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 1-9 of 9 results.

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.

Links

Crossrefs

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

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
Edited by N. J. A. Sloane, Apr 29 2009, incorporating comments from Omar E. Pol, M. F. Hasler, Rob Pratt, Jaume Oliver Lafont, Franklin T. Adams-Watters, R. J. Mathar, David W. Wilson, David Applegate, Benoit Jubin and others.
Further edited by N. J. A. Sloane, Jan 28 2010

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.

Links

Crossrefs

Cf. A169708 (first differences), A147562, A147582, A169648, A169649, A169709, A169710, A246333, A246334, A246335, A246336, A253098 (partial sums).
Cf. also A162795, A150552, A160104, A170903, A048645, A187200.
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

A162795 Total number of toothpicks in the toothpick structure A139250 that are parallel to the initial toothpick, after n odd rounds.

Original entry on oeis.org

1, 5, 9, 21, 25, 37, 53, 85, 89, 101, 117, 149, 165, 201, 261, 341, 345, 357, 373, 405, 421, 457, 517, 597, 613, 649, 709, 793, 853, 965, 1173, 1365, 1369, 1381, 1397, 1429, 1445, 1481, 1541, 1621, 1637, 1673, 1733, 1817, 1877, 1989, 2197, 2389, 2405, 2441, 2501
Offset: 1

Views

Author

Omar E. Pol, Jul 14 2009

Keywords

Comments

Partial sums of A162793.
Also, total number of ON cells at stage n of the two-dimensional cellular automaton defined as follows: replace every "vertical" toothpick of length 2 with a centered unit square "ON" cell, so we have a cellular automaton which is similar to both A147562 and A169707 (this is the "one-step bishop" version). For the "one-step rook" version we use toothpicks of length sqrt(2), then rotate the structure 45 degrees and then replace every toothpick with a unit square "ON" cell. For the illustration of the sequence as a cellular automaton we now have three versions: the original version with toothpicks, the one-step rook version and one-step bishop version. Note that the last two versions refer to the standard ON cells in the same way as the two versions of A147562 and the two versions of A169707. It appears that the graph of this sequence lies between the graphs of A147562 and A169707. Also, it appears that this sequence shares infinitely many terms with both A147562 and A169707, see Formula section and Example section. - Omar E. Pol, Feb 20 2015
It appears that this is also a bisection (the odd terms) of A255747.

Examples

			From _Omar E. Pol_, Feb 18 2015: (Start)
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, 53, 85;
   89,101,117,149,165,201,261,341;
  345,357,373,405,421,457,517,597,613,649,709,793,853,965,1173,1365;
  ...
The right border gives the positive terms of A002450.
(End)
It appears that T(j,k) = A147562(j,k) = A169707(j,k), if k is a power of 2, for example: it appears that the three mentioned triangles only share the elements of the columns 1, 2, 4, 8, 16, ... - _Omar E. Pol_, Feb 20 2015

Links

Crossrefs

Cf. A002450, A048645, A139250, A139251, A147562, A153000, A159791, A159792, A160164, A160552, A162793, A162794, A162796, A162797, A169707, A255263, A255264, A255747.

Formula

It appears that a(n) = A147562(n) = A169707(n), if n is a term of A048645, otherwise A147562(n) < a(n) < A169707(n). - Omar E. Pol, Feb 20 2015
It appears that a(n) = (A169707(2n) - 1)/4 = A255747(2n-1). - Omar E. Pol, Mar 07 2015
a(n) = 1 + 4*A255737(n-1). - Omar E. Pol, Mar 08 2015

Extensions

More terms from N. J. A. Sloane, Dec 28 2009

A256264 Partial sums of A256263.

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, 205, 234, 269, 310, 341, 342, 345, 350, 357, 362, 373, 390, 405, 410, 421, 438, 461, 490, 525, 566, 597, 602, 613, 630, 653, 682, 717, 758, 805, 858, 917, 982, 1053, 1130, 1213, 1302, 1365
Offset: 0

Views

Author

Omar E. Pol, Mar 30 2015

Keywords

Comments

First differs from A255747 at a(27).

Examples

			Also, written as an irregular triangle in which the row lengths are the terms of A011782, 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, 205, 234, 269,310,341;
...
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 at least shares with the triangles from the following sequences; A151920, A255737, A255747, A256249, the positive elements of the columns k, if k is a power of 2.
From _Omar E. Pol, Jan 02 2016: (Start)
Illustration of initial terms in the fourth quadrant of the square grid:
---------------------------------------------------------------------------
n    a(n)                 Compact diagram
---------------------------------------------------------------------------
0     0     _
1     1    |_|_ _
2     2      |_| |
3     5      |_ _|_ _ _ _
4     6          |_| | | |
5     9          |_ _| | |
6    14          |_ _ _| |
7    21          |_ _ _ _|_ _ _ _ _ _ _ _
8    22                  |_| | | |_ _  | |
9    25                  |_ _| | |_  | | |
10   30                  |_ _ _| | | | | |
11   37                  |_ _ _ _| | | | |
12   42                  | | |_ _ _| | | |
13   53                  | |_ _ _ _ _| | |
14   70                  |_ _ _ _ _ _ _| |
15   85                  |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
16   86                                  |_| | | |_ _  | |_ _ _ _ _ _  | |
17   89                                  |_ _| | |_  | | |_ _ _ _ _  | | |
18   94                                  |_ _ _| | | | | |_ _ _ _  | | | |
19  101                                  |_ _ _ _| | | | |_ _ _  | | | | |
20  106                                  | | |_ _ _| | | |_ _  | | | | | |
21  117                                  | |_ _ _ _ _| | |_  | | | | | | |
22  134                                  |_ _ _ _ _ _ _| | | | | | | | | |
23  149                                  |_ _ _ _ _ _ _ _| | | | | | | | |
24  154                                  | | | | | | |_ _ _| | | | | | | |
25  165                                  | | | | | |_ _ _ _ _| | | | | | |
26  182                                  | | | | |_ _ _ _ _ _ _| | | | | |
27  205                                  | | | |_ _ _ _ _ _ _ _ _| | | | |
28  234                                  | | |_ _ _ _ _ _ _ _ _ _ _| | | |
29  269                                  | |_ _ _ _ _ _ _ _ _ _ _ _ _| | |
30  310                                  |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
31  341                                  |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.
a(n) is also the total number of cells in the first n regions of the diagram. A256263(n) gives the number of cells in the n-th region of the diagram.
(End)

Links

Crossrefs

Cf. A002450, A011782, A047849, A139250, A151920, A255737, A255747, A256249, A256258, A256260, A256261, A256263, A256265.

Programs

  • Mathematica
    Accumulate@Flatten@Join[{0}, NestList[Join[#, Range[Length[#] - 1]*6 - 1, {2 #[[-1]] + 1}] &, {1}, 5]] (* Ivan Neretin, Feb 14 2017 *)

Formula

a(n) = (A256260(n+1) - 1)/4.

A256249 Partial sums of A006257 (Josephus problem).

Original entry on oeis.org

0, 1, 2, 5, 6, 9, 14, 21, 22, 25, 30, 37, 46, 57, 70, 85, 86, 89, 94, 101, 110, 121, 134, 149, 166, 185, 206, 229, 254, 281, 310, 341, 342, 345, 350, 357, 366, 377, 390, 405, 422, 441, 462, 485, 510, 537, 566, 597, 630, 665, 702, 741, 782, 825, 870, 917, 966, 1017, 1070, 1125, 1182, 1241, 1302, 1365, 1366, 1369, 1374
Offset: 0

Views

Author

Omar E. Pol, Mar 20 2015

Keywords

Comments

Also total number of ON states after n generations in one of the four wedges of the one-step rook version (or in one of the four quadrants of the one-step bishop version) of the cellular automaton of A256250.
A006257 gives the number of cells turned ON at n-th stage.
First differs from A255747 at a(11).
First differs from A169779 at a(10).
It appears that the odd terms (a bisection) give A256250.

Examples

			Written as an irregular triangle T(n,k), k >= 1, in which the row lengths are the terms of A011782 the sequence begins:
   0;
   1;
   2,  5;
   6,  9, 14, 21;
  22, 25, 30, 37, 46, 57, 70, 85;
  86, 89, 94,101,110,121,134,149,166,185,206,229,254,281,310,341;
  ...
Right border, a(2^m-1), gives A002450(m) for m >= 0.
a(2^m-2) = A203241(m) for m >= 2.
It appears that this triangle at least shares with the triangles from the following sequences; A151920, A255737, A255747, the positive elements of the columns k, if k is a power of 2.
From _Omar E. Pol_, Jan 03 2016: (Start)
Illustration of initial terms in the fourth quadrant of the square grid:
---------------------------------------------------------------------------
n    a(n)                 Compact diagram
---------------------------------------------------------------------------
0     0     _
1     1    |_|_ _
2     2      |_| |
3     5      |_ _|_ _ _ _
4     6          |_| | | |
5     9          |_ _| | |
6    14          |_ _ _| |
7    21          |_ _ _ _|_ _ _ _ _ _ _ _
8    22                  |_| | | | | | | |
9    25                  |_ _| | | | | | |
10   30                  |_ _ _| | | | | |
11   37                  |_ _ _ _| | | | |
12   46                  |_ _ _ _ _| | | |
13   57                  |_ _ _ _ _ _| | |
14   70                  |_ _ _ _ _ _ _| |
15   85                  |_ _ _ _ _ _ _ _|
.
a(n) is also the total number of cells in the first n regions of the diagram. A006257(n) gives the number of cells in the n-th region of the diagram.
(End)

Links

Crossrefs

Cf. A002450, A006257, A011782, A139250, A151920, A169779, A255737, A255747, A256250, A256251, A266540.

Programs

Formula

a(n) = (A256250(n+1) - 1)/4.

A256265 Sums of two successive terms of A256264.

Original entry on oeis.org

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, 387, 439, 503, 579, 651, 683, 687, 695, 707, 719, 735, 763, 795, 815, 831, 859, 899, 951, 1015, 1091, 1163, 1199, 1215, 1243, 1283, 1335, 1399, 1475, 1563, 1663, 1775, 1899, 2035, 2183, 2343, 2515, 2667, 2731
Offset: 1

Views

Author

Omar E. Pol, Apr 10 2015

Keywords

Comments

First differs from A139250 (the toothpick sequence) at a(27).
It appears that both sequences share infinitely many terms, for example: a(1)..a(26), a(31)..a(42), a(47)..a(50), a(63)..a(74), a(79)..a(82), etc.
The main entry for this sequence is A256263.

Crossrefs

Cf. A007583, A139250, A255747, A256260, A256261, A256263, A256264.

Formula

a(n) = A256264(n) + A256264(n-1).

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.

Links

Crossrefs

Partial sums give A266530.
Cf. A004171, A011782, A139250, A160552, A255747, A266509, A266539.

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 *)

A266530 Partial sums of A266529.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 7, 10, 11, 12, 15, 18, 23, 28, 35, 42, 43, 44, 47, 50, 55, 60, 67, 74, 79, 84, 95, 106, 123, 140, 155, 170, 171, 172, 175, 178, 183, 188, 195, 202, 207, 212, 223, 234, 251, 268, 283, 298, 303, 308, 319, 330, 347, 364, 383, 402, 423, 444, 483, 522, 571, 620, 651, 682, 683, 684, 687, 690, 695, 700
Offset: 1

Views

Author

Omar E. Pol, Jan 02 2016

Keywords

Comments

Also the toothpick sequence A139250 and twice the terms of A255747 interleaved.
It appears that this sequence has a fractal (or fractal-like) behavior.
First differs from A266510 at a(55), with which it shares infinitely many terms.
First differs from A266540 at a(25), with which it shares infinitely many terms.

Links

Crossrefs

Cf. A139250, A160552, A255747, A266510, A266529, A266540.

Programs

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

Formula

a(2n-1) = A139250(n).
a(2n) = 2 * A255747(n-1).
a(n) = (a(n-1) + a(n+1))/2, if n is an odd number greater than 1.

A255737 Total number of toothpicks in the toothpick structure of A153000 that are parallel to the initial toothpick, after n odd rounds.

Original entry on oeis.org

0, 1, 2, 5, 6, 9, 13, 21, 22, 25, 29, 37, 41, 50, 65, 85, 86, 89, 93, 101, 105, 114, 129, 149, 153, 162, 177, 198, 213, 241, 293, 341, 342, 345, 349, 357, 361, 370, 385, 405, 409, 418, 433, 454, 469, 497, 549, 597, 601, 610, 625, 646, 661, 689, 741, 790, 805, 833, 885, 941, 994, 1085, 1253, 1365, 1366, 1369
Offset: 0

Views

Author

Omar E. Pol, Mar 07 2015

Keywords

Comments

Total number of toothpicks in the first quadrant of the toothpick structure of A139250 that are parallel to the initial toothpick, after n odd rounds.
Written as an irregular triangle in which the row lengths are the terms of A011782 the right border gives A002450.

Links

Crossrefs

Cf. A002450, A011782, A139250, A151920, A153000, A162795, A255747.

Formula

a(n) = (A162795(n+1) - 1)/4.
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