cp's OEIS Frontend

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

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A189980 a(n) is the number of incongruent two-color bracelets of n beads, 10 from them are black (A005515), having a diameter of symmetry.

Original entry on oeis.org

1, 1, 6, 6, 21, 21, 56, 56, 126, 126, 252, 252, 462, 462, 792, 792, 1287, 1287, 2002, 2002, 3003, 3003, 4368, 4368, 6188, 6188, 8568, 8568, 11628, 11628, 15504, 15504, 20349, 20349, 26334, 26334, 33649, 33649
Offset: 10

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Author

Vladimir Shevelev, May 03 2011

Keywords

Comments

For n >= 11, a(n-1) is the number of incongruent two-color bracelets of n beads, 11 from them are black (A032282), having a diameter of symmetry.

Crossrefs

Programs

Formula

a(n) = binomial(floor(n/2), 5). [Typo fixed by Colin Barker, Feb 07 2013]
a(n+6) = A194005(n, n-5). - Johannes W. Meijer, Aug 15 2011
G.f.: x^10/((x-1)^6*(x+1)^5). - Colin Barker, Feb 07 2013

Extensions

Data added and link corrected by Johannes W. Meijer, Aug 15 2011

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

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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

A143654 Array T(n,k) read by rows: number of binary bracelets with n beads, k of them 0, with 00 prohibited, (n >= 2, 0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 4, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 7, 8, 3, 1, 1, 1, 5, 8, 10, 5, 1, 1, 1, 6, 10, 16, 10, 4, 1, 1, 1, 6, 12, 20, 16, 7, 1, 1, 1, 7, 14, 29, 26, 16, 4, 1, 1, 1, 7, 16, 35, 38, 26, 8, 1, 1, 1, 8, 19, 47, 57, 50
Offset: 2

Views

Author

Washington Bomfim, Aug 28 2008

Keywords

Comments

The initial columns give A057427, A057427, A004526, A069905, A005232, A032279, A005513, A032280, A005514, A032281, A005515, A032282, A005516. Row sums give A129526.
A binary bracelet with n beads, k of them 0, with 00 prohibited has from 0 to floor(n/2) beads 0, i.e., 0 <= k <= floor(n/2). If n is even, the bracelet 0101...01 with n/2 beads of each kind does not have 00 and we cannot change any 1 of it to a 0. If n is odd we cannot change a 1 to a 0 in the bracelet 0101...011 with (n-1)/2 beads 0.
The number of binary bracelets with n beads, 0 <= k <= floor(n/2) of them 0 with 00 prohibited, is equal to the number of binary bracelets with n-k beads, k of them 0. See below.
Let B be a binary bracelet with n-k beads, k of them 0. If we insert one 1 (circularly) after a 0 of B, we obtain a bracelet with n-k+1 beads, k of them 0.
If we do this insertion k times, each time after a distinct 0 of B, we obtain a bracelet with n = n-k+k beads, k of them 0, with 00 prohibited.
On the contrary, Let B be a binary bracelet with n beads, k of them 0, with 00 prohibited. If we remove from B one 1 that is after a 0, we obtain a bracelet of n-1 beads, k of them 0. (If not and we undo the removal, the configuration obtained cannot be a bracelet and this is absurd.) If we repeat this removal k times, after each distinct bead 0, we obtain a bracelet with n-k beads, k of them 0.

Examples

			Array begins
1 1
1 1
1 1 1
1 1 1
1 1 2 1
1 1 2 1
1 1 3 2 1
1 1 3 3 1
1 1 4 4 3 1
...
A129526(10) = A057427(10) + A057427(9) + A004526(8) + A069905(7) + A005232(6) +
A032279(5) = 1+1+4+4+3+1 = 14.
		

Crossrefs

Showing 1-3 of 3 results.