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.

Previous Showing 11-19 of 19 results.

A089177 Triangle read by rows: T(n,k) (n >= 0, 0 <= k <= 1+log_2(floor(n))) giving number of non-squashing partitions of n into k parts.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 4, 4, 1, 1, 5, 6, 2, 1, 6, 9, 4, 1, 7, 12, 6, 1, 8, 16, 10, 1, 1, 9, 20, 14, 2, 1, 10, 25, 20, 4, 1, 11, 30, 26, 6, 1, 12, 36, 35, 10, 1, 13, 42, 44, 14, 1, 14, 49, 56, 20, 1, 15, 56, 68, 26, 1, 16, 64, 84, 36, 1, 1, 17, 72, 100, 46, 2, 1, 18, 81, 120, 60, 4, 1
Offset: 0

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Author

N. J. A. Sloane, Dec 08 2003

Keywords

Comments

T(n,k) = A181322(n,k) - A181322(n,k-1) for n>0. - Alois P. Heinz, Jan 25 2014

Examples

			Triangle begins:
  1;
  1, 1;
  1, 2,  1;
  1, 3,  2;
  1, 4,  4,  1;
  1, 5,  6,  2;
  1, 6,  9,  4;
  1, 7, 12,  6;
  1, 8, 16, 10,  1;
		

Crossrefs

Cf. A078121, A089178. Columns give A002620, A008804, A088932, A088954. Row sums give A000123.

Programs

  • Maple
    T:= proc(n) option remember;
         `if`(n=0, 1, zip((x, y)-> x+y, [T(n-1)], [0, T(floor(n/2))], 0)[])
        end:
    seq(T(n), n=0..25);  # Alois P. Heinz, Apr 01 2012
  • Mathematica
    row[0] = {1}; row[1] = {1, 1}; row[n_] := row[n] = Plus @@ PadRight[ {row[n-1], Join[{0}, row[Floor[n/2]]]} ]; Table[row[n], {n, 0, 25}] // Flatten (* Jean-François Alcover, Jan 31 2014 *)

Formula

Row 0 = {1}, row 1 = {1 1}; for n >=2, row n = row n-1 + (row floor(n/2) shifted one place right).
G.f. for column k (k >= 2): x^(2^(k-2))/((1-x)*Product_{j=0..k-2} (1-x^(2^j))). [corrected by Jason Yuen, Jan 12 2025]
Conjecture: let R(n,x) be the n-th reversed row polynomial, then R(n,x) = Sum_{k=0..A000523(A053645(n)) + 1} T(A053645(n),k)*R(2^(A000523(n)-k),x) for n > 0, n != 2^m with R(0,x) = 1 where R(2^m,x) is the (m+1)-th row polynomial of A078121. - Mikhail Kurkov, Jun 28 2025

Extensions

More terms from Alford Arnold, May 22 2004

A014557 Multiplicity of K_3 in K_n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 2, 4, 8, 12, 20, 28, 40, 52, 70, 88, 112, 136, 168, 200, 240, 280, 330, 380, 440, 500, 572, 644, 728, 812, 910, 1008, 1120, 1232, 1360, 1488, 1632, 1776, 1938, 2100, 2280, 2460, 2660, 2860, 3080, 3300, 3542, 3784, 4048, 4312, 4600, 4888, 5200
Offset: 0

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Author

Keywords

Comments

The multiplicity of triangles in K_n is defined to be the minimum number of monochromatic copies of K_3 that occur in any 2-coloring of the edges of K_n. - Allan Bickle, Mar 04 2023
Twice A008804 (up to offset).
From Alexander Adamchuk, Nov 29 2006: (Start)
n divides a(n) for n = {1,2,3,4,5,8,10,13,14,16,17,20,22,25,26,28,29,32,34,37,38,40,41,44,46,49,50,52,53,56,58,61,62,64,65,68,70,73,74,76,77,80,82,85,86,88,89,92,94,97,98,100,...}.
Prime p divides a(p) for p = {2,3,5,13,17,29,37,41,53,61,73,89,97,101,109,113,137,149,157,173,181,193,197,...} = (2,3) and all primes from A002144: Pythagorean primes: primes of form 4n+1.
(n+1) divides a(n) for n = {1,2,3,4,5,19,27,43,51,67,75,91,99,...}.
(p+1) divides a(p) for prime p = {2,3,5,19,43,67,139,163,211,283,307,331,379,499,523,547,571,619,643,691,739,787,811,859,883,907,...} = {2,5} and all primes from A141373: Primes of the form 3x^2+16y^2.
(n-1) divides a(n) for n = {2,3,4,5,21,29,45,53,69,77,93,101,...}.
(p-1) divides a(p) for prime p = {2,3,5,29,53,101,149,173,197,269,293,317,389,461,509,557,653,677,701,773,797,821,941,..} = {2,3} and all primes from A107003: Primes of the form 5x^2+2xy+5y^2, with x and y any integer.
(n-2) divides a(n) for n = {3,4,5,12,16,24,28,36,40,48,52,60,64,72,76,84,88,96,100,...} = {3,5} and 4*A032766: Numbers congruent to 0 or 1 mod 3.
(n+3) divides a(n) for n = {1,2,3,4,5,9,11,18,32,39}.
(n-3) divides a(n) for n = {4,5,7,9,23,31,47,55,71,79,95,103,119,127,143,151,167,175,...}.
(p+3) divides a(p) for prime p = {5,7,23,31,47,71,79,103,127,151,167,191,199,...} = {5} and all primes from A007522: Primes of form 8n+7.
(n-4) divides a(n) for n = {5,6,8,11,12,14,15,18,20,23,24,26,27,30,32,35,36,38,39,42,44,47,48,50,...}.
(p-4) divides a(p) for prime p = {5,11,23,47,59,71,83,107,131,167,179,191,...} = {5} and all primes from A068231: Primes congruent to 11 (mod 12).
(n+5) divides a(n) for n = {1,2,3,4,5,30,31,45,58,145}.
(n-5) divides a(n) for n = {6,7,9,10,20,25,33,49,57,73,81,97,105,...}.
(p-5) divides a(p) for prime p = {7,73,97,193,241,313,337,409,433,457,577,601,673,769,937,...} = {7} and all primes from A107008: Primes of the form x^2+24y^2. (End)

Examples

			Any 2-coloring of the edges of K_6 produces at least two monochromatic triangles.  Having colors induce K_3,3 and 2K_3 shows this is attained, so a(6) = 2.
		

Crossrefs

Programs

  • Magma
    [n*(n-1)*(n-2)/6 - Floor((n/2)*Floor(((n-1)/2)^2)): n in [1..20]]; // G. C. Greubel, Oct 06 2017
  • Maple
    A049322 := proc(n) local u; if n mod 2 = 0 then u := n/2; RETURN(u*(u-1)*(u-2)/3); elif n mod 4 = 1 then u := (n-1)/4; RETURN(u*(u-1)*(4*u+1)*2/3); else u := (n-3)/4; RETURN(u*(u+1)*(4*u-1)*2/3); fi; end;
  • Mathematica
    Table[Binomial[n,3] - Floor[n/2*Floor[((n-1)/2)^2]],{n,0,100}] (* Alexander Adamchuk, Nov 29 2006 *)
  • PARI
    x='x+O('x^99); concat(vector(6), Vec(2*x^6/((x-1)^4*(x+1)^2*(x^2+1)))) \\ Altug Alkan, Apr 08 2016
    

Formula

a(n) = binomial(n,3) - floor(n/2 * floor(((n-1)/2)^2)). - Alexander Adamchuk, Nov 29 2006
G.f.: 2*x^6/((x-1)^4*(x+1)^2*(x^2+1)). - Colin Barker, Nov 28 2012
E.g.f.: ((x - 3)*x^2*cosh(x) - 6*sin(x) + (6 + 3*x - 3*x^2 + x^3)*sinh(x))/24. - Stefano Spezia, May 15 2023

Extensions

Entry revised by N. J. A. Sloane, Mar 22 2004

A088954 G.f.: 1/((1-x)^2*(1-x^2)*(1-x^4)*(1-x^8)*(1-x^16)).

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166, 202, 238, 284, 330, 390, 450, 524, 598, 692, 786, 900, 1014, 1154, 1294, 1460, 1626, 1827, 2028, 2264, 2500, 2780, 3060, 3384, 3708, 4088, 4468, 4904, 5340, 5844, 6348, 6920, 7492, 8148, 8804, 9544
Offset: 0

Views

Author

N. J. A. Sloane, Dec 02 2003

Keywords

Comments

a(n) is the number of partitions of 2*n into powers of 2 less than or equal to 2^5. First differs from A000123 at n=32. - Alois P. Heinz, Apr 02 2012

Crossrefs

See A000027, A002620, A008804, A088932, A000123 for similar sequences.
Column k=5 of A181322.

Programs

  • Maple
    f := proc(n,k) option remember; if k > n then RETURN(0); fi; if k= 0 then if n=0 then RETURN(1) else RETURN(0); fi; fi; if k = 1 then RETURN(1); fi; if n mod 2 = 1 then RETURN(f(n-1,k)); fi; f(n-1,k)+f(n/2,k-1); end; # present sequence is f(2m,6)
    GFF := k->x^(2^(k-2))/((1-x)*mul((1-x^(2^j)),j=0..k-2)); # present g.f. is GFF(6)/x^16
    a:= proc(n) local m, r; m:= iquo(n, 16, 'r'); r:= r+1; [1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166][r] +(((((128/5*m +8*(15+r))*m +(228 +[0, 32, 68, 104, 144, 184, 228, 272, 320, 368, 420, 472, 528, 584, 644, 704][r]))*m +(172 +[0, 43, 98, 153, 223, 293, 378, 463, 566, 669, 790, 911, 1053, 1195, 1358, 1521][r]))*m +(247/5 +[0, 22, 55, 88, 138, 188, 255, 322, 415, 508, 627, 746, 900, 1054, 1243, 1432][r]))*m)/3 end: seq(a(n), n=0..60); # Alois P. Heinz, Apr 17 2009
  • Mathematica
    CoefficientList[Series[1/((1-x)^2(1-x^2)(1-x^4)(1-x^8)(1-x^16)),{x,0,70}],x] (* or *) LinearRecurrence[{2,0,-2,2,-2,0,2,0,-2,0,2,-2,2,0,-2,2,-2,0,2,-2,2,0,-2,0,2,0,-2,2,-2,0,2,-1},{1,2,4,6,10,14,20,26,36,46,60,74,94,114,140,166,202,238,284,330,390,450,524,598,692,786,900,1014,1154,1294,1460,1626},70](* Harvey P. Dale, Feb 12 2013 *)

Formula

a(0)=1, a(1)=2, a(2)=4, a(3)=6, a(4)=10, a(5)=14, a(6)=20, a(7)=26, a(8)=36, a(9)=46, a(10)=60, a(11)=74, a(12)=94, a(13)=114, a(14)=140, a(15)=166, a(16)=202, a(17)=238, a(18)=284, a(19)=330, a(20)=390, a(21)=450, a(22)=524, a(23)=598, a(24)=692, a(25)=786, a(26)=900, a(27)=1014, a(28)=1154, a(29)=1294, a(30)=1460, a(31)=1626, a(n)=2*a(n-1)-2*a(n-3)+ 2*a(n-4)- 2*a(n-5)+ 2*a(n-7)-2*a(n-9)+2*a(n-11)-2*a(n-12)+2*a(n-13)-2*a(n-15)+2*a(n-16)-2*a(n-17)+ 2*a(n-19)- 2*a(n-20)+ 2*a(n-21)-2*a(n-23)+2*a(n-25)-2*a(n-27)+2*a(n-28)-2*a(n-29)+ 2*a(n-31)-a(n-32). - Harvey P. Dale, Feb 12 2013

A008643 Molien series for group of 4 X 4 upper triangular matrices over GF(2).

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 6, 10, 10, 14, 14, 20, 20, 26, 26, 35, 35, 44, 44, 56, 56, 68, 68, 84, 84, 100, 100, 120, 120, 140, 140, 165, 165, 190, 190, 220, 220, 250, 250, 286, 286, 322, 322, 364, 364, 406, 406, 455, 455, 504, 504, 560, 560, 616, 616, 680, 680, 744
Offset: 0

Views

Author

Keywords

Comments

Number of partitions of n into parts 1, 2, 4 and 8. - Ilya Gutkovskiy, May 24 2017

References

  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 105.

Crossrefs

Cf. A008804, A088932 (partial sums).

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 65); Coefficients(R!( (&*[1/(1-x^(2^j)): j in [0..3]]) )); // G. C. Greubel, Feb 01 2020
    
  • Maple
    a:= proc(n) local m, r; m := iquo(n, 8, 'r'); r:= iquo(r,2)+1; ([11, 17, 26, 35][r]+ (9+ 3*r+ 4*m) *m) *m/3+ [1, 2, 4, 6][r] end: seq(a(n), n=0..100);  # Alois P. Heinz, Oct 06 2008
  • Mathematica
    CoefficientList[1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8)) + O[x]^65, x] (* Jean-François Alcover, May 29 2015 *)
    LinearRecurrence[{1,1,-1,1,-1,-1,1,1,-1,-1,1,-1,1,1,-1}, {1,1,2,2,4,4,6,6,10,10,14,14,20,20,26}, 65] (* Ray Chandler, Jul 15 2015 *)
  • PARI
    my(x='x+O('x^65)); Vec(1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8))) \\ G. C. Greubel, Feb 01 2020
    
  • PARI
    my(b(m) = (m^3 + 12*m^2 + (44 - 3*(m%2))*m + 48)\48); vector(59,n,b((n-1)\2)) \\ Hoang Xuan Thanh, Aug 14 2025
    
  • Sage
    def A077952_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( 1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8)) ).list()
    A077952_list(65) # G. C. Greubel, Feb 01 2020

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8)).
a(n) = floor(((n+14)*(3*(n+1)*(-1)^n + 2*n^2 + 17*n + 57) + 24*(floor(n/2) + 1)*(-1)^floor(n/2))/768). - Tani Akinari, Jun 16 2013
a(n) ~ 1/384*n^3. - Ralf Stephan, Apr 29 2014

A032250 "DHK[ n ](2n)" (bracelet, identity, unlabeled, n parts, evaluated at 2n) transform of 1,1,1,1,...

Original entry on oeis.org

1, 1, 1, 2, 10, 29, 113, 368, 1316, 4490, 15907, 55866, 199550, 714601, 2583575, 9385280, 34311304, 126018592, 465044951, 1722987050, 6407739430, 23909854891, 89493459541, 335911158480, 1264104712300
Offset: 1

Views

Author

Keywords

Comments

From Petros Hadjicostas, Feb 24 2019: (Start)
Let ff(k, x) = x^k/2 * ( (1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1 -x^2)^floor(k/2 + 1) ) be Herbert Kociemba's formula for the g.f. of the number of all bracelets with k black beads and n-k white beads that have no reflection symmetry.
Let gg(k, x) be the generating function of the number of all aperiodic bracelets with k black beads and n-k white beads that have no reflection symmetry.
We conjecture that gg(k, x)= Sum_{d|k} mu(d)*ff(k/d, x^d).
For n >= 3, a(n) is the coefficient of x^(2*n) of the Taylor expansion of gg(n, x) around x=0. [Bower has special definitions for DHK[1] and DHK[2].]
(End)

Crossrefs

Programs

  • Maple
    # This is a crude program that assumes the above conjecture is true (which was later proved in Hadjicostas (2019)). It is only valid for n >= 3 because of Bower's special definition of DHK[k] for the cases k=1 and k=2.
    with(NumberTheory);
    ff := proc (k, x) (1/2)*x^k*(add(phi(n)/(1-x^n)^(k/n), n in Divisors(k))/k-(x+1)/(1-x^2)^floor((1/2)*k+1)); end proc;
    gg := proc (k, x) add(Moebius(d)*ff(k/d, x^d), d in Divisors(k)); end proc;
    vv := proc (n) simplify(subs(x = 0, diff(gg(n, x), x$(2*n)))/factorial(2*n)); end proc;
    for i from 3 to 100 do print(i, vv(i)); end do; # Petros Hadjicostas, Feb 24 2019

A308401 Number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of 6 white beads and n-6 black beads.

Original entry on oeis.org

3, 6, 16, 30, 56, 91, 150, 224, 336, 477, 672, 912, 1233, 1617, 2112, 2700, 3432, 4290, 5340, 6552, 8008, 9678, 11648, 13888, 16503, 19448, 22848, 26658, 31008, 35853, 41346, 47424, 54264, 61803, 70224, 79464, 89733, 100947, 113344, 126840, 141680, 157780, 175416, 194480, 215280, 237708
Offset: 9

Views

Author

Petros Hadjicostas, May 24 2019

Keywords

Comments

Bracelets that have no reflection symmetry are also known as chiral bracelets.
Here, for n >= 6, a(n) is also the number of dihedral compositions of n with 6 parts that have no reflection symmetry. Taking the MacMahon conjugates of these dihedral compositions, we see that a(n) is also the number of dihedral compositions of n into n-6 parts that have no reflection symmetry.
A cyclic composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation. Such compositions were first studied extensively by Sommerville (1909).
A dihedral composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation or a reversal of order. Such compositions were studied, for example, by Knopfmacher and Robbins (2013).
Given a bracelet of length n with k white beads and n-k black beads, we may get the corresponding dihedral composition using MacMahon's correspondence: start with a white bead and count that bead and the black beads that follow (in one direction), and call that b_1; then start with the next white bead and count that one and the black beads that follow, and call that b_2; repeat this process until you reach the k-th white bead and count that one and the black beads that follow, and call that b_k. The corresponding dihedral composition is b_1 + b_2 + ... + b_k.
If in the previous paragraph (given a bracelet of length n with k white beads and n-k black beads), we replace the white beads with black beads and the black beads with white beads, we get a dihedral composition of n into n-k parts: c_1 + c_2 + ... + c_{n-k}. These two dihedral compositions (which correspond to the same bracelet) are called "conjugate" compositions. See p. 273 in Sommerville (1909) for an explanation of "conjugate" compositions in the context of cyclic compositions.
Symmetric cyclic compositions of a positive integer n were first studied by Sommerville (1909, pp. 301-304). It can be proved that the study of necklaces with reflection symmetry using beads of two colors is equivalent to the study of symmetric cyclic compositions of a positive integer. Clearly all the necklaces with reflection symmetry are all the bracelets (turnover necklaces) with reflection symmetry. See also the comments for sequences A119963, A292200, and A295925.

Examples

			Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry (and thus, a(9) = 3).
Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric (and thus, a(9) = 3).
We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries (and thus, a(9) = 3). For example, dihedral composition 1 is symmetric because we can draw an axis of symmetry through one of the 1s and 4. In addition, dihedral composition 5 is symmetric because we may draw an axis of symmetry through the numbers 2 and 3.
		

Crossrefs

Programs

  • PARI
    a(n) = (1/12)* (sumdiv(gcd(n, 6), d,  eulerphi(d)*binomial((n/d) - 1, (6/d) - 1))) - (1/2)*binomial(floor(n/2), 3); \\ Michel Marcus, May 28 2019
    
  • PARI
    Vec(x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2) + O(x^50)) \\ Colin Barker, Jun 02 2019

Formula

G.f.: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)) with k = 6. (This formula is due to Herbert Kociemba.)
a(n) = A005513(n) - A058187(n-6) = A005513(n) - binomial(floor(n/2), 3) for n >= 6.
a(n) = -(1/2)*binomial(floor(n/2), 3) + (1/12)* Sum_{d|gcd(n, 6)} phi(d)*binomial((n/d) - 1, (6/d) - 1) for n >= 6. (This is a modification of formulas found in Gupta (1979) and Shevelev (2004).)
From Colin Barker, May 26 2019: (Start)
G.f.: x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4) + a(n-5) + 4*a(n-6) - 3*a(n-7) - 3*a(n-8) + 4*a(n-9) + a(n-10) - a(n-11) - 3*a(n-12) + a(n-13) + 2*a(n-14) - a(n-15) for n > 23. (End)

A308583 Triangle read by rows: T(n,k) = number of aperiodic chiral bracelets (turnover necklaces with no reflection symmetry and period n) with n beads, k of which are white and n - k are black, for n >= 1 and 1 <= k <= n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 3, 4, 4, 3, 0, 0, 0, 0, 0, 4, 6, 10, 6, 4, 0, 0, 0, 0, 0, 5, 10, 16, 16, 10, 5, 0, 0, 0, 0, 0, 7, 14, 28, 29, 28, 14, 7, 0, 0, 0, 0, 0, 8, 20, 42, 56, 56, 42, 20, 8, 0, 0, 0, 0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0, 0, 0, 12, 35, 90, 150, 197, 197, 150, 90, 35, 12, 0, 0, 0, 0, 0, 14, 44, 126, 222, 340, 368, 340, 222, 126, 44, 14, 0, 0, 0
Offset: 1

Views

Author

Petros Hadjicostas, Jun 08 2019

Keywords

Comments

For k = 1, 4, or a prime, the columns of this triangular array are exactly the same as the corresponding columns for the triangular array A180472. In other words, all chiral bracelets with n beads, k of which are white and n - k are black, are aperiodic if k = 1, 4, or a prime.
Note that, T(n, k) is also the number of aperiodic dihedral compositions of n with k parts and no reflection symmetry. Since T(n, k) = T(n, n - k), T(n, k) is also the number of aperiodic dihedral compositions of n with n - k parts and no reflection symmetry.

Examples

			The triangle begins (with rows for n >= 1 and columns for k >= 1) as follows:
  0;
  0, 0;
  0, 0,  0;
  0, 0,  0,  0;
  0, 0,  0,  0,  0;
  0, 0,  1,  0,  0,  0;
  0, 0,  1,  1,  0,  0,   0;
  0, 0,  2,  2,  2,  0,   0,  0;
  0, 0,  3,  4,  4,  3,   0,  0,  0;
  0, 0,  4,  6, 10,  6,   4,  0,  0,  0;
  0, 0,  5, 10, 16, 16,  10,  5,  0,  0,  0;
  0, 0,  7, 14, 28, 29,  28, 14,  7,  0,  0, 0;
  0, 0,  8, 20, 42, 56,  56, 42, 20,  8,  0, 0, 0;
  0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0;
  ...
Notice, for example, that T(14, 6) = 90 <> 91 = A180472(14, 6). Out of the 91 chiral bracelets with 6 W and 8 B beads, only WWBWBBBWWBWBBB is periodic.
Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry. Because chiral bracelets 2, 3, and 6 are aperiodic as well, we have T(9, 3) = 3 = T(9, 6).
Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric. In addition, chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and so (again) T(9, 3) = 3.
We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries. Chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and hence T(9, 6) = 3.
		

Crossrefs

Cf. A032239 (row sums for n >= 3), A180472.
Cf. A001399 (column k = 3 with a different offset), A008804 (column k = 4 with a different offset), A032246 (column k = 5), A032247 (column k = 6), A032248 (column k = 7), A032249 (column k = 8).

Formula

T(n, k) = Sum_{d|gcd(n,k)} mu(d) * A180472(n/d, k/d) for 1 <= k <= n.
T(n, k) = T(n, n - k) for 1 <= k <= n - 1.
T(n, k) = (1/(2*k)) * Sum_{d|gcd(n, k)} mu(d) * (binomial(n/d - 1, k/d - 1) - k * binomial(floor(b(n, k, d)/2), floor(k/(2*d)))) for 1 <= k <= n, where b(n, k, d) = (n/d) + ((-1)^(k/d) - 1)/2.
T(n, k) = (1/(2*n)) * Sum_{d|gcd(n, k)} mu(d) * (binomial(n/d, k/d) - n * binomial(floor(b(n, k, d)/2), floor(k/(2*d)))) for 1 <= k <= n, where b(n, k, d) = (n/d) + ((-1)^(k/d) - 1)/2.
G.f. for column k >= 1: (x^k/(2*k)) * Sum_{d|k} mu(d) * (1/(1 - x^d)^(k/d) - k * (1 + x^d)/(1 - x^(2*d))^floor((k/(2*d)) + 1)).
Bivariate g.f.: Sum_{n,k >= 1} T(n, k)*x^n*y^k = (1/2) * Sum_{d >= 1} mu(d) * (1 - (1 + x^d) * (1 + x^d*y^d) / (1 - x^(2*d) * (1 + y^(2*d)))) - (1/2) * Sum_{d >= 1} (mu(d)/d) * log(1 - x^d * (1 + y^d)).

A002985 Number of trees in an n-node wheel.

Original entry on oeis.org

1, 1, 1, 2, 3, 6, 11, 20, 36, 64, 108, 179, 292, 464, 727, 1124, 1714, 2585, 3866, 5724, 8418, 12290, 17830, 25713, 36898, 52664, 74837, 105873, 149178, 209364, 292793, 407990, 566668, 784521, 1082848, 1490197, 2045093, 2798895, 3820629, 5202085
Offset: 1

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Author

Keywords

Comments

This is the number of nonequivalent spanning trees of the n-wheel graph up to isomorphism of the trees.

Examples

			All trees that span a wheel on 5 nodes are equivalent to one of the following:
      o         o         o
      |         | \     /   \
   o--o--o   o--o  o   o--o  o
      |         |           /
      o         o         o
		

References

  • F. Harary, P. E. O'Neil, R. C. Read and A. J. Schwenk, The number of trees in a wheel, in D. J. A. Welsh and D. R. Woodall, editors, Combinatorics. Institute of Mathematics and Its Applications. Southend-on-Sea, England, 1972, pp. 155-163.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    terms = 40;
    A003293[n_] := SeriesCoefficient[Product[(1-x^k)^(-Ceiling[k/2]), {k, 1, terms}], {x, 0, n}];
    A008804[n_] := SeriesCoefficient[1/((1-x)^4 (1+x)^2 (1+x^2)), {x, 0, n}];
    a[n_] := A003293[n-1] - A008804[n-3];
    Array[a, terms] (* Jean-François Alcover, Sep 02 2019 *)
  • PARI
    \\ here b(n) is A003293 and d(n) is A008804.
    b(n)={polcoeff( prod(k=1, n, (1-x^k+x*O(x^n))^-ceil(k/2)), n)}
    d(n)={(84+12*(-1)^n+6*I*((-I)^n-I^n)+(85+3*(-1)^n)*n+24*n^2+2*n^3)/96}
    a(n)=b(n-1)-d(n-3); \\ Andrew Howroyd, Oct 09 2017

Formula

a(n) = A003293(n-1) - A008804(n-3). - Andrew Howroyd, Oct 09 2017

Extensions

Terms a(32) and beyond from Andrew Howroyd, Oct 09 2017

A073463 Triangle of number of partitions of 2n into powers of 2 where the largest part is 2^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 0, 1, 4, 4, 1, 0, 1, 5, 6, 2, 0, 0, 1, 6, 9, 4, 0, 0, 0, 1, 7, 12, 6, 0, 0, 0, 0, 1, 8, 16, 10, 1, 0, 0, 0, 0, 1, 9, 20, 14, 2, 0, 0, 0, 0, 0, 1, 10, 25, 20, 4, 0, 0, 0, 0, 0, 0, 1, 11, 30, 26, 6, 0, 0, 0, 0, 0, 0, 0, 1, 12, 36, 35, 10, 0, 0, 0, 0, 0, 0, 0, 0, 1, 13, 42, 44
Offset: 0

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Author

Henry Bottomley, Aug 02 2002

Keywords

Comments

In the recurrence T(n,k)=T(n-1,k)+T([n/2],k-1): T(n-1,k) represents the partitions where the smallest part is 1 and T([n/2],k-1) those where it is not.

Examples

			Rows start:
  1;
  1, 1;
  1, 2, 1;
  1, 3, 2, 0;
  1, 4, 4, 1, 0;
  1, 5, 6, 2, 0, 0;
  ...
		

Crossrefs

Columns include A000012, A000027, A002620, A008804. Subsequent columns start like A000123 (offset). Row sums are A000123.

Formula

T(n, k) = T(n-1, k)+T([n/2], k-1) starting with T(n, 0)=1 and T(0, k)=0 for k>0.
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