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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.

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

A005836 Numbers whose base-3 representation contains no 2.

Original entry on oeis.org

0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121, 243, 244, 246, 247, 252, 253, 255, 256, 270, 271, 273, 274, 279, 280, 282, 283, 324, 325, 327, 328, 333, 334, 336, 337, 351, 352
Offset: 1

Views

Author

N. J. A. Sloane, Jeffrey Shallit

Keywords

Comments

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.
This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001
In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010
Complement of A074940. - Reinhard Zumkeller, Mar 23 2003
Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003
Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003
A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.
Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006
Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009
Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011
Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011
It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011
Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015
Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019

Examples

			12 is a term because 12 = 110_3.
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
   0
   1
   3,  4
   9, 10, 12, 13
  27, 28, 30, 31, 36, 37, 39, 40
  81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121
... - _Philippe Deléham_, Jun 06 2015

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E10, pp. 317-323.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A039966 (characteristic function).
Cf. A002426, A004793, A005823, A007088, A007089, A032924, A033042-A033052, A054591, A055246, A062548, A065361, A074940, A081601, A081603, A081611, A083096, A089118, A121153, A170943, A185256.
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Row 3 of array A104257.
Summary of increasing sequences avoiding arithmetic progressions of specified lengths (the second of each pair is obtained by adding 1 to the first):
3-term AP: A005836 (>=0), A003278 (>0);
4-term AP: A005839 (>=0), A005837 (>0);
5-term AP: A020654 (>=0), A020655 (>0);
6-term AP: A020656 (>=0), A005838 (>0);
7-term AP: A020657 (>=0), A020658 (>0);
8-term AP: A020659 (>=0), A020660 (>0);
9-term AP: A020661 (>=0), A020662 (>0);
10-term AP: A020663 (>=0), A020664 (>0).
See also A000452.

Programs

  • Haskell
    a005836 n = a005836_list !! (n-1)
a005836_list = filter ((== 1) . a039966) [0..]
-- Reinhard Zumkeller, Jun 09 2012, Sep 29 2011
  • Julia
    function a(n)
    m, r, b = n, 0, 1
    while m > 0
        m, q = divrem(m, 2)
        r += b * q
        b *= 3
    end
r end; [a(n) for n in 0:57] |> println # Peter Luschny, Jan 03 2021
  • Maple
    t := (j, n) -> add(binomial(n,k)^j, k=0..n):
for i from 1 to 400 do
    if(t(4,i) mod 3 <>0) then print(i) fi
od; # Gary Detlefs, Nov 28 2011
# alternative Maple program:
a:= proc(n) option remember: local k, m:
if n=1 then 0 elif n=2 then 1 elif n>2 then k:=floor(log[2](n-1)): m:=n-2^k: procname(m)+3^k: fi: end proc:
seq(a(n), n=1.. 20); # Paul Weisenhorn, Mar 22 2020
# third Maple program:
a:= n-> `if`(n=1, 0, irem(n-1, 2, 'q')+3*a(q+1)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 26 2022
  • Mathematica
    Table[FromDigits[IntegerDigits[k, 2], 3], {k, 60}]
Select[Range[0, 400], DigitCount[#, 3, 2] == 0 &] (* Harvey P. Dale, Jan 04 2012 *)
Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)
FromDigits[#,3]&/@Tuples[{0,1},7] (* Harvey P. Dale, May 10 2019 *)
  • PARI
    A=vector(100);for(n=2,#A,A[n]=if(n%2,3*A[n\2+1],A[n-1]+1));A \\ Charles R Greathouse IV, Jul 24 2012
  • PARI
    is(n)=while(n,if(n%3>1,return(0));n\=3);1 \\ Charles R Greathouse IV, Mar 07 2013
  • PARI
    a(n) = fromdigits(binary(n-1),3);  \\ Gheorghe Coserea, Jun 15 2018
  • Python
    def A005836(n):
    return int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015

Formula

a(n) = A005823(n)/2 = A003278(n)-1 = A033159(n)-2 = A033162(n)-3.
Numbers n such that the coefficient of x^n is > 0 in prod (k >= 0, 1 + x^(3^k)). - Benoit Cloitre, Jul 29 2003
a(n+1) = Sum_{k=0..m} b(k)* 3^k and n = Sum( b(k)* 2^k ).
a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0.
a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2). - Ralf Stephan, Apr 27 2003
G.f.: (x/(1-x)) * Sum_{k>=0} 3^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003
a(n) = Sum_{k = 1..n-1} (1 + 3^A007814(k)) / 2. - Philippe Deléham, Jul 09 2005
From Reinhard Zumkeller, Mar 02 2008: (Start)
A081603(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 2 then y else if x mod 3 = 2 then f(y+1, y+1) else f(floor(x/3), y). (End)
With offset a(0) = 0: a(n) = Sum_{k>=0} A030308(n,k)*3^k. - Philippe Deléham, Oct 15 2011
a(2^n) = A003462(n). - Philippe Deléham, Jun 06 2015
We have liminf_{n->infinity} a(n)/n^(log(3)/log(2)) = 1/2 and limsup_{n->infinity} a(n)/n^(log(3)/log(2)) = 1. - Gheorghe Coserea, Sep 13 2015
a(2^k+m) = a(m) + 3^k with 1 <= m <= 2^k and 1 <= k, a(1)=0, a(2)=1. - Paul Weisenhorn, Mar 22 2020
Sum_{n>=2} 1/a(n) = 2.682853110966175430853916904584699374821677091415714815171756609672281184705... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022
A065361(a(n)) = n-1. - Rémy Sigrist, Feb 06 2023
a(n) ≍ n^k, where k = log 3/log 2 = 1.5849625007. (I believe the constant varies from 1/2 to 1.) - Charles R Greathouse IV, Mar 29 2024

Extensions

Offset corrected by N. J. A. Sloane, Mar 02 2008
Edited by the Associate Editors of the OEIS, Apr 07 2009

A139251 First differences of toothpicks numbers A139250.

Original entry on oeis.org

0, 1, 2, 4, 4, 4, 8, 12, 8, 4, 8, 12, 12, 16, 28, 32, 16, 4, 8, 12, 12, 16, 28, 32, 20, 16, 28, 36, 40, 60, 88, 80, 32, 4, 8, 12, 12, 16, 28, 32, 20, 16, 28, 36, 40, 60, 88, 80, 36, 16, 28, 36, 40, 60, 88, 84, 56, 60, 92, 112, 140, 208, 256, 192, 64, 4, 8, 12, 12, 16, 28, 32, 20, 16, 28
Offset: 0

Views

Author

Omar E. Pol, Apr 24 2008

Keywords

Comments

Number of toothpicks added to the toothpick structure at the n-th step (see A139250).
It appears that if n is equal to 1 plus a power of 2 with positive exponent then a(n) = 4. (For proof see the second Applegate link.)
It appears that there is a relation between this sequence, even superperfect numbers, Mersenne primes and even perfect numbers. Conjecture: The sum of the toothpicks added to the toothpick structure between the stage A061652(k) and the stage A000668(k) is equal to the k-th even perfect number, for k >= 1. For example: A000396(1) = 2+4 = 6. A000396(2) = 4+4+8+12 = 28. A000396(3) = 16+4+8+12+12+16+28+32+20+16+28+36+40+60+88+80 = 496. - Omar E. Pol, May 04 2009
Concerning this conjecture, see David Applegate's comments on the conjectures in A153006. - N. J. A. Sloane, May 14 2009
In the triangle (See example lines), the sum of row k is equal to A006516(k), for k >= 1. - Omar E. Pol, May 15 2009
Equals (1, 2, 2, 2, ...) convolved with A160762: (1, 0, 2, -2, 2, 2, 2, -6, ...). - Gary W. Adamson, May 25 2009
Convolved with the Jacobsthal sequence A001045 = A160704: (1, 3, 9, 19, 41, ...). - Gary W. Adamson, May 24 2009
It appears that the sums of two successive terms of A160552 give the positive terms of this sequence. - Omar E. Pol, Feb 19 2015
From Omar E. Pol, Feb 28 2019: (Start)
The study of the toothpick automaton on triangular grid (A296510), and other C.A. of the same family, reveals that some cellular automata that have recurrent periods can be represented in general by irregular triangles (of first differences) whose row lengths are the terms of A011782 multiplied by k, where k >= 1, is the length of an internal cycle. This internal cycle is called "word" of a cellular automaton. For example: A160121 has word "a", so k = 1. This sequence has word "ab", so k = 2. A296511 has word "abc", so k = 3. A299477 has word "abcb" so k = 4. A299479 has word "abcbc", so k = 5.
The structure of this triangle (with word "ab" and k = 2) for the nonzero terms is as follows:
a,b;
a,b;
a,b,a,b;
a,b,a,b,a,b,a,b;
a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b;
...
The row lengths are the terms of A011782 multiplied by 2, equaling the column 2 of the square array A296612: 2, 2, 4, 8, 16, ...
This arrangement has the property that the odd-indexed columns (a) contain numbers of the toothpicks that are parallel to initial toothpick, and the even-indexed columns (b) contain numbers of the toothpicks that are orthogonal to the initial toothpick (see the third triangle in the Example section).
An associated sound to the animation could be (tick, tock), (tick, tock), ..., the same as the ticking clock sound.
For further information about the "word" of a cellular automaton see A296612. (End)

Examples

			From _Omar E. Pol_, Dec 16 2008: (Start)
Triangle begins:
1;
2;
4,4;
4,8,12,8;
4,8,12,12,16,28,32,16;
4,8,12,12,16,28,32,20,16,28,36,40,60,88,20,32;
(End)
From _David Applegate_, Apr 29 2009: (Start)
The layout of the triangle was adjusted to reveal that the columns become constant as shown below:
. 0;
. 1;
. 2,4;
. 4,4,8,12;
. 8,4,8,12,12,16,28,32;
.16,4,8,12,12,16,28,32,20,16,28,36,40,60,88,80;
.32,4,8,12,12,16,28,32,20,16,28,36,40,60,88,80,36,16,28,36,40,60,88,84,56,...
...
The row sums give A006516.
(End)
From _Omar E. Pol_, Feb 28 2018: (Start)
Also the nonzero terms can write as an irregular triangle in which the row lengths are the terms of A011782 multiplied by 2 as shown below:
1,2;
4,4;
4,8,12,8;
4,8,12,12,16,28,32,16;
4,8,12,12,16,28,32,20,16,28,36,40,60,88,20,32;
...
(End)

Links

Crossrefs

Equals 2*A152968 and 4*A152978 (if we ignore the first couple of terms).
See A147646 for the limiting behavior of the rows. See also A006516.
Row lengths in A011782.
Cf. A139250, A139252, A139253, A152980, A153000, A153001, A000396, A000668, A061652, A153006.
Cf. A006516, A153007, A159790, A001045, A160704, A160762, A147582, A296612.
Cf. A160121 (word "a"), A296511 (word "abc"), A299477 (word "abcb"), A299479 (word "abcbc").

Programs

  • Maple
    G := (x/(1+2*x)) * (1 + 2*x*mul(1+x^(2^k-1)+2*x^(2^k),k=0..20)); # N. J. A. Sloane, May 20 2009, Jun 05 2009
# A139250 is T, A139251 is a.
a:=[0,1,2,4]; T:=[0,1,3,7]; M:=10;
for k from 1 to M do
a:=[op(a),2^(k+1)];
T:=[op(T),T[nops(T)]+a[nops(a)]];
for j from 1 to 2^(k+1)-1 do
a:=[op(a), 2*a[j+1]+a[j+2]];
T:=[op(T),T[nops(T)]+a[nops(a)]];
od: od: a; T;
# N. J. A. Sloane, Dec 25 2009
  • Mathematica
    CoefficientList[Series[((x - x^2)/((1 - x) (1 + 2 x))) (1 + 2 x Product[1 + x^(2^k - 1) + 2 x^(2^k), {k, 0, 20}]), {x, 0, 60}], x] (* Vincenzo Librandi, Aug 22 2014 *)

Formula

Recurrence from N. J. A. Sloane, Jul 20 2009: a(0) = 0; a(2^i)=2^i for all i; otherwise write n=2^i+j, 0 < j < 2^i, then a(n) = 2a(j)+a(j+1). Proof: This is a simplification of the following recurrence of David Applegate. QED
Recurrence from David Applegate, Apr 29 2009: (Start)
Write n=2^(i+1)+j, where 0 <= j < 2^(i+1). Then, for n > 3:
for j=0, a(n) = 2*a(n-2^i) (= n = 2^(i+1))
for 1 <= j <= 2^i - 1, a(n) = a(n-2^i)
for j=2^i, a(n) = a(n-2^i)+4 (= 2^(i+1)+4)
for 2^i+1 <= j <= 2^(i+1)-2, a(n) = 2*a(n-2^i) + a(n-2^i+1)
for j=2^(i+1)-1, a(n) = 2*a(n-2^i) + a(n-2^i+1)-4
and a(n) = 2^(n-1) for n=1,2,3. (End)
G.f.: (x/(1+2*x)) * (1 + 2*x*Product_{k>=0} (1 + x^(2^k-1) + 2*x^(2^k))). - N. J. A. Sloane, May 20 2009, Jun 05 2009
With offset 0 (which would be more natural, but offset 1 is now entrenched): a(0) = 1, a(1) = 2; for i >= 1, a(2^i) = 4; otherwise write n = 2^i +j, 0 < j < 2^i, then a(n) = 2 * Sum_{ k >= 0 } 2^(wt(j+k)-k)*binomial(wt(j+k),k). - N. J. A. Sloane, Jun 03 2009
It appears that a(n) = A187221(n+1)/2. - Omar E. Pol, Mar 08 2011
It appears that a(n) = A160552(n-1) + A160552(n), n >= 1. - Omar E. Pol, Feb 18 2015

Extensions

Partially edited by Omar E. Pol, Feb 28 2019

A001316 Gould's sequence: a(n) = Sum_{k=0..n} (binomial(n,k) mod 2); number of odd entries in row n of Pascal's triangle (A007318); a(n) = 2^A000120(n).

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32, 32, 64, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also called Dress's sequence.
This sequence might be better called Glaisher's sequence, since James Glaisher showed that odd binomial coefficients are counted by 2^A000120(n) in 1899. - Eric Rowland, Mar 17 2017 [However, the name "Gould's sequence" is deeply entrenched in the literature. - N. J. A. Sloane, Mar 17 2017] [Named after the American mathematician Henry Wadsworth Gould (b. 1928). - Amiram Eldar, Jun 19 2021]
All terms are powers of 2. The first occurrence of 2^k is at n = 2^k - 1; e.g., the first occurrence of 16 is at n = 15. - Robert G. Wilson v, Dec 06 2000
a(n) is the highest power of 2 dividing binomial(2n,n) = A000984(n). - Benoit Cloitre, Jan 23 2002
Also number of 1's in n-th row of triangle in A070886. - Hans Havermann, May 26 2002. Equivalently, number of live cells in generation n of a one-dimensional cellular automaton, Rule 90, starting with a single live cell. - Ben Branman, Feb 28 2009. Ditto for Rule 18. - N. J. A. Sloane, Aug 09 2014. This is also the odd-rule cellular automaton defined by OddRule 003 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015
Also number of numbers k, 0<=k<=n, such that (k OR n) = n (bitwise logical OR): a(n) = #{k : T(n,k)=n, 0<=k<=n}, where T is defined as in A080098. - Reinhard Zumkeller, Jan 28 2003
To construct the sequence, start with 1 and use the rule: If k >= 0 and a(0),a(1),...,a(2^k-1) are the first 2^k terms, then the next 2^k terms are 2*a(0),2*a(1),...,2*a(2^k-1). - Benoit Cloitre, Jan 30 2003
Also, numerator((2^k)/k!). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
The odd entries in Pascal's triangle form the Sierpiński Gasket (a fractal). - Amarnath Murthy, Nov 20 2004
Row sums of Sierpiński's Gasket A047999. - Johannes W. Meijer, Jun 05 2011
Fixed point of the morphism "1" -> "1,2", "2" -> "2,4", "4" -> "4,8", ..., "2^k" -> "2^k,2^(k+1)", ... starting with a(0) = 1; 1 -> 12 -> 1224 -> = 12242448 -> 122424482448488(16) -> ... . - Philippe Deléham, Jun 18 2005
a(n) = number of 1's of stage n of the one-dimensional cellular automaton with Rule 90. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 01 2006
a(33)..a(63) = A117973(1)..A117973(31). - Stephen Crowley, Mar 21 2007
Or the number of solutions of the equation: A000120(x) + A000120(n-x) = A000120(n). - Vladimir Shevelev, Jul 19 2009
For positive n, a(n) equals the denominator of the permanent of the n X n matrix consisting entirely of (1/2)'s. - John M. Campbell, May 26 2011
Companions to A001316 are A048896, A105321, A117973, A151930 and A191488. They all have the same structure. We observe that for all these sequences a((2*n+1)*2^p-1) = C(p)*A001316(n), p >= 0. If C(p) = 2^p then a(n) = A001316(n), if C(p) = 1 then a(n) = A048896(n), if C(p) = 2^p+2 then a(n) = A105321(n+1), if C(p) = 2^(p+1) then a(n) = A117973(n), if C(p) = 2^p-2 then a(n) = (-1)*A151930(n) and if C(p) = 2^(p+1)+2 then a(n) = A191488(n). Furthermore for all a(2^p - 1) = C(p). - Johannes W. Meijer, Jun 05 2011
a(n) = number of zeros in n-th row of A219463 = number of ones in n-th row of A047999. - Reinhard Zumkeller, Nov 30 2012
This is the Run Length Transform of S(n) = {1,2,4,8,16,...} (cf. A000079). The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product). - N. J. A. Sloane, Sep 05 2014
A105321(n+1) = a(n+1) + a(n). - Reinhard Zumkeller, Nov 14 2014
a(n) = A261363(n,n) = number of distinct terms in row n of A261363 = number of odd terms in row n+1 of A261363. - Reinhard Zumkeller, Aug 16 2015
From Gary W. Adamson, Aug 26 2016: (Start)
A production matrix for the sequence is lim_{k->infinity} M^k, the left-shifted vector of M:
1, 0, 0, 0, 0, ...
2, 0, 0, 0, 0, ...
0, 1, 0, 0, 0, ...
0, 2, 0, 0, 0, ...
0, 0, 1, 0, 0, ...
0, 0, 2, 0, 0, ...
0, 0, 0, 1, 0, ...
...
The result is equivalent to the g.f. of Apr 06 2003: Product_{k>=0} (1 + 2*z^(2^k)). (End)
Number of binary palindromes of length n for which the first floor(n/2) symbols are themselves a palindrome (Ji and Wilf 2008). - Jeffrey Shallit, Jun 15 2017

Examples

			Has a natural structure as a triangle:
  1,
  2,
  2,4,
  2,4,4,8,
  2,4,4,8,4,8,8,16,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32,64,
  ...
The rows converge to A117973.
From _Omar E. Pol_, Jun 07 2009: (Start)
Also, triangle begins:
   1;
   2,2;
   4,2,4,4;
   8,2,4,4,8,4,8,8;
  16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16;
  32,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32;
  64,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,...
(End)
G.f. = 1 + 2*x + 2*x^2 + 4*x^3 + 2*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 2*x^8 + ... - _Michael Somos_, Aug 26 2015

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, p. 75ff.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 145-151.
  • James W. L. Glaisher, On the residue of a binomial-theorem coefficient with respect to a prime modulus, Quarterly Journal of Pure and Applied Mathematics, Vol. 30 (1899), pp. 150-156.
  • H. W. Gould, Exponential Binomial Coefficient Series. Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sep 1961.
  • Olivier Martin, Andrew M. Odlyzko, and Stephen Wolfram, Algebraic properties of cellular automata, Comm. Math. Physics, Vol. 93 (1984), pp. 219-258. Reprinted in Theory and Applications of Cellular Automata, S Wolfram, Ed., World Scientific, 1986, pp. 51-90 and in Cellular Automata and Complexity: Collected Papers of Stephen Wolfram, Addison-Wesley, 1994, pp. 71-113
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991, page 383.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Andrew Wuensche, Exploring Discrete Dynamics, Luniver Press, 2011. See Fig. 2.3.

Links

Crossrefs

Equals left border of triangle A166548. - Gary W. Adamson, Oct 16 2009
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
For partial sums see A006046. For first differences see A151930.
This is the numerator of 2^n/n!, while A049606 gives the denominator.
Cf. A051638, A048967, A007318, A094959, A048896, A117973, A008977, A139541, A048883, A102376, A038573, A159913, A000079, A166548, A006047, A089898, A105321, A061142.
Cf. A047999, A261363, A261366.
If we subtract 1 from the terms we get a pair of essentially identical sequences, A038573 and A159913.
A163000 and A163577 count binomial coefficients with 2-adic valuation 1 and 2. A275012 gives a measure of complexity of these sequences. - Eric Rowland, Mar 15 2017
Cf. A286575 (run-length transform), A368655 (binomial transform), also A037445.

Programs

  • Haskell
    import Data.List (transpose)
a001316 = sum . a047999_row  -- Reinhard Zumkeller, Nov 24 2012
a001316_list = 1 : zs where
   zs = 2 : (concat $ transpose [zs, map (* 2) zs])
-- Reinhard Zumkeller, Aug 27 2014, Sep 16 2011
(Sage, Python)
from functools import cache
@cache
def A001316(n):
    if n <= 1: return n+1
    return A001316(n//2) << n%2
print([A001316(n) for n in range(88)])  # Peter Luschny, Nov 19 2012
  • Maple
    A001316 := proc(n) local k; add(binomial(n,k) mod 2, k=0..n); end;
S:=[1]; S:=[op(S),op(2*s)]; # repeat ad infinitum!
a := n -> 2^add(i,i=convert(n,base,2)); # Peter Luschny, Mar 11 2009
  • Mathematica
    Table[ Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ], {n, 0, 100} ]
Nest[ Join[#, 2#] &, {1}, 7] (* Robert G. Wilson v, Jan 24 2006 and modified Jul 27 2014 *)
Map[Function[Apply[Plus,Flatten[ #1]]], CellularAutomaton[90,{{1},0},100]] (* Produces counts of ON cells. N. J. A. Sloane, Aug 10 2009 *)
ArrayPlot[CellularAutomaton[90, {{1}, 0}, 20]] (* Illustration of first 20 generations. - N. J. A. Sloane, Aug 14 2014 *)
Table[2^(RealDigits[n - 1, 2][[1]] // Total), {n, 1, 100}] (* Gabriel C. Benamy, Dec 08 2009 *)
CoefficientList[Series[Exp[2*x], {x, 0, 100}], x] // Numerator (* Jean-François Alcover, Oct 25 2013 *)
Count[#,?OddQ]&/@Table[Binomial[n,k],{n,0,90},{k,0,n}] (* _Harvey P. Dale, Sep 22 2015 *)
2^DigitSum[Range[0, 100], 2] (* Paolo Xausa, Jul 31 2025 *)
  • PARI
    {a(n) = if( n<0, 0, numerator(2^n / n!))};
  • PARI
    A001316(n)=1<M. F. Hasler, May 03 2009
  • PARI
    a(n)=2^hammingweight(n) \\ Charles R Greathouse IV, Jan 04 2013
  • Python
    def A001316(n):
    return 2**bin(n)[2:].count("1") # Indranil Ghosh, Feb 06 2017
  • Python
    def A001316(n): return 1<Karl-Heinz Hofmann, Aug 01 2025
  • Python
    import numpy # (version >= 2.0.0)
n_up_to = 2**22
A000079 = 1 << numpy.arange(n_up_to.bit_length())
A001316 = A000079[numpy.bitwise_count(numpy.arange(n_up_to))]
print(A001316[0:100]) # Karl-Heinz Hofmann, Aug 01 2025
  • Scheme
    (define (A001316 n) (let loop ((n n) (z 1)) (cond ((zero? n) z) ((even? n) (loop (/ n 2) z)) (else (loop (/ (- n 1) 2) (* z 2)))))) ;; Antti Karttunen, May 29 2017

Formula

a(n) = 2^A000120(n).
a(0) = 1; for n > 0, write n = 2^i + j where 0 <= j < 2^i; then a(n) = 2*a(j).
a(n) = 2*a(n-1)/A006519(n) = A000079(n)*A049606(n)/A000142(n).
a(n) = A038573(n) + 1.
G.f.: Product_{k>=0} (1+2*z^(2^k)). - Ralf Stephan, Apr 06 2003
a(n) = Sum_{i=0..2*n} (binomial(2*n, i) mod 2)*(-1)^i. - Benoit Cloitre, Nov 16 2003
a(n) mod 3 = A001285(n). - Benoit Cloitre, May 09 2004
a(n) = 2^n - 2*Sum_{k=0..n} floor(binomial(n, k)/2). - Paul Barry, Dec 24 2004
a(n) = Product_{k=0..log_2(n)} 2^b(n, k), b(n, k) = coefficient of 2^k in binary expansion of n. - Paul D. Hanna
Sum_{k=0..n-1} a(k) = A006046(n).
a(n) = n/2 + 1/2 + (1/2)*Sum_{k=0..n} (-(-1)^binomial(n,k)). - Stephen Crowley, Mar 21 2007
G.f. for a(n)/A156769(n): (1/2)*z^(1/2)*sinh(2*z^(1/2)). - Johannes W. Meijer, Feb 20 2009
Equals infinite convolution product of [1,2,0,0,0,0,0,0,0] aerated (A000079 - 1) times, i.e., [1,2,0,0,0,0,0,0,0] * [1,0,2,0,0,0,0,0,0] * [1,0,0,0,2,0,0,0,0]. - Mats Granvik, Gary W. Adamson, Oct 02 2009
a(n) = f(n, 1) with f(x, y) = if x = 0 then y otherwise f(floor(x/2), y*(1 + x mod 2)). - Reinhard Zumkeller, Nov 21 2009
a(n) = 2^(number of 1's in binary form of (n-1)). - Gabriel C. Benamy, Dec 08 2009
a((2*n+1)*2^p-1) = (2^p)*a(n), p >= 0. - Johannes W. Meijer, Jun 05 2011
a(n) = A000120(A001317(n)). - Reinhard Zumkeller, Nov 24 2012
a(n) = A226078(n,1). - Reinhard Zumkeller, May 25 2013
a(n) = lcm(n!, 2^n) / n!. - Daniel Suteu, Apr 28 2017
a(n) = A061142(A005940(1+n)). - Antti Karttunen, May 29 2017
a(0) = 1, a(2*n) = a(n), a(2*n+1) = 2*a(n). - Daniele Parisse, Feb 15 2024
a(n*m) <= a(n)^A000120(m). - Joe Amos, Mar 27 2025

Extensions

Additional comments from Henry Bottomley, Mar 12 2001
Further comments from N. J. A. Sloane, May 30 2009

A048720 Multiplication table {0..i} X {0..j} of binary polynomials (polynomials over GF(2)) interpreted as binary vectors, then written in base 10; or, binary multiplication without carries.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 4, 3, 0, 0, 4, 6, 6, 4, 0, 0, 5, 8, 5, 8, 5, 0, 0, 6, 10, 12, 12, 10, 6, 0, 0, 7, 12, 15, 16, 15, 12, 7, 0, 0, 8, 14, 10, 20, 20, 10, 14, 8, 0, 0, 9, 16, 9, 24, 17, 24, 9, 16, 9, 0, 0, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 0, 0, 11, 20, 27, 32, 27, 20, 27, 32, 27, 20, 11, 0
Offset: 0

Views

Author

Antti Karttunen, Apr 26 1999

Keywords

Comments

Essentially same as A091257 but computed starting from offset 0 instead of 1.
Each polynomial in GF(2)[X] is encoded as the number whose binary representation is given by the coefficients of the polynomial, e.g., 13 = 2^3 + 2^2 + 2^0 = 1101_2 encodes 1*X^3 + 1*X^2 + 0*X^1 + 1*X^0 = X^3 + X^2 + X^0. - Antti Karttunen and Peter Munn, Jan 22 2021
To listen to this sequence, I find instrument 99 (crystal) works well with the other parameters defaulted. - Peter Munn, Nov 01 2022

Examples

			Top left corner of array:
  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 ...
  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ...
  0  2  4  6  8 10 12 14 16 18 20 22 24 26 28 30 ...
  0  3  6  5 12 15 10  9 24 27 30 29 20 23 18 17 ...
  ...
From _Antti Karttunen_ and _Peter Munn_, Jan 23 2021: (Start)
Multiplying 10 (= 1010_2) and 11 (= 1011_2), in binary results in:
     1011
  *  1010
  -------
   c1011
  1011
  -------
  1101110  (110 in decimal),
and we see that there is a carry-bit (marked c) affecting the result.
In carryless binary multiplication, the second part of the process (in which the intermediate results are summed) looks like this:
    1011
  1011
  -------
  1001110  (78 in decimal).
(End)

Links

Crossrefs

Cf. A051776 (Nim-product), A091257 (subtable).
Carryless multiplication in other bases: A325820 (3), A059692 (10).
Ordinary {0..i} * {0..j} multiplication table: A004247 and its differences from this: A061858 (which lists further sequences related to presence/absence of carry in binary multiplication).
Carryless product of the prime factors of n: A234741.
Binary irreducible polynomials ("X-primes"): A014580, factorization table: A256170, table of "X-powers": A048723, powers of 3: A001317, rearranged subtable with distinct terms (comparable to A054582): A277820.
See A014580 for further sequences related to the difference between factorization into GF(2)[X] irreducibles and ordinary prime factorization of the integer encoding.
Row/column 3: A048724 (even bisection of A003188), 5: A048725, 6: A048726, 7: A048727; main diagonal: A000695.
Associated additive operation: A003987.
Equivalent sequences, as compared with standard integer multiplication: A048631 (factorials), A091242 (composites), A091255 (gcd), A091256 (lcm), A280500 (division).
See A091202 (and its variants) and A278233 for maps from/to ordinary multiplication.
See A115871, A115872 and A277320 for tables related to cross-domain congruences.

Programs

  • Maple
    trinv := n -> floor((1+sqrt(1+8*n))/2); # Gives integral inverses of the triangular numbers
# Binary multiplication of nn and mm, but without carries (use XOR instead of ADD):
Xmult := proc(nn,mm) local n,m,s; n := nn; m := mm; s := 0; while (n > 0) do if(1 = (n mod 2)) then s := XORnos(s,m); fi; n := floor(n/2); # Shift n right one bit. m := m*2; # Shift m left one bit. od; RETURN(s); end;
  • Mathematica
    trinv[n_] := Floor[(1 + Sqrt[1 + 8*n])/2];
Xmult[nn_, mm_] := Module[{n = nn, m = mm, s = 0}, While[n > 0, If[1 == Mod[n, 2], s = BitXor[s, m]]; n = Floor[n/2]; m = m*2]; Return[s]];
a[n_] := Xmult[(trinv[n] - 1)*((1/2)*trinv[n] + 1) - n, n - (trinv[n]*(trinv[n] - 1))/2];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 16 2015, updated Mar 06 2016 after Maple *)
  • PARI
    up_to = 104;
A048720sq(b,c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2);
A048720list(up_to) = { my(v = vector(1+up_to), i=0); for(a=0, oo, for(col=0, a, i++; if(i > up_to, return(v)); v[i] = A048720sq(col, a-col))); (v); };
v048720 = A048720list(up_to);
A048720(n) = v048720[1+n]; \\ Antti Karttunen, Feb 15 2021

Formula

a(n) = Xmult( (((trinv(n)-1)*(((1/2)*trinv(n))+1))-n), (n-((trinv(n)*(trinv(n)-1))/2)) );
T(2b, c)=T(c, 2b)=T(b, 2c)=2T(b, c); T(2b+1, c)=T(c, 2b+1)=2T(b, c) XOR c - Henry Bottomley, Mar 16 2001
For n >= 0, A003188(2n) = T(n, 3); A003188(2n+1) = T(n, 3) XOR 1, where XOR is the bitwise exclusive-or operator, A003987. - Peter Munn, Feb 11 2021

A047999 Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 2.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

Restored the alternative spelling of Sierpinski to facilitate searching for this triangle using regular-expression matching commands in ASCII. - N. J. A. Sloane, Jan 18 2016
Also triangle giving successive states of cellular automaton generated by "Rule 60" and "Rule 102". - Hans Havermann, May 26 2002
Also triangle formed by reading triangle of Eulerian numbers (A008292) mod 2. - Philippe Deléham, Oct 02 2003
Self-inverse when regarded as an infinite lower triangular matrix over GF(2).
Start with [1], repeatedly apply the map 0 -> [00/00], 1 -> [10/11] [Allouche and Berthe]
Also triangle formed by reading triangles A011117, A028338, A039757, A059438, A085881, A086646, A086872, A087903, A104219 mod 2. - Philippe Deléham, Jun 18 2005
J. H. Conway writes (in Math Forum): at least the first 31 rows give odd-sided constructible polygons (sides 1, 3, 5, 15, 17, ... see A001317). The 1's form a Sierpiński sieve. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
When regarded as an infinite lower triangular matrix, its inverse is a (-1,0,1)-matrix with zeros undisturbed and the nonzero entries in every column form the Prouhet-Thue-Morse sequence (1,-1,-1,1,-1,1,1,-1,...) A010060 (up to relabeling). - David Callan, Oct 27 2006
Triangle read by rows: antidiagonals of an array formed by successive iterates of running sums mod 2, beginning with (1, 1, 1, ...). - Gary W. Adamson, Jul 10 2008
T(n,k) = A057427(A143333(n,k)). - Reinhard Zumkeller, Oct 24 2010
The triangle sums, see A180662 for their definitions, link Sierpiński’s triangle A047999 with seven sequences, see the crossrefs. The Kn1y(n) and Kn2y(n), y >= 1, triangle sums lead to the Sierpiński-Stern triangle A191372. - Johannes W. Meijer, Jun 05 2011
Used to compute the total Steifel-Whitney cohomology class of the Real Projective space. This was an essential component of the proof that there are no product operations without zero divisors on R^n for n not equal to 1, 2, 4 or 8 (real numbers, complex numbers, quaternions, Cayley numbers), proved by Bott and Milnor. - Marcus Jaiclin, Feb 07 2012
T(n,k) = A134636(n,k) mod 2. - Reinhard Zumkeller, Nov 23 2012
T(n,k) = 1 - A219463(n,k), 0 <= k <= n. - Reinhard Zumkeller, Nov 30 2012
From Vladimir Shevelev, Dec 31 2013: (Start)
Also table of coefficients of polynomials s_n(x) of degree n which are defined by formula s_n(x) = Sum_{i=0..n} (binomial(n,i) mod 2)*x^k. These polynomials we naturally call Sierpiński's polynomials. They also are defined by the recursion: s_0(x)=1, s_(2*n+1)(x) = (x+1)*s_n(x^2), n>=0, and s_(2*n)(x) = s_n(x^2), n>=1.
Note that: s_n(1) = A001316(n),
s_n(2) = A001317(n),
s_n(3) = A100307(n),
s_n(4) = A001317(2*n),
s_n(5) = A100308(n),
s_n(6) = A100309(n),
s_n(7) = A100310(n),
s_n(8) = A100311(n),
s_n(9) = A100307(2*n),
s_n(10) = A006943(n),
s_n(16) = A001317(4*n),
s_n(25) = A100308(2*n), etc.
The equality s_n(10) = A006943(n) means that sequence A047999 is obtained from A006943 by the separation by commas of the digits of its terms. (End)
Comment from N. J. A. Sloane, Jan 18 2016: (Start)
Take a diamond-shaped region with edge length n from the top of the triangle, and rotate it by 45 degrees to get a square S_n. Here is S_6:
[1, 1, 1, 1, 1, 1]
[1, 0, 1, 0, 1, 0]
[1, 1, 0, 0, 1, 1]
[1, 0, 0, 0, 1, 0]
[1, 1, 1, 1, 0, 0]
[1, 0, 1, 0, 0, 0].
Then (i) S_n contains no square (parallel to the axes) with all four corners equal to 1 (cf. A227133); (ii) S_n can be constructed by using the greedy algorithm with the constraint that there is no square with that property; and (iii) S_n contains A064194(n) 1's. Thus A064194(n) is a lower bound on A227133(n). (End)
See A123098 for a multiplicative encoding of the rows, i.e., product of the primes selected by nonzero terms; e.g., 1 0 1 => 2^1 * 3^0 * 5^1. - M. F. Hasler, Sep 18 2016
From Valentin Bakoev, Jul 11 2020: (Start)
The Sierpinski's triangle with 2^n rows is a part of a lower triangular matrix M_n of dimension 2^n X 2^n. M_n is a block matrix defined recursively: M_1= [1, 0], [1, 1], and for n>1, M_n = [M_(n-1), O_(n-1)], [M_(n-1), M_(n-1)], where M_(n-1) is a block matrix of the same type, but of dimension 2^(n-1) X 2^(n-1), and O_(n-1) is the zero matrix of dimension 2^(n-1) X 2^(n-1). Here is how M_1, M_2 and M_3 look like:
1 0 1 0 0 0 1 0 0 0 0 0 0 0
1 1 1 1 0 0 1 1 0 0 0 0 0 0 - It is seen the self-similarity of the
1 0 1 0 1 0 1 0 0 0 0 0 matrices M_1, M_2, ..., M_n, ...,
1 1 1 1 1 1 1 1 0 0 0 0 analogously to the Sierpinski's fractal.
1 0 0 0 1 0 0 0
1 1 0 0 1 1 0 0
1 0 1 0 1 0 1 0
1 1 1 1 1 1 1 1
M_n can also be defined as M_n = M_1 X M_(n-1) where X denotes the Kronecker product. M_n is an important matrix in coding theory, cryptography, Boolean algebra, monotone Boolean functions, etc. It is a transformation matrix used in computing the algebraic normal form of Boolean functions. Some properties and links concerning M_n can be seen in LINKS. (End)
Sierpinski's gasket has fractal (Hausdorff) dimension log(A000217(2))/log(2) = log(3)/log(2) = 1.58496... (and cf. A020857). This gasket is the first of a family of gaskets formed by taking the Pascal triangle (A007318) mod j, j >= 2 (see CROSSREFS). For prime j, the dimension of the gasket is log(A000217(j))/log(j) = log(j(j + 1)/2)/log(j) (see Reiter and Bondarenko references). - Richard L. Ollerton, Dec 14 2021

Examples

			Triangle begins:
              1,
             1,1,
            1,0,1,
           1,1,1,1,
          1,0,0,0,1,
         1,1,0,0,1,1,
        1,0,1,0,1,0,1,
       1,1,1,1,1,1,1,1,
      1,0,0,0,0,0,0,0,1,
     1,1,0,0,0,0,0,0,1,1,
    1,0,1,0,0,0,0,0,1,0,1,
   1,1,1,1,0,0,0,0,1,1,1,1,
  1,0,0,0,1,0,0,0,1,0,0,0,1,
  ...

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004).
  • John W. Milnor and James D. Stasheff, Characteristic Classes, Princeton University Press, 1974, pp. 43-49 (sequence appears on p. 46).
  • H.-O. Peitgen, H. Juergens and D. Saupe: Chaos and Fractals (Springer-Verlag 1992), p. 408.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

Links

Crossrefs

Sequences based on the triangles formed by reading Pascal's triangle mod m: (this sequence) (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
Other versions: A090971, A038183.
Cf. A007318, A054431, A001317, A008292, A083093, A034931, A034930, A008975, A034932, A166360, A249133, A064194, A227133.
From Johannes W. Meijer, Jun 05 2011: (Start)
A106344 is a skew version of this triangle.
Triangle sums (see the comments): A001316 (Row1; Related to Row2), A002487 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A007306 (Kn3, Kn4), A060632 (Fi1, Fi2), A120562 (Ca1, Ca2), A112970 (Gi1, Gi2), A127830 (Ze3, Ze4). (End)

Programs

  • Haskell
    import Data.Bits (xor)
a047999 :: Int -> Int -> Int
a047999 n k = a047999_tabl !! n !! k
a047999_row n = a047999_tabl !! n
a047999_tabl = iterate (\row -> zipWith xor ([0] ++ row) (row ++ [0])) [1]
-- Reinhard Zumkeller, Dec 11 2011, Oct 24 2010
  • Magma
    A047999:= func< n,k | BitwiseAnd(n-k, k) eq 0 select 1 else 0 >;
[A047999(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 03 2024
  • Maple
    # Maple code for first M rows (here M=10) - N. J. A. Sloane, Feb 03 2016
ST:=[1,1,1]; a:=1; b:=2; M:=10;
for n from 2 to M do ST:=[op(ST),1];
for i from a to b-1 do ST:=[op(ST), (ST[i+1]+ST[i+2]) mod 2 ]; od:
ST:=[op(ST),1];
a:=a+n; b:=a+n; od:
ST; # N. J. A. Sloane
# alternative
A047999 := proc(n,k)
    modp(binomial(n,k),2) ;
end proc:
seq(seq(A047999(n,k),k=0..n),n=0..12) ; # R. J. Mathar, May 06 2016
  • Mathematica
    Mod[ Flatten[ NestList[ Prepend[ #, 0] + Append[ #, 0] &, {1}, 13]], 2] (* Robert G. Wilson v, May 26 2004 *)
rows = 14; ca = CellularAutomaton[60, {{1}, 0}, rows-1]; Flatten[ Table[ca[[k, 1 ;; k]], {k, 1, rows}]] (* Jean-François Alcover, May 24 2012 *)
Mod[#,2]&/@Flatten[Table[Binomial[n,k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 26 2019 *)
A047999[n_,k_]:= Boole[BitAnd[n-k,k]==0];
Table[A047999[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 03 2025 *)
  • PARI
    \\ Recurrence for Pascal's triangle mod p, here p = 2.
p = 2; s=13; T=matrix(s,s); T[1,1]=1;
for(n=2,s, T[n,1]=1; for(k=2,n, T[n,k] = (T[n-1,k-1] + T[n-1,k])%p ));
for(n=1,s,for(k=1,n,print1(T[n,k],", "))) \\ Gerald McGarvey, Oct 10 2009
  • PARI
    A011371(n)=my(s);while(n>>=1,s+=n);s
T(n,k)=A011371(n)==A011371(k)+A011371(n-k) \\ Charles R Greathouse IV, Aug 09 2013
  • PARI
    T(n,k)=bitand(n-k,k)==0 \\ Charles R Greathouse IV, Aug 11 2016
  • Python
    def A047999_T(n,k):
    return int(not ~n & k) # Chai Wah Wu, Feb 09 2016

Formula

Lucas's Theorem is that T(n,k) = 1 if and only if the 1's in the binary expansion of k are a subset of the 1's in the binary expansion of n; or equivalently, k AND NOT n is zero, where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 09 2016 and N. J. A. Sloane, Feb 10 2016
Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
T(n,k) = T(n-1,k-1) XOR T(n-1,k), 0 < k < n; T(n,0) = T(n,n) = 1. - Reinhard Zumkeller, Dec 13 2009
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 2 = |T(n-1,k-1) - T(n-1,k)|, 0 < k < n; T(n,0) = T(n,n) = 1. - Rick L. Shepherd, Feb 23 2018
From Vladimir Shevelev, Dec 31 2013: (Start)
For polynomial {s_n(x)} we have
s_0(x)=1; for n>=1, s_n(x) = Product_{i=1..A000120(n)} (x^(2^k_i) + 1),
if the binary expansion of n is n = Sum_{i=1..A000120(n)} 2^k_i;
G.f. Sum_{n>=0} s_n(x)*z^n = Product_{k>=0} (1 + (x^(2^k)+1)*z^(2^k)) (0
Let x>1, t>0 be real numbers. Then
Sum_{n>=0} 1/s_n(x)^t = Product_{k>=0} (1 + 1/(x^(2^k)+1)^t);
Sum_{n>=0} (-1)^A000120(n)/s_n(x)^t = Product_{k>=0} (1 - 1/(x^(2^k)+1)^t).
In particular, for t=1, x>1, we have
Sum_{n>=0} (-1)^A000120(n)/s_n(x) = 1 - 1/x. (End)
From Valentin Bakoev, Jul 11 2020: (Start)
(See my comment about the matrix M_n.) Denote by T(i,j) the number in the i-th row and j-th column of M_n (0 <= i, j < 2^n). When i>=j, T(i,j) is the j-th number in the i-th row of the Sierpinski's triangle. For given i and j, we denote by k the largest integer of the type k=2^m and k
T(i,0) = T(i,i) = 1, or
T(i,j) = 0 if i < j, or
T(i,j) = T(i-k,j), if j < k, or
T(i,j) = T(i-k,j-k), if j >= k.
Thus, for given i and j, T(i,j) can be computed in O(log_2(i)) steps. (End)

Extensions

Additional links from Lekraj Beedassy, Jan 22 2004

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.

Links

Crossrefs

First differences of A001045.
See A151575 for a signed version.
Bisections: A047849, A020988.
Cf. A151548, A139250, A151555, A153006.

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

A006516 a(n) = 2^(n-1)*(2^n - 1), n >= 0.

Original entry on oeis.org

0, 1, 6, 28, 120, 496, 2016, 8128, 32640, 130816, 523776, 2096128, 8386560, 33550336, 134209536, 536854528, 2147450880, 8589869056, 34359607296, 137438691328, 549755289600, 2199022206976, 8796090925056, 35184367894528, 140737479966720, 562949936644096
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n) is also the number of different lines determined by pair of vertices in an n-dimensional hypercube. The number of these lines modulo being parallel is in A003462. - Ola Veshta (olaveshta(AT)my-deja.com), Feb 15 2001
Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001
a(n) is the number of n-letter words formed using four distinct letters, one of which appears an odd number of times. - Lekraj Beedassy, Jul 22 2003 [See, e.g., the Balakrishnan reference, problems 2.67 and 2.68, p. 69. - Wolfdieter Lang, Jul 16 2017]
Number of 0's making up the central triangle in a Pascal's triangle mod 2 gasket. - Lekraj Beedassy, May 14 2004
m-th triangular number, where m is the n-th Mersenne number, i.e., a(n)=A000217(A000225(n)). - Lekraj Beedassy, May 25 2004
Number of walks of length 2n+1 between two nodes at distance 3 in the cycle graph C_8. - Herbert Kociemba, Jul 02 2004
The sequence of fractions a(n+1)/(n+1) is the 3rd binomial transform of (1, 0, 1/3, 0, 1/5, 0, 1/7, ...). - Paul Barry, Aug 05 2005
Number of monic irreducible polynomials of degree 2 in GF(2^n)[x]. - Max Alekseyev, Jan 23 2006
(A007582(n))^2 + a(n)^2 = A007582(2n). E.g., A007582(3) = 36, a(3) = 28; A007582(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006
The sequence 6*a(n), n>=1, gives the number of edges of the Hanoi graph H_4^{n} with 4 pegs and n>=1 discs. - Daniele Parisse, Jul 28 2006
8*a(n) is the total border length of the 4*n masks used when making an order n regular DNA chip, using the bidimensional Gray code suggested by Pevzner in the book "Computational Molecular Biology." - Bruno Petazzoni (bruno(AT)enix.org), Apr 05 2007
If we start with 1 in binary and at each step we prepend 1 and append 0, we construct this sequence: 1 110 11100 1111000 etc.; see A109241(n-1). - Artur Jasinski, Nov 26 2007
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which x does not equal y. - Ross La Haye, Jan 02 2008
Wieder calls these "conjoint usual 2-combinations." The set of "conjoint strict k-combinations" is the subset of conjoint usual k-combinations where the empty set and the set itself are excluded from possible selection. These numbers C(2^n - 2,k), which for k = 2 (i.e., {x,y} of the power set of a set) give {1, 0, 1, 15, 91, 435, 1891, 7875, 32131, 129795, 521731, ...}. - Ross La Haye, Jan 15 2008
If n is a member of A000043 then a(n) is also a perfect number (A000396). - Omar E. Pol, Aug 30 2008
a(n) is also the number whose binary representation is A109241(n-1), for n>0. - Omar E. Pol, Aug 31 2008
From Daniel Forgues, Nov 10 2009: (Start)
If we define a spoof-perfect number as:
A spoof-perfect number is a number that would be perfect if some (one or more) of its odd composite factors were wrongly assumed to be prime, i.e., taken as a spoof prime.
And if we define a "strong" spoof-perfect number as:
A "strong" spoof-perfect number is a spoof-perfect number where sigma(n) does not reveal the compositeness of the odd composite factors of n which are wrongly assumed to be prime, i.e., taken as a spoof prime.
The odd composite factors of n which are wrongly assumed to be prime then have to be obtained additively in sigma(n) and not multiplicatively.
Then:
If 2^n-1 is odd composite but taken as a spoof prime then 2^(n-1)*(2^n - 1) is an even spoof perfect number (and moreover "strong" spoof-perfect).
For example:
a(8) = 2^(8-1)*(2^8 - 1) = 128*255 = 32640 (where 255 (with factors 3*5*17) is taken as a spoof prime);
sigma(a(8)) = (2^8 - 1)*(255 + 1) = 255*256 = 2*(128*255) = 2*32640 = 2n is spoof-perfect (and also "strong" spoof-perfect since 255 is obtained additively);
a(11) = 2^(11-1)*(2^11 - 1) = 1024*2047 = 2096128 (where 2047 (with factors 23*89) is taken as a spoof prime);
sigma(a(11)) = (2^11 - 1)*(2047 + 1) = 2047*2048 = 2*(1024*2047) = 2*2096128 = 2n is spoof-perfect (and also "strong" spoof-perfect since 2047 is obtained additively).
I did a Google search and didn't find anything about the distinction between "strong" versus "weak" spoof-perfect numbers. Maybe some other terminology is used.
An example of an even "weak" spoof-perfect number would be:
n = 90 = 2*5*9 (where 9 (with factors 3^2) is taken as a spoof prime);
sigma(n) = (1+2)*(1+5)*(1+9) = 3*(2*3)*(2*5) = 2*(2*5*(3^2)) = 2*90 = 2n is spoof-perfect (but is not "strong" spoof-perfect since 9 is obtained multiplicatively as 3^2 and is thus revealed composite).
Euler proved:
If 2^k - 1 is a prime number, then 2^(k-1)*(2^k - 1) is a perfect number and every even perfect number has this form.
The following seems to be true (is there a proof?):
If 2^k - 1 is an odd composite number taken as a spoof prime, then 2^(k-1)*(2^k - 1) is a "strong" spoof-perfect number and every even "strong" spoof-perfect number has this form?
There is only one known odd spoof-perfect number (found by Rene Descartes) but it is a "weak" spoof-perfect number (cf. 'Descartes numbers' and 'Unsolved problems in number theory' links below). (End)
a(n+1) = A173787(2*n+1,n); cf. A020522, A059153. - Reinhard Zumkeller, Feb 28 2010
Also, row sums of triangle A139251. - Omar E. Pol, May 25 2010
Starting with "1" = (1, 1, 2, 4, 8, ...) convolved with A002450: (1, 5, 21, 85, 341, ...); and (1, 3, 7, 15, 31, ...) convolved with A002001: (1, 3, 12, 48, 192, ...). - Gary W. Adamson, Oct 26 2010
a(n) is also the number of toothpicks in the corner toothpick structure of A153006 after 2^n - 1 stages. - Omar E. Pol, Nov 20 2010
The number of n-dimensional odd theta functions of half-integral characteristic. (Gunning, p.22) - Michael Somos, Jan 03 2014
a(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) is the nearest triangular number below 2^(2n-1); cf. A007582, A233327. - Antti Karttunen, Feb 26 2014
a(n) is the sum of all the remainders when all the odd numbers < 2^n are divided by each of the powers 2,4,8,...,2^n. - J. M. Bergot, May 07 2014
Let b(m,k) = number of ways to form a sequence of m selections, without replacement, from a circular array of m labeled cells, such that the first selection of a cell whose adjacent cells have already been selected (a "first connect") occurs on the k-th selection. b(m,k) is defined for m >=3, and for 3 <= k <= m. Then b(m,k)/2m ignores rotations and reflection. Let m=n+2, then a(n) = b(m,m-1)/2m. Reiterated, a(n) is the (m-1)th column of the triangle b(m,k)/2m, whose initial rows are (1), (1 2), (2 6 4), (6 18 28 8), (24 72 128 120 16), (120 360 672 840 496 32), (720 2160 4128 5760 5312 2016 64); see A249796. Note also that b(m,3)/2m = n!, and b(m,m)/2m = 2^n. Proofs are easy. - Tony Bartoletti, Oct 30 2014
Beginning at a(1) = 1, this sequence is the sum of the first 2^(n-1) numbers of the form 4*k + 1 = A016813(k). For example, a(4) = 120 = 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29. - J. M. Bergot, Dec 07 2014
a(n) is the number of edges in the (2^n - 1)-dimensional simplex. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete plane graph in 2^n points. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete parallelotope graph in n dimensions. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of lattices L in Z^n such that the quotient group Z^n / L is C_4. - Álvar Ibeas, Nov 26 2015
a(n) gives the quadratic coefficient of the polynomial ((x + 1)^(2^n) + (x - 1)^(2^n))/2, cf. A201461. - Martin Renner, Jan 14 2017
Let f(x)=x+2*sqrt(x) and g(x)=x-2*sqrt(x). Then f(4^n*x)=b(n)*f(x)+a(n)*g(x) and g(4^n*x)=a(n)*f(x)+b(n)*g(x), where b is A007582. - Luc Rousseau, Dec 06 2018
For n>=1, a(n) is the covering radius of the first order Reed-Muller code RM(1,2n). - Christof Beierle, Dec 22 2021
a(n) =

Examples

			G.f. = x + 6*x^2 + 28*x^3 + 120*x^4 + 496*x^5 + 2016*x^6 + 8128*x^7 + 32640*x^8 + ...

References

  • V. K. Balakrishnan, Theory and problems of Combinatorics, "Schaum's Outline Series", McGraw-Hill, 1995, p. 69.
  • Martin Gardner, Mathematical Carnival, "Pascal's Triangle", p. 201, Alfred A. Knopf NY, 1975.
  • Richard K. Guy, Unsolved problems in number theory, (p. 72).
  • Ross Honsberger, Mathematical Gems, M.A.A., 1973, p. 113.
  • Clifford A. Pickover, Wonders of Numbers, Chap. 55, Oxford Univ. Press NY 2000.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Equals A006095(n+1) - A006095(n). In other words, A006095 gives the partial sums.
Cf. A000108, A007582, A010036, A016290, A003462, A079598, A134346, A048473, A151821, A010684.
Cf. A000043, A000396. - Omar E. Pol, Aug 30 2008
Cf. A109241, A139251, A153006. - Omar E. Pol, Aug 31 2008, May 25 2010, Nov 20 2010
Cf. A002450, A002001. - Gary W. Adamson, Oct 26 2010
Cf. A049072, A000384, A201461, A005059 (binomial transform, and special 5-letter words), A065442, A211705.
Cf. A171476.

Programs

  • GAP
    List([0..25],n->2^(n-1)*(2^n-1)); # Muniru A Asiru, Dec 06 2018
  • Haskell
    a006516 n = a006516_list !! n
a006516_list = 0 : 1 :
    zipWith (-) (map (* 6) $ tail a006516_list) (map (* 8) a006516_list)
-- Reinhard Zumkeller, Oct 25 2013
  • Magma
    [2^(n-1)*(2^n - 1): n in [0..30]]; // Vincenzo Librandi, Oct 31 2014
  • Maple
    GBC := proc(n,k,q) local i; mul( (q^(n-i)-1)/(q^(k-i)-1),i=0..k-1); end; # define q-ary Gaussian binomial coefficient [ n,k ]_q
[ seq(GBC(n+1,2,2)-GBC(n,2,2), n=0..30) ]; # produces A006516
A006516:=1/(4*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation
seq(binomial(2^n, 2), n=0..19); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    Table[2^(n - 1)(2^n - 1), {n, 0, 30}] (* or *) LinearRecurrence[{6, -8}, {0, 1}, 30] (* Harvey P. Dale, Jul 15 2011 *)
  • Maxima
    A006516(n):=2^(n-1)*(2^n - 1)$ makelist(A006516(n),n,0,30); /* Martin Ettl, Nov 15 2012 */
  • PARI
    a(n)=(1<Charles R Greathouse IV, Jun 10 2011
  • PARI
    vector(100, n, n--; 2^(n-1)*(2^n-1)) \\ Altug Alkan, Oct 06 2015
  • Python
    for n in range(0, 30): print(2**(n-1)*(2**n - 1), end=', ') # Stefano Spezia, Dec 06 2018
  • Sage
    [lucas_number1(n,6,8) for n in range(24)]  # Zerinvary Lajos, Apr 22 2009
  • Sage
    [(4**n - 2**n) / 2 for n in range(24)]  # Zerinvary Lajos, Jun 05 2009

Formula

G.f.: x/((1 - 2*x)*(1 - 4*x)).
E.g.f. for a(n+1), n>=0: 2*exp(4*x) - exp(2*x).
a(n) = 2^(n-1)*Stirling2(n+1,2), n>=0, with Stirling2(n,m)=A008277(n,m).
Second column of triangle A075497.
a(n) = Stirling2(2^n,2^n-1) = binomial(2^n,2). - Ross La Haye, Jan 12 2008
a(n+1) = 4*a(n) + 2^n. - Philippe Deléham, Feb 20 2004
Convolution of 4^n and 2^n. - Ross La Haye, Oct 29 2004
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} 4^(n-j)*binomial(j,k). - Paul Barry, Aug 05 2005
a(n+2) = 6*a(n+1) - 8*a(n), a(1) = 1, a(2) = 6. - Daniele Parisse, Jul 28 2006 [Typo corrected by Yosu Yurramendi, Aug 06 2008]
Row sums of triangle A134346. Also, binomial transform of A048473: (1, 5, 17, 53, 161, ...); double bt of A151821: (1, 4, 8, 16, 32, 64, ...) and triple bt of A010684: (1, 3, 1, 3, 1, 3, ...). - Gary W. Adamson, Oct 21 2007
a(n) = 3*Stirling2(n+1,4) + Stirling2(n+2,3). - Ross La Haye, Jun 01 2008
a(n) = (4^n - 2^n)/2.
a(n) = A153006(2^n-1). - Omar E. Pol, Nov 20 2010
Sum_{n>=1} 1/a(n) = 2 * (A065442 - 1) = A211705 - 2. - Amiram Eldar, Dec 24 2020
a(n) = binomial(2*n+2, n+1) - Catalan(n+2). - N. J. A. Sloane, Apr 01 2021
a(n) = A171476(n-1), for n >= 1, and a(0) = 0. - Wolfdieter Lang, Jul 27 2022

A007583 a(n) = (2^(2*n + 1) + 1)/3.

Original entry on oeis.org

1, 3, 11, 43, 171, 683, 2731, 10923, 43691, 174763, 699051, 2796203, 11184811, 44739243, 178956971, 715827883, 2863311531, 11453246123, 45812984491, 183251937963, 733007751851, 2932031007403, 11728124029611, 46912496118443, 187649984473771, 750599937895083
Offset: 0

Views

Author

Simon Plouffe

Keywords

Comments

Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)-w(k), v(k+1)=u(k)-v(k)+w(k), w(k+1)=-u(k)+v(k)+w(k); let M(k)=Max(u(k),v(k),w(k)); then a(n)=M(2n)=M(2n-1). - Benoit Cloitre, Mar 25 2002
Also the number of words of length 2n generated by the two letters s and t that reduce to the identity 1 by using the relations ssssss=1, tt=1 and stst=1. The generators s and t along with the three relations generate the dihedral group D6=C2xD3. - Jamaine Paddyfoot (jay_paddyfoot(AT)hotmail.com) and John W. Layman, Jul 08 2002
Binomial transform of A025192. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_6. Example: a(1)=3 because in the cycle ABCDEF we have three walks of length 3 between A and B: ABAB, ABCB and AFAB. - Emeric Deutsch, Apr 01 2004
Numbers of the form 1 + Sum_{i=1..m} 2^(2*i-1). - Artur Jasinski, Feb 09 2007
Prime numbers of the form 1+Sum[2^(2n-1)] are in A000979. Numbers x such that 1+Sum[2^(2n-1)] is prime for n=1,2,...,x is A127936. - Artur Jasinski, Feb 09 2007
Related to A024493(6n+1), A131708(6n+3), A024495(6n+5). - Paul Curtz, Mar 27 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010
Number of toothpicks in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Feb 28 2011
Numbers whose binary representation is "10" repeated (n-1) times with "11" appended on the end, n >= 1. For example 171 = 10101011 (2). - Omar E. Pol, Nov 22 2012
a(n) is the smallest number for which A072219(a(n)) = 2*n+1. - Ramasamy Chandramouli, Dec 22 2012
An Engel expansion of 2 to the base b := 4/3 as defined in A181565, with the associated series expansion 2 = b + b^2/3 + b^3/(3*11) + b^4/(3*11*43) + .... Cf. A007051. - Peter Bala, Oct 29 2013
The positive integer solution (x,y) of 3*x - 2^n*y = 1, n>=0, with smallest x is (a(n/2), 2) if n is even and (a((n-1)/2), 1) if n is odd. - Wolfdieter Lang, Feb 15 2014
The smallest positive number that requires at least n additions and subtractions of powers of 2 to be formed. See Puzzling StackExchange link. - Alexander Cooke Jul 16 2023

References

  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (1.77), page 10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. also A006054, A006356, A005578, A075680.
Partial sums of A081294.
Cf. A002450, A004171, A007051, A083065, A083066, A083884.
Cf. A000978, A000979, A124400, A124401, A127936, A127955, A127956, A127957, A127958.
Cf. location of records in A007302.

Programs

  • GAP
    List([0..25], n-> (2^(2*n+1) + 1)/3); # G. C. Greubel, Dec 25 2019
  • Haskell
    a007583 = (`div` 3) . (+ 1) . a004171
-- Reinhard Zumkeller, Jan 09 2013
  • Magma
    [(2^(2*n+1) + 1)/3: n in [0..30] ]; // Vincenzo Librandi, Apr 28 2011
  • Maple
    a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-1 od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 22 2008, with correction by K. Spage, Aug 20 2014
A007583 := proc(n)
    (2^(2*n+1)+1)/3 ;
end proc: # R. J. Mathar, Feb 19 2015
  • Mathematica
    (* From Michael De Vlieger, Aug 22 2016 *)
Table[(2^(2n+1) + 1)/3, {n, 0, 23}]
Table[1 + 2Sum[4^k, {k, 0, n-1}], {n, 0, 23}]
NestList[4# -1 &, 1, 23]
Table[Sum[Binomial[n+k, 2k]/2^(k-n), {k, 0, n}], {n, 0, 23}]
CoefficientList[Series[(1-2x)/(1-5x+4x^2), {x, 0, 23}], x] (* End *)
  • PARI
    a(n)=sum(k=-n\3,n\3,binomial(2*n+1,n+1+3*k))
  • PARI
    a=1; for(n=1,23, print1(a,", "); a=bitor(a,3*a)) \\ K. Spage, Aug 20 2014
  • PARI
    Vec((1-2*x)/(1-5*x+4*x^2) + O(x^30)) \\ Altug Alkan, Dec 08 2015
  • PARI
    apply( {A007583(n)=2<<(2*n)\/3}, [0..25]) \\ M. F. Hasler, Nov 30 2021
  • Sage
    [(2^(2*n+1) + 1)/3 for n in (0..25)] # G. C. Greubel, Dec 25 2019

Formula

a(n) = 2*A002450(n) + 1.
From Wolfdieter Lang, Apr 24 2001: (Start)
a(n) = Sum_{m = 0..n} A060920(n, m) = A002450(n+1) - 2*A002450(n).
G.f.: (1-2*x)/(1-5*x+4*x^2). (End)
a(n) = Sum_{k = 0..n} binomial(n+k, 2*k)/2^(k - n).
a(n) = 4*a(n-1) - 1, n > 0.
From Paul Barry, Mar 17 2003: (Start)
a(n) = 1 + 2*Sum_{k = 0..n-1} 4^k;
a(n) = A001045(2n+1). (End)
a(n) = A020988(n-1) + 1 = A039301(n+1) - 1 = A083584(n-1) + 2. - Ralf Stephan, Jun 14 2003
a(0) = 1; a(n+1) = a(n) * 4 - 1. - Regis Decamps (decamps(AT)users.sf.net), Feb 04 2004 (correction to lead index by K. Spage, Aug 20 2014)
a(n) = Sum_{i + j + k = n; 0 <= i, j, k <= n} (n+k)!/i!/j!/(2*k)!. - Benoit Cloitre, Mar 25 2004
a(n) = 5*a(n-1) - 4*a(n-2). - Emeric Deutsch, Apr 01 2004
a(n) = 4^n - A001045(2*n). - Paul Barry, Apr 17 2004
a(n) = 2*(A001045(n))^2 + (A001045(n+1))^2. - Paul Barry, Jul 15 2004
a(n) = left and right terms in M^n * [1 1 1] where M = the 3X3 matrix [1 1 1 / 1 3 1 / 1 1 1]. M^n * [1 1 1] = [a(n) A002450(n+1) a(n)] E.g. a(3) = 43 since M^n * [1 1 1] = [43 85 43] = [a(3) A002450(4) a(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = A072197(n) - A020988(n). - Creighton Dement, Dec 31 2004
a(n) = A139250(2^n). - Omar E. Pol, Feb 28 2011
a(n) = A193652(2*n+1). - Reinhard Zumkeller, Aug 08 2011
a(n) = Sum_{k = -floor(n/3)..floor(n/3)} binomial(2*n, n+3*k)/2. - Mircea Merca, Jan 28 2012
a(n) = 2^(2*(n+1)) - A072197(n). - Vladimir Pletser, Apr 12 2014
a(n) == 2*n + 1 (mod 3). Indeed, from Regis Decamps' formula (Feb 04 2004) we have a(i+1) - a(i) == -1 (mod 3), i= 0, 1, ..., n - 1. Summing, we have a(n) - 1 == -n (mod 3), and the formula follows. - Vladimir Shevelev, May 20 2015
For n > 0 a(n) = A133494(0) + 2 * (A133494(n) + Sum_{x = 1..n - 1}Sum_{k = 0..x - 1}(binomial(x - 1, k)*(A133494(k+1) + A133494(n-x+k)))). - J. Conrad, Dec 06 2015
a(n) = Sum_{k = 0..2n} (-2)^k == 1 + Sum_{k = 1..n} 2^(2k-1). - Bob Selcoe, Aug 21 2016
E.g.f.: (1 + 2*exp(3*x))*exp(x)/3. - Ilya Gutkovskiy, Aug 21 2016
A075680(a(n)) = 1, for n > 0. - Ralf Stephan, Jun 17 2025

A073200 Number of simple Catalan bijections of type B.

Original entry on oeis.org

0, 1, 0, 3, 1, 0, 2, 2, 1, 0, 7, 3, 3, 1, 0, 8, 4, 2, 3, 1, 0, 6, 6, 8, 2, 3, 1, 0, 4, 5, 7, 7, 2, 3, 1, 0, 5, 7, 6, 6, 8, 2, 3, 1, 0, 17, 8, 5, 8, 7, 7, 2, 2, 1, 0, 18, 9, 4, 4, 6, 8, 7, 3, 3, 1, 0, 20, 10, 22, 5, 5, 5, 8, 4, 2, 2, 1, 0, 21, 14, 21, 17, 4, 4, 6, 5, 8, 3, 3, 1, 0
Offset: 0

Views

Author

Antti Karttunen, Jun 25 2002

Keywords

Comments

Each row is a permutation of nonnegative integers induced by a Catalan bijection (constructed as explained below) acting on the parenthesizations/plane binary trees as encoded and ordered by A014486/A063171.
The construction process is akin to the constructive mapping of primitive recursive functions to N: we have two basic primitives, A069770 (row 0) and A072796 (row 1), of which the former swaps the left and the right subtree of a binary tree and the latter exchanges the positions of the two leftmost subtrees of plane general trees, unless the tree's degree is less than 2, in which case it just fixes it. From then on, the even rows are constructed recursively from any other Catalan bijection in this table, using one of the five allowed recursion types:
0 - Apply the given Catalan bijection and then recurse down to both subtrees of the new binary tree obtained. (last decimal digit of row number = 2)
1 - First recurse down to both subtrees of the old binary tree and only after that apply the given Catalan bijection. (last digit = 4)
2 - Apply the given Catalan bijection and then recurse down to the right subtree of the new binary tree obtained. (last digit = 6)
3 - First recurse down to the right subtree of old binary tree and only after that apply the given Catalan bijection. (last digit = 8)
4 - First recurse down to the left subtree of old binary tree, after that apply the given Catalan bijection and then recurse down to the right subtree of the new binary tree. (last digit = 0)
The odd rows > 2 are compositions of the rows 0, 1, 2, 4, 6, 8, ... (i.e. either one of the primitives A069770 or A072796, or one of the recursive compositions) at the left hand side and any Catalan bijection from the same array at the right hand side. See the scheme-functions index-for-recursive-sgtb and index-for-composed-sgtb how to compute the positions of the recursive and ordinary compositions in this table.

Links

  • A. Karttunen, Gatomorphisms (With the complete source and explanation)

Crossrefs

Four other tables giving the corresponding cycle-counts: A073201, counts of the fixed elements: A073202, the lengths of the largest cycles: A073203, the LCM's of all the cycles: A073204. The ordinary compositions are encoded using the N X N -> N bijection A054238 (which in turn uses the bit-interleaving function A000695).
The first 21 rows of this table:.
Row 0: A069770. Row 1: A072796. Row 2: A057163. Row 3: A073269, Row 4: A057163 (duplicate), Row 5: A073270, Row 6: A069767, Row 7: A001477 (identity perm.), Row 8: A069768, Row 9: A073280.
Row 10: A069770 (dupl.), Row 11: A072796 (dupl.), Row 12: A057511, Row 13: A073282, Row 14: A057512, Row 15: A073281, Row 16: A057509, Row 17: A073280 (dupl.), Row 18: A057510, Row 19: A073283, Row 20: A073284.
Other Catalan bijection-induced EIS-permutations which occur in this table. Only the first known occurrence is given. Involutions are marked with *, others paired with their inverse:.
Row 164: A057164*, Row 168: A057508*, Row 179: A072797*.
Row 41: A073286 - Row 69: A073287. Row 105: A073290 - Row 197: A073291. Row 416: A073288 - Row 696: A073289.
Row 261: A057501 - Row 521: A057502. Row 2618: A057503 - Row 5216: A057504. Row 2614: A057505 - Row 5212: A057506.
Row 10435: A073292 - Row ...: A073293. Row 17517: A057161 - Row ...: A057162.
For a more practical enumeration system of (some) Catalan automorphisms see table A089840 and its various "recursive derivations".
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