cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 14 results. Next

A000011 Number of n-bead necklaces (turning over is allowed) where complements are equivalent.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 8, 9, 18, 23, 44, 63, 122, 190, 362, 612, 1162, 2056, 3914, 7155, 13648, 25482, 48734, 92205, 176906, 337594, 649532, 1246863, 2405236, 4636390, 8964800, 17334801, 33588234, 65108062, 126390032, 245492244, 477353376, 928772650, 1808676326, 3524337980
Offset: 0

Views

Author

Keywords

Comments

a(n) is also the number of minimal fibrations of a bidirectional n-cycle over the 2-bouquet up to precompositions with automorphisms of the n-cycle and postcomposition with automorphisms of the 2-bouquet. (Boldi et al.) - Sebastiano Vigna, Jan 08 2018
For n >= 3, also the number of distinct planar embeddings of the n-sunlet graph. - Eric W. Weisstein, May 21 2024

Examples

			From Jason Orendorff (jason.orendorff(AT)gmail.com), Jan 09 2009: (Start)
The binary bracelets for small n are:
  n: bracelets
  0: (the empty bracelet)
  1: 0
  2: 00, 01
  3: 000, 001
  4: 0000, 0001, 0011, 0101
  5: 00000, 00001, 00011, 00101
  6: 000000, 000001, 000011, 000101, 000111, 001001, 001011, 010101
(End)
		

References

  • 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).

Crossrefs

Column 2 of A320748.
Cf. A000013. Bisections give A000117 and A092668.
The 8 sequences in Table 8 of Fujita (2017) are A053656, A000011, A256216, A256217, A123045, A283846, A283847, A283848.

Programs

  • Maple
    with(numtheory): A000011 := proc(n) local s,d; if n = 0 then RETURN(1) else s := 2^(floor(n/2)); for d in divisors(n) do s := s+(phi(2*d)*2^(n/d))/(2*n); od; RETURN(s/2); fi; end;
  • Mathematica
    a[n_] := Fold[ #1 + EulerPhi[2#2]2^(n/#2)/(2n) &, 2^Floor[n/2], Divisors[n]]/2
    a[ n_] := If[ n < 1, Boole[n == 0], 2^Quotient[n, 2] / 2 + DivisorSum[ n, EulerPhi[2 #] 2^(n/#) &] / (4 n)]; (* Michael Somos, Dec 19 2014 *)
  • PARI
    {a(n) = if( n<1, n==0, 2^(n\2) / 2 + sumdiv(n, k, eulerphi(2*k) * 2^(n/k)) / (4*n))}; /* Michael Somos, Jun 03 2002 */

Formula

a(n) = (A000013(n) + 2^floor(n/2))/2.

Extensions

Better description from Christian G. Bower
More terms from David W. Wilson, Jan 13 2000

A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.

Original entry on oeis.org

1, 1, 4, 4, 10, 10, 20, 20, 35, 35, 56, 56, 84, 84, 120, 120, 165, 165, 220, 220, 286, 286, 364, 364, 455, 455, 560, 560, 680, 680, 816, 816, 969, 969, 1140, 1140, 1330, 1330, 1540, 1540, 1771, 1771, 2024, 2024, 2300, 2300, 2600, 2600, 2925, 2925, 3276, 3276
Offset: 0

Views

Author

Henry Bottomley, Nov 20 2000

Keywords

Comments

For n >= i, i = 6,7, a(n - i) is the number of incongruent two-color bracelets of n beads, i of which are black (cf. A005513, A032280), having a diameter of symmetry. The latter means the following: if we imagine (0,1)-beads as points (with the corresponding labels) dividing a circumference of a bracelet into n identical parts, then a diameter of symmetry is a diameter (connecting two beads or not) such that a 180-degree turn of one of two sets of points around it (obtained by splitting the circumference by this diameter) leads to the coincidence of the two sets (including their labels). - Vladimir Shevelev, May 03 2011
From Johannes W. Meijer, May 20 2011: (Start)
The Kn11, Kn12, Kn13, Fi1 and Ze1 triangle sums, see A180662 for their definitions, of the Connell-Pol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Fi1(n) = a(n-1) + 5*a(n-2) + a(n-3) + 5*a(n-4).
The Kn11, Kn12, Kn13, Kn21, Kn22, Kn23, Fi1, Fi2, Ze1 and Ze2 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above. (End)
The number of quadruples of integers [x, u, v, w] that satisfy x > u > v > w >= 0, n + 5 = x + u. - Michael Somos, Feb 09 2015
Also, this sequence is the fourth column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018

Crossrefs

Cf. A057884. Sum of 2 consecutive terms gives A006918, whose sum of 2 consecutive terms gives A002623, whose sum of 2 consecutive terms gives A000292, which is this sequence without the duplication. Continuing to sum 2 consecutive terms gives A000330, A005900, A001845, A008412 successively.

Programs

  • Haskell
    a058187 n = a058187_list !! n
    a058187_list = 1 : f 1 1 [1] where
       f x y zs = z : f (x + y) (1 - y) (z:zs) where
         z = sum $ zipWith (*) [1..x] [x,x-1..1]
    -- Reinhard Zumkeller, Dec 21 2011
    
  • Maple
    A058187:= proc(n) option remember; A058187(n):= binomial(floor(n/2)+3, 3) end: seq(A058187(n), n=0..51); # Johannes W. Meijer, May 20 2011
  • Mathematica
    a[n_]:= Length @ FindInstance[{x>u, u>v, v>w, w>=0, x+u==n+5}, {x, u, v, w}, Integers, 10^9]; (* Michael Somos, Feb 09 2015 *)
    With[{tetra=Binomial[Range[30]+2,3]},Riffle[tetra,tetra]] (* Harvey P. Dale, Mar 22 2015 *)
  • PARI
    {a(n) = binomial(n\2+3, 3)}; /* Michael Somos, Jun 07 2005 */
    
  • Sage
    [binomial((n//2)+3, 3) for n in (0..60)] # G. C. Greubel, Feb 18 2022

Formula

a(n) = A006918(n+1) - a(n-1).
a(2*n) = a(2*n+1) = A000292(n) = (n+1)*(n+2)*(n+3)/6.
a(n) = (2*n^3 + 21*n^2 + 67*n + 63)/96 + (n^2 + 7*n + 11)(-1)^n/32. - Paul Barry, Aug 19 2003
a(n) = A108299(n-3,n)*(-1)^floor(n/2) for n > 2. - Reinhard Zumkeller, Jun 01 2005
Euler transform of finite sequence [1, 3]. - Michael Somos, Jun 07 2005
G.f.: 1 / ((1 - x) * (1 - x^2)^3) = 1 / ((1 + x)^3 * (1 - x)^4). a(n) = -a(-7-n) for all n in Z.
a(n) = binomial(floor(n/2) + 3, 3). - Vladimir Shevelev, May 03 2011
a(-n) = -a(n-7); a(n) = A000292(A008619(n)). - Guenther Schrack, Sep 13 2018
Sum_{n>=0} 1/a(n) = 3. - Amiram Eldar, Aug 18 2022

A052307 Triangle read by rows: T(n,k) = number of bracelets (reversible necklaces) with n beads, k of which are black and n - k are white.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 1, 3, 4, 4, 3, 1, 1, 1, 1, 4, 5, 8, 5, 4, 1, 1, 1, 1, 4, 7, 10, 10, 7, 4, 1, 1, 1, 1, 5, 8, 16, 16, 16, 8, 5, 1, 1, 1, 1, 5, 10, 20, 26, 26, 20, 10, 5, 1, 1, 1, 1, 6, 12, 29, 38, 50, 38, 29, 12, 6, 1, 1, 1, 1, 6, 14, 35, 57, 76, 76, 57, 35, 14, 6, 1, 1
Offset: 0

Views

Author

Christian G. Bower, Nov 15 1999

Keywords

Comments

Equivalently, T(n,k) is the number of orbits of k-element subsets of the vertices of a regular n-gon under the usual action of the dihedral group D_n, or under the action of Euclidean plane isometries. Note that each row of the table is symmetric and unimodal. - Austin Shapiro, Apr 20 2009
Also, the number of k-chords in n-tone equal temperament, up to (musical) transposition and inversion. Example: there are 29 tetrachords, 38 pentachords, 50 hexachords in the familiar 12-tone equal temperament. Called "Forte set-classes," after Allen Forte who first cataloged them. - Jon Wild, May 21 2004

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   1,  1;
   1,  1,  1;
   1,  1,  1,  1;
   1,  1,  2,  1,  1;
   1,  1,  2,  2,  1,  1;
   1,  1,  3,  3,  3,  1,  1;
   1,  1,  3,  4,  4,  3,  1,  1;
   1,  1,  4,  5,  8,  5,  4,  1,  1;
   1,  1,  4,  7, 10, 10,  7,  4,  1,  1;
   1,  1,  5,  8, 16, 16, 16,  8,  5,  1,  1;
   1,  1,  5, 10, 20, 26, 26, 20, 10,  5,  1,  1;
   1,  1,  6, 12, 29, 38, 50, 38, 29, 12,  6,  1,  1;
   ...
		

References

  • Martin Gardner, "New Mathematical Diversions from Scientific American" (Simon and Schuster, New York, 1966), pages 245-246.
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Programs

  • Maple
    A052307 := proc(n,k)
            local hk,a,d;
            if k = 0 then
                    return 1 ;
            end if;
            hk := k mod 2 ;
            a := 0 ;
            for d in numtheory[divisors](igcd(k,n)) do
                    a := a+ numtheory[phi](d)*binomial(n/d-1,k/d-1) ;
            end do:
            %/k + binomial(floor((n-hk)/2),floor(k/2)) ;
            %/2 ;
    end proc: # R. J. Mathar, Sep 04 2011
  • Mathematica
    Table[If[m*n===0,1,1/2*If[EvenQ[n], If[EvenQ[m], Binomial[n/2, m/2], Binomial[(n-2)/2, (m-1)/2 ]], If[EvenQ[m], Binomial[(n-1)/2, m/2], Binomial[(n-1)/2, (m-1)/2]]] + 1/2*Fold[ #1 +(EulerPhi[ #2]*Binomial[n/#2, m/#2])/n &, 0, Intersection[Divisors[n], Divisors[m]]]], {n,0,12}, {m,0,n}] (* Wouter Meeussen, Aug 05 2002, Jan 19 2009 *)
  • PARI
    B(n,k)={ if(n==0, return(1)); GCD = gcd(n, k); S = 0;
    for(d = 1, GCD, if((k%d==0)&&(n%d==0), S+=eulerphi(d)*binomial(n/d,k/d)));
    return (binomial(floor(n/2)- k%2*(1-n%2), floor(k/2))/2 + S/(2*n)); }
    n=0;k=0; for(L=0, 8645, print(L, " ", B(n,k)); k++; if(k>n, k=0; n++))
    /* Washington Bomfim, Jun 30 2012 */
    
  • Python
    from sympy import binomial as C, totient, divisors, gcd
    def T(n, k): return 1 if n==0 else C((n//2) - k%2 * (1 - n%2), (k//2))/2 + sum(totient(d)*C(n//d, k//d) for d in divisors(gcd(n, k)))/(2*n)
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 23 2017

Formula

T(0,0) = 1. If n > 0, T(n,k) = binomial(floor(n/2) - (k mod 2) * (1 - (n mod 2)), floor(k/2)) / 2 + Sum_{d|n, d|k} (phi(d)*binomial(n/d, k/d)) / (2*n). - Washington Bomfim, Jun 30 2012 [edited by Petros Hadjicostas, May 29 2019]
From Freddy Barrera, Apr 21 2019: (Start)
T(n,k) = (1/2) * (A119963(n,k) + A047996(n,k)).
T(n,k) = T(n, n-k) for each k < n (Theorem 2 of H. Gupta). (End)
G.f. for column k >= 1: (x^k/2) * ((1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m) + (1 + x)/(1 - x^2)^floor((k/2) + 1)). (This formula is due to Herbert Kociemba.) - Petros Hadjicostas, May 25 2019
Bivariate o.g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = (1/2) * ((x + 1) * (x*y + 1) / (1 - x^2 * (y^2 + 1)) + 1 - Sum_{d >= 1} (phi(d)/d) * log(1 - x^d * (1 + y^d))). - Petros Hadjicostas, Jun 13 2019

A005232 Expansion of (1-x+x^2)/((1-x)^2*(1-x^2)*(1-x^4)).

Original entry on oeis.org

1, 1, 3, 4, 8, 10, 16, 20, 29, 35, 47, 56, 72, 84, 104, 120, 145, 165, 195, 220, 256, 286, 328, 364, 413, 455, 511, 560, 624, 680, 752, 816, 897, 969, 1059, 1140, 1240, 1330, 1440, 1540, 1661, 1771, 1903, 2024, 2168, 2300, 2456, 2600, 2769, 2925, 3107, 3276
Offset: 0

Views

Author

Keywords

Comments

Also number of n X 2 binary matrices under row and column permutations and column complementations (if offset is 0).
Also Molien series for certain 4-D representation of dihedral group of order 8.
With offset 4, number of bracelets (turnover necklaces) of n-bead of 2 colors with 4 red beads. - Washington Bomfim, Aug 27 2008
From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 4 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=4 (see our comment to A032279). (End)
Number of 2 X 2 matrices with nonnegative integer values totaling n under row and column permutations. - Gabriel Burns, Nov 08 2016
From Petros Hadjicostas, Jan 12 2019: (Start)
By "necklace", Vladimir Shevelev (above) means "turnover necklace", i.e., a bracelet. Zagaglia Salvi (1999) also uses this terminology: she calls a bracelet "necklace" and a necklace "cycle".
According to Cyvin et al. (1997), the sequence (a(n): n >= 0) consists of "the total numbers of isomers of polycyclic conjugated hydrocarbons with q + 1 rings and q internal carbons in one ring (class Q_q)", where q = 4 and n is the hydrogen content (i.e., we count certain isomers of C_{n+2*q} H_n with q = 4 and n >= 0). (End)

Examples

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 10*x^5 + 16*x^6 + 20*x^7 + 29*x^8 + ...
There are 8 4 X 2 matrices up to row and column permutations and column complementations:
  [1 1] [1 0] [1 0] [0 1] [0 1] [0 1] [0 1] [0 0]
  [1 1] [1 1] [1 0] [1 0] [1 0] [1 0] [0 1] [0 1]
  [1 1] [1 1] [1 1] [1 1] [1 0] [1 0] [1 0] [1 0]
  [1 1] [1 1] [1 1] [1 1] [1 1] [1 0] [1 0] [1 1].
There are 8 2 X 2 matrices of nonnegative integers totaling 4 up to row and column permutations:
  [4 0] [3 1] [2 2] [2 1] [2 1] [3 0] [2 0] [1 1]
  [0 0] [0 0] [0 0] [0 1] [1 0] [1 0] [2 0] [1 1].
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Row n=2 of A343875.
Column k=4 of A052307.

Programs

  • Maple
    A005232:=-(-1-z-2*z**3+2*z**2+z**7-2*z**6+2*z**4)/(z**2+1)/(1+z)**2/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence apart from an initial 1
  • Mathematica
    k = 4; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    CoefficientList[ Series[(1 - x + x^2)/((1 - x)^2(1 - x^2)(1 - x^4)), {x, 0, 51}], x] (* Robert G. Wilson v, Mar 29 2006 *)
    LinearRecurrence[{2,0,-2,2,-2,0,2,-1},{1,1,3,4,8,10,16,20},60] (* Harvey P. Dale, Oct 24 2012 *)
    k=4 (* Number of red beads in bracelet problem *); CoefficientList[Series[(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)
  • PARI
    {a(n) = (n^3 + 9*n^2 + (32-9*(n%2))*n + [48, 15, 36, 15][n%4+1]) / 48}; \\ Michael Somos, Feb 01 2007
    
  • PARI
    {a(n) = my(s=1); if( n<-5, n = -6-n; s=-1); if( n<0, 0, s * polcoeff( (1 - x + x^2) / ((1 - x)^2 * (1 - x^2) * (1 - x^4)) + x * O(x^n), n))}; \\ Michael Somos, Feb 01 2007
    
  • PARI
    a(n) = round((n^3 +9*n^2 +(32-9*(n%2))*n)/48 +0.6) \\ Washington Bomfim, Jul 17 2008
    
  • PARI
    a(n) = ceil((n+1)*(2*n^2+16*n+39+9*(-1)^n)/96) \\ Tani Akinari, Aug 23 2013
    
  • Python
    a=lambda n: sum((k//2+1)*((n-k)//2+1) for k in range((n-1)//2+1))+(n+1)%2*(((n//4+1)*(n//4+2))//2)  # Gabriel Burns, Nov 08 2016

Formula

G.f.: (1+x^3)/((1-x)*(1-x^2)^2*(1-x^4)).
G.f.: (1/8)*(1/(1-x)^4+3/(1-x^2)^2+2/(1-x)^2/(1-x^2)+2/(1-x^4)). - Vladeta Jovovic, Aug 05 2000
Euler transform of length 6 sequence [ 1, 2, 1, 1, 0, -1 ]. - Michael Somos, Feb 01 2007
a(2n+1) = A006918(2n+2)/2;
a(2n) = (A006918(2n+1) + A008619(n))/2.
a(n) = -a(-6 - n) for all n in Z. - Michael Somos, Feb 05 2011
From Vladimir Shevelev, Apr 22 2011: (Start)
if n == 0 (mod 4), then a(n) = n*(n^2-3*n+8)/48;
if n == 1, 3 (mod 4), then a(n) = (n^2-1)*(n-3)/48;
if n == 2 (mod 4), then a(n) = (n-2)*(n^2-n+6)/48. (End)
a(n) = 2*a(n-1) - 2*a(n-3) + 2*a(n-4) - 2*a(n-5) + 2*a(n-7) - a(n-8) with a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 4, a(4) = 8, a(5) = 10, a(6) = 16, a(7) = 20. - Harvey P. Dale, Oct 24 2012
a(n) = ((n+3)*(2*n^2+12*n+19+9*(-1)^n) + 6*(-1)^((2*n-1+(-1)^n)/4)*(1+(-1)^n))/96. - Luce ETIENNE, Mar 16 2015
a(n) = |A128498(n)| + |A128498(n-3)|. - R. J. Mathar, Jun 11 2019

Extensions

Sequence extended by Christian G. Bower

A006840 Number of 2n-bead black-white reversible complementable necklaces with n black beads.

Original entry on oeis.org

1, 1, 2, 3, 7, 13, 35, 85, 257, 765, 2518, 8359, 28968, 101340, 361270, 1297879, 4707969, 17179435, 63068876, 232615771, 861725794, 3204236779, 11955836258, 44748176653, 167959144032, 632058070310, 2384235077576, 9013628451275
Offset: 0

Views

Author

Keywords

References

  • J. A. Hoskins, C. E. Praeger and A. P. Street, Balanced twills with bounded float length, Congress. Numerantium, 40 (1983), 77-89.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of A259341.
Cf. A045629.

Programs

  • Mathematica
    b[n_] := (1/(2*n))*DivisorSum[n, EulerPhi[n/#]*Binomial[2*# - 1, # - 1] + EulerPhi[2*(n/#)]*2^(# - 1) &]; a[0] = 1; a[n_] := (b[n] + 2^(n-2) + Binomial[n - Mod[n, 2], Quotient[n, 2]]/2)/2; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Oct 08 2017, after Andrew Howroyd *)
  • PARI
    \\ here b is A045629
    b(n) = (1/(2*n)) * sumdiv(n, d, eulerphi(n/d)*binomial(2*d-1, d-1) + eulerphi(2*n/d)*2^(d-1));
    a(n) = if(n==0, 1, (b(n) + 2^(n-2) + binomial(n-n%2, n\2)/2) / 2); \\ Andrew Howroyd, Sep 27 2017

Formula

If n is odd, a(n) = (1/2) * (A045629 + (1/2) * C(n-1, (n-1)/2) + 2^(n-2)); if n is even, a(n) = (1/2) * (A045629 + (1/2) * C(n, n/2) + 2^(n-2)). - Christian G. Bower

Extensions

More terms from David W. Wilson

A005514 Number of n-bead bracelets (turnover necklaces) with 8 red beads and n-8 black beads.

Original entry on oeis.org

1, 1, 5, 10, 29, 57, 126, 232, 440, 750, 1282, 2052, 3260, 4950, 7440, 10824, 15581, 21879, 30415, 41470, 56021, 74503, 98254, 127920, 165288, 211276, 268228, 337416, 421856, 523260, 645456, 790704, 963793, 1167645, 1408185
Offset: 8

Views

Author

Keywords

Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 8 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=8 (see our comment at A032279). (End)
From Petros Hadjicostas, Jul 14 2018: (Start)
Let (c(n): n >= 1) be a sequence of nonnegative integers and let C(x) = Sum_{n>=1} c(n)*x^n be its g.f. Let k be a positive integer. Let a_k = (a_k(n): n >= 1) be the output sequence of the DIK[k] transform of sequence (c(n): n >= 1), and let A_k(x) = Sum_{n>=1} a_k(n)*x^n be its g.f. See Bower's web link below. It can be proved that, when k is even, A_k(x) = ((1/k)*Sum_{d|k} phi(d)*C(x^d)^(k/d) + (1/2)*C(x^2)^((k/2)-1)*(C(x)^2 + C(x^2)))/2.
For this sequence, k=8, c(n) = 1 for all n >= 1, and C(x) = x/(1-x). Thus, a(n) = a_8(n) for all n >= 1. Since a_k(n) = 0 for 1 <= n <= k-1, the offset of this sequence is n = k = 8. Applying the formula for the g.f. of DIK[8] of (c(n): n >= 1) with C(x) = x/(1-x) and k=8, we get Herbert Kociemba's formula below.
Here, a(n) is defined to be the number of n-bead bracelets of two colors with 8 red beads and n-8 black beads. But it is also the number of dihedral compositions of n with 8 parts. (This statement is equivalent to Vladimir Shevelev's statement above that a(n) is the "number of non-equivalent necklaces of 8 beads each of them painted by one of n colors." By "necklaces", he means "turnover necklaces". See the second paragraph of Section 2 in his 2004 paper in the Indian Journal of Pure and Applied Mathematics.)
Two cyclic compositions of n (with k = 8 parts) belong to the same equivalence class corresponding to a dihedral composition of n if and only if one can be obtained from the other by a rotation or reversal of order. (End)

Examples

			From _Petros Hadjicostas_, Jul 14 2018: (Start)
Every n-bead bracelet of two colors such that 8 beads are red and n-8 are black can be transformed into a dihedral composition of n with 8 parts in the following way. Start with one R bead and go in one direction (say clockwise) until you reach the next R bead. Continue this process until you come back to the original R bead.
Let b_i be the number of beads from R bead i until you reach the last B bead before R bead i+1 (or R bead 1). Here, b_i = 1 iff there are no B beads between R bead i and R bead i+1 (or R bead 8 and R bead 1). Then b_1 + b_2 + ... + b_8 = n, and we get a dihedral composition of n. (Of course, b_2 + b_3 + ... + b_8 + b_1 and b_8 + b_7 + ... + b_1 belong to the same equivalence class of the dihedral composition b_1 + ... + b_8.)
For example, a(10) = 5, and we have the following bracelets with 8 R beads and 2 B beads. Next to the bracelets we list the corresponding dihedral compositions of n with k=8 parts (they must be viewed on a circle):
RRRRRRRRBB <-> 1+1+1+1+1+1+1+3
RRRRRRRBRB <-> 1+1+1+1+1+1+2+2
RRRRRRBRRB <-> 1+1+1+1+1+2+1+2
RRRRRBRRRB <-> 1+1+1+1+2+1+1+2
RRRRBRRRRB <-> 1+1+1+2+1+1+1+2
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Programs

  • Mathematica
    k = 8; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    k=8;CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

S. J. Cyvin et al. (1997) give a g.f. (See equation (18) on p. 870 of their paper. Their g.f. is the same as the one given by V. Jovovic below except for the extra x^8.) - Petros Hadjicostas, Jul 14 2018
G.f.: (x^8/16)*(1/(1 - x)^8 + 4/(1 - x^8) + 5/(1 - x^2)^4 + 2/(1 - x^4)^2 + 4/(1 - x)^2/(1 - x^2)^3) = x^8*(2*x^10 - 3*x^9 + 7*x^8 - 6*x^7 + 7*x^6 - 2*x^5 + 2*x^4 - 2*x^3 + 5*x^2 - 3*x + 1)/(1 - x)^8/(1 + x)^4/(1 + x^2)^2/(1 + x^4). - Vladeta Jovovic, Jul 17 2002
a(n) = ((n+4)/32)*s(n,0,8) + ((n-4)/32)*s(n,4,8) + (48*C(n-1,7) + (n+1)*(n-2)*(n-4)*(n-6))/768, if n is even >= 8; a(n) = (48*C(n-1,7) + (n-1)*(n-3)*(n-5)*(n-7))/768, if n odd >= 8, where s(n,k,d)=1, if n == k (mod d), and 0 otherwise. - Vladimir Shevelev, Apr 23 2011
G.f.: k=8, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. - Herbert Kociemba, Nov 05 2016 [edited by Petros Hadjicostas, Jul 18 2018]
From Petros Hadjicostas, Jul 14 2018: (Start)
a(n) = (A032193(n) + A119963(n, 8))/2 = (A032193(n) + C(floor(n/2), 4))/2 for n >= 8.
The sequence (a(n): n >= 8) is the output sequence of Bower's "DIK[ 8 ]" (bracelet, indistinct, unlabeled, 8 parts) transform of 1, 1, 1, 1, ...
(End)

Extensions

Sequence extended and description corrected by Christian G. Bower
Name edited by Petros Hadjicostas, Jul 20 2018

A005515 Number of n-bead bracelets (turnover necklaces) of two colors with 10 red beads and n-10 black beads.

Original entry on oeis.org

1, 1, 6, 14, 47, 111, 280, 600, 1282, 2494, 4752, 8524, 14938, 25102, 41272, 65772, 102817, 156871, 235378, 346346, 502303, 716859, 1010256, 1404624, 1931540, 2625658, 3534776, 4711448, 6226148, 8156396, 10603704, 13679696, 17527595, 22304765, 28209566, 35459694
Offset: 10

Views

Author

Keywords

Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent (turnover) necklaces of 10 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=10 (see our comment to A032279). (End)

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Column k=10 of A052307.

Programs

  • Mathematica
    k = 10; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    k=10;CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

From Vladimir Shevelev, Apr 23 2011: (Start)
Put s(n,k,d) = 1, if n == k (mod d), and s(n,k,d) = 0, otherwise. Then a(n) = n*s(n,0,5)/25 + (384*C(n-1,9) + (n+1)*(n-2)*(n-4)*(n-6)*(n-8))/7680, if n is even; a(n) = (n-5)*s(n,0,5)/25 + (384*C(n-1,9) + (n-1)*(n-3)*(n-5)*(n-7)*(n-9))/7680, if n is odd. (End)
From Herbert Kociemba, Nov 04 2016: (Start)
G.f.: (1/20)*x^10*(1/(-1+x)^10 + 10/((-1+x)^6*(1+x)^5) + 1/(1-x^2)^5 + 4/(-1+x^5)^2 - 4/(-1+x^10)).
G.f.: k=10, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. [edited by Petros Hadjicostas, Jan 10 2019] (End)

Extensions

Sequence extended and description corrected by Christian G. Bower
Name edited by Petros Hadjicostas, Jan 10 2019

A005516 Number of n-bead bracelets (turnover necklaces) with 12 red beads.

Original entry on oeis.org

1, 1, 7, 19, 72, 196, 561, 1368, 3260, 7105, 14938, 29624, 56822, 104468, 186616, 322786, 544802, 896259, 1444147, 2278640, 3532144, 5380034, 8070400, 11926928, 17393969, 25042836, 35638596, 50152013, 69855536
Offset: 12

Views

Author

Keywords

Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent (turnover) necklaces of 12 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=12 (see our comment to A032279). (End)

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=12 of A052307.

Programs

  • Mathematica
    k = 12; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    k=12;CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[k/2+1])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

Let s(n,k,d) = 1, if n==k (mod d), s(n,k,d) = 0, otherwise. Then a(n) = s(n,0,12)/6 + (n-6)*s(n,0,6)/72 + (n-4)*(n-8)*s(n,0,4)/384 + (n-3)*(n-6)*(n-9)*s(n,0,3)/1944 + (3840*C(n-1,11) + (n+1)*(n-2)*(n-4)*(n-6)*(n-8)*(n-10))/92160, if n is even; a(n) = (n-3)*(n-6)*(n-9)*s(n,0,3)/1944 + (3840*C(n-1,11) + (n-1)*(n-3)*(n-5)*(n-7)*(n-9)*(n-11))/92160, if n is odd. - Vladimir Shevelev, Apr 23 2011
From Herbert Kociemba, Nov 04 2016: (Start)
G.f.: 1/2*x^12*((1+x)/(1-x^2)^7 + 1/12*(1/(-1+x)^12 + 1/(-1+x^2)^6 + 2/(-1+x^3)^4 - 2/(-1+x^4)^3 + 2/(-1+x^6)^2 - 4/(-1+x^12))).
G.f.: k=12, x^k*((1/k)*(Sum_{d|k} phi(d)*(1 - x^d)^(-k/d)) + (1 + x)/(1 -x^2)^floor((k+2)/2))/2. (End)

Extensions

Sequence extended and description corrected by Christian G. Bower

A321306 The number of connected weighted cubic graphs with weight n on 6 vertices.

Original entry on oeis.org

2, 2, 7, 12, 26, 41, 76, 113, 183, 264, 393, 543, 768, 1024, 1385, 1801, 2355, 2989, 3811, 4740, 5911, 7234, 8857, 10680, 12883, 15336, 18254, 21496, 25293, 29491, 34361, 39713, 45860, 52598, 60260, 68627, 78079, 88354, 99882, 112385, 126316, 141379, 158082, 176080
Offset: 6

Views

Author

R. J. Mathar, Nov 03 2018

Keywords

Comments

Each vertex of the 2 simple cubic graphs is assigned an integer number (weight) >=1. The weight of the graph is the sum of the weights of the vertices.
The cycle indices of the permutation group of vertex permutations of the two cubic graphs on 6 vertices are ( +t[1]^6 +3*t[1]^2*t[2]^2 +2*t[3]^2 +4*t[2]^3 +2*t[6])/12 and +( +t[1]^6 +6*t[1]^4*t[2] +9*t[1]^2*t[2]^2 +4*t[1]^3*t[3] +12*t[1]*t[2]*t[3] +6*t[2]^3 +18*t[2]*t[4] +12*t[6] +4*t[3]^2)/72 . The ordinary generating function of the sequence is obtained by adding the two cycle indices and setting t[i] -> x^i/(1-x^i).

Examples

			a(6)=2 because there are 2 cubic graphs (see A002851), and if the weight is the same as the number of vertices, there is one case for each.
		

Crossrefs

Cf. A026810 (4 vertices), A321307 (8 vertices), A005513.

Formula

G.f.: (x^10 +3*x^8 -x^7 +4*x^6 +4*x^4 +3*x^2 -2*x+2) *x^6/((-1+x)^6 *(1+x)^3 *(1+x^2) *(x^2+x+1)^2 *(x^2-x+1)).

Extensions

Terms a(36) and beyond from Andrew Howroyd, Apr 27 2020

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)
Showing 1-10 of 14 results. Next