)*(a(n-1) + a(n-2)) - a(n-3) for all n in Z. - Michael Somos, Mar 18 2022

A153006 Toothpick sequence starting at the outside corner of an infinite square from which protrudes a half toothpick.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Dec 17 2008, Dec 19 2008, Apr 28 2009

Keywords

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

Crossrefs

Programs

  • 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

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

Extensions

Edited by N. J. A. Sloane, Dec 19 2008

A084247 a(n) = -a(n-1) + 2*a(n-2), a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 0, 4, -4, 12, -20, 44, -84, 172, -340, 684, -1364, 2732, -5460, 10924, -21844, 43692, -87380, 174764, -349524, 699052, -1398100, 2796204, -5592404, 11184812, -22369620, 44739244, -89478484, 178956972, -357913940, 715827884, -1431655764, 2863311532
Offset: 0

Views

Author

Paul Barry, May 23 2003

Keywords

Comments

Row sums of triangle in A112555. - Philippe Deléham, Sep 21 2009

Crossrefs

Programs

  • Magma
    [(4-(-2)^n)/3: n in [0..40]]; // Vincenzo Librandi, Aug 13 2011
    
  • Mathematica
    LinearRecurrence[{-1,2},{1,2},40] (* Harvey P. Dale, Nov 05 2016 *)
  • PARI
    Vec((1+3*x)/((1-x)*(1+2*x)) + O(x^40)) \\ Michel Marcus, Feb 25 2016
    
  • SageMath
    [(4-(-2)^n)/3 for n in range(41)] # G. C. Greubel, Apr 24 2023

Formula

Binomial transform of A084246.
a(n) = A077925(n-1) + 1.
a(n) = (4 - (-2)^n)/3.
G.f.: (1+3*x)/((1-x)*(1+2*x)).
E.g.f.: (4*exp(x) - exp(-2*x))/3.
a(n) = (-1)^(n+1)*(A078008(n+1) - A078008(n)). - Paul Curtz, Jun 30 2008, Feb 24 2021
a(n) = -2*a(n-1) + 4 . - Philippe Deléham, Sep 15 2009
a(n+1) = 2*A151575(n). - Philippe Deléham, Sep 21 2009
a(n) = A077925(n) + 3*A077925(n-1). - R. J. Mathar, Feb 24 2021

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

A151555 G.f.: (1 + 2x) * Product_{n>=1} (1 + x^(2^n-1) + 2*x^(2^n)).

Original entry on oeis.org

1, 3, 4, 5, 5, 10, 12, 9, 5, 10, 13, 15, 20, 32, 32, 17, 5, 10, 13, 15, 20, 32, 33, 23, 20, 33, 41, 50, 72, 96, 80, 33, 5, 10, 13, 15, 20, 32, 33, 23, 20, 33, 41, 50, 72, 96, 81, 39, 20, 33, 41, 50, 72, 97, 89, 66, 73, 107, 132, 172, 240, 272, 192, 65, 5, 10, 13, 15, 20, 32, 33, 23
Offset: 0

Views

Author

N. J. A. Sloane, May 20 2009

Keywords

Comments

From Gary W. Adamson, May 25 2009: (Start)
Convolved with A078008 signed (A151575) [1, 0, 2, -2, 6, -10, 22, -42, 86, -170, ...]
equals the toothpick sequence A153006: (1, 3, 6, 9, 13, 20, 28, ...). (End)
If A151550 is written as a triangle then the rows converge to this sequence. - N. J. A. Sloane, Jun 16 2009

Examples

			From _Omar E. Pol_, Jun 19 2009: (Start)
May be written as a triangle:
1;
3;
4,5;
5,10,12,9;
5,10,13,15,20,32,32,17;
5,10,13,15,20,32,33,23,20,33,41,50,72,96,80,33;
5,10,13,15,20,32,33,23,20,33,41,50,72,96,81,39,20,33,41,50,72,97,89,66,73,...
(End)
		

Crossrefs

A238275 a(n) = (4*7^n - 1)/3.

Original entry on oeis.org

1, 9, 65, 457, 3201, 22409, 156865, 1098057, 7686401, 53804809, 376633665, 2636435657, 18455049601, 129185347209, 904297430465, 6330082013257, 44310574092801, 310174018649609, 2171218130547265, 15198526913830857, 106389688396816001, 744727818777712009
Offset: 0

Views

Author

Philippe Deléham, Feb 21 2014

Keywords

Comments

Sum of n-th row of triangle of powers of 7: 1; 1 7 1; 1 7 49 7 1; 1 7 49 343 49 7 1; ...
Number of cubes in the crystal structure cubic carbon CCC(n+1), defined in the Baig et al. and in the Gao et al. references. - Emeric Deutsch, May 28 2018

Examples

			a(0) = 1;
a(1) = 1 + 7 + 1 = 9;
a(2) = 1 + 7 + 49 + 7 + 1 = 65;
a(3) = 1 + 7 + 49 + 343 + 49 + 7 + 1 = 457; etc.
		

Crossrefs

Cf. Similar sequences: A151575, A000012, A040000, A005408, A033484, A048473, A020989, A057651, A061801, this sequence, A238276, A138894, A090843, A199023.

Programs

Formula

G.f.: (1+x)/((1-x)*(1-7*x)).
a(n) = 7*a(n-1) + 2, a(0) = 1.
a(n) = 8*a(n-1) - 7*a(n-2), a(0) = 1, a(1) = 9.
a(n) = Sum_{k=0..n} A112468(n,k)*8^k.
E.g.f.: exp(x)*(4*exp(6*x) - 1)/3. - Stefano Spezia, Feb 12 2025

A238276 a(n) = (9*8^n - 2)/7.

Original entry on oeis.org

1, 10, 82, 658, 5266, 42130, 337042, 2696338, 21570706, 172565650, 1380525202, 11044201618, 88353612946, 706828903570, 5654631228562, 45237049828498, 361896398627986, 2895171189023890, 23161369512191122, 185290956097528978, 1482327648780231826
Offset: 0

Views

Author

Philippe Deléham, Feb 21 2014

Keywords

Comments

Sum of n-th row of triangle of powers of 8: 1; 1 8 1; 1 8 64 8 1; 1 8 64 512 64 8 1; ...

Examples

			a(0) = 1;
a(1) = 1 + 8 + 1 = 10;
a(2) = 1 + 8 + 64 + 8 + 1 = 82;
a(3) = 1 + 8 + 64 + 512 + 64 + 8 + 1 = 658; etc.
		

Crossrefs

Cf. Similar sequences: A151575, A000012, A040000, A005408, A033484, A048473, A020989, A057651, A061801, A238275, this sequence, A138894, A090843, A199023.

Programs

Formula

G.f.: (1+x)/((1-x)*(1-8*x)).
a(n) = 8*a(n-1) + 2, a(0) = 1.
a(n) = 9*a(n-1) - 8*a(n-2), a(0) = 1, a(1) = 10.
a(n) = Sum_{k=0..n} A112468(n,k)*9^k.

Extensions

Corrected by Vincenzo Librandi, Feb 23 2014

A238303 Triangle T(n,k), 0<=k<=n, read by rows given by T(n,0) = 1, T(n,k) = 2 if k>0.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 0

Views

Author

Philippe Deléham, Feb 24 2014

Keywords

Comments

Row sums are A005408(n).
Diagonals sums are A109613(n).
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A000012(n), A005408(n), A036563(n+2), A058481(n+1), A083584(n), A137410(n), A233325(n), A233326(n), A233328(n), A211866(n+1), A165402(n+1) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A151575(n), A000012(n), A040000(n), A005408(n), A033484(n), A048473(n), A020989(n), A057651(n), A061801(n), A238275(n), A238276(n), A138894(n), A090843(n), A199023(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively.
Sum_{k=0..n} T(n,k)^x = A000027(n+1), A005408(n), A016813(n), A017077(n) for x = 0, 1, 2, 3 respectively.
Sum_{k=0..n} k*T(n,k) = A002378(n).
Sum_{k=0..n} A000045(k)*T(n,k) = A019274(n+2).
Sum_{k=0..n} A000142(k)*T(n,k) = A066237(n+1).

Examples

			Triangle begins:
1;
1, 2;
1, 2, 2;
1, 2, 2, 2;
1, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2;
...
		

Crossrefs

Cf. Diagonals: A040000.
Cf. Columns: A000012, A007395.
First differences of A001614.

Programs

Formula

T(n,0) = A000012(n) = 1, T(n+k,k) = A007395(n) = 2 for k>0.

Extensions

Data section extended to a(104) by Antti Karttunen, Jan 19 2025

A113679 Expansion of (1-x-2x^2)/(1-x).

Original entry on oeis.org

1, 0, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2
Offset: 0

Views

Author

Paul Barry, Nov 04 2005

Keywords

Comments

From Gary W. Adamson, Mar 06 2012: (Start)
Signed (1, 0, -2, 2, -2, 2, ...) and convolved with the Toothpick sequence A139250 = A151548: (1, 3, 5, 7, 5, 11, ...). The inverse of (1, 0, -2, 2, -2, ...) = A151575: (1, 0, 2, -2, 6, -10, 22, ...).
The unsigned sequence convolved with:
(1, 2, 3, ...) = A002523, (n^2 + 1). Convolved with:
(A001045) = .... A097064: (1, 1, 5, 9, 21, 41, ...).
(End)

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-x-2x^2)/(1-x),{x,0,80}],x] (* or *) Join[{1,0}, PadRight[{},80,-2]] (* Harvey P. Dale, Mar 05 2012 *)

Formula

a(n) = C(0, n) + 2*C(1, n) - 2.
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