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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A214609 Table of numbers of distinct bracelets (reversible necklaces) with n beads corresponding to one partition P of n. Each part p of P corresponds to p beads of a distinct color.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 2, 2, 1, 1, 12, 6, 4, 2, 2, 1, 1, 60, 30, 16, 11, 10, 6, 3, 3, 3, 1, 1, 360, 180, 90, 48, 60, 30, 18, 10, 15, 9, 4, 3, 3, 1, 1, 2520, 1260, 630, 318, 171, 420, 210, 108, 70, 38, 105, 54, 33, 19, 8, 21, 12, 5, 4, 4, 1, 1, 20160, 10080, 5040, 2520, 1272, 3360, 1680, 840, 432, 560, 280, 94, 840, 420, 216, 140, 76, 38, 168, 84, 48, 28, 10, 28, 16, 7, 4, 4, 1, 1
Offset: 1

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Author

Washington Bomfim, Jul 22 2012

Keywords

Examples

			Table begins
   1
   1,1
   1,1,1
   3,2,2,1,1
  12,6,4,2,2,1,1
   ...
Line number 4 is 3,2,2,1,1 because three bracelets, (0 1 2 3), (0 1 3 2), and (0 2 1 3) correspond to partition [1 1 1 1]. (The colors are denoted by 0,1,2, and 3.) Bracelets (0 0 1 2), and (0 1 0 2) which have two beads of color 0, one of color 1, and one of color 2, correspond to [2 1 1]. (0 0 1 1), and (0 1 0 1) => [2 2]; (0 0 0 1) => [3 1], and (0 0 0 0) => [4].
From _Marko Riedel_, Jan 23 2025 (Start)
The ordering of the partitions used here is graded reflected lexicographic illustrated below with n=5:
  1,1,1,1,1 => 12
  1,1,1,2 => 6
  1,2,2 => 4
  1,1,3 => 2
  2,3 => 2
  1,4 => 1
  5 => 1 (End)
		

Crossrefs

Cf. A000041 (row lengths), A213939 (another version with partitions found in a different order), A005654, A005656, A141783, A000010, A380401.
Row sums give A213943.
Cf. A080576 (graded reflected lexicographic order).

Programs

  • PARI
    N; L = 0; max_n = 25;
    x = vector(max_n+1); P = vector(max_n+1);          \\ P - partition of n
    gcdP(t) = {if(t == 2, return(P[2])); GCD = gcd(P[2], P[3]);
    for(J = 4, t, GCD = gcd(GCD, P[J])); return(GCD) }
    x_P_div_d(t, d) = for(J = 2, t, x[J] = P[J]/d);
    F(a, t)= { b = x[2]!; for(J = 3, t, b *= x[J]!); return(a!/b) }
    Sum(t) = { S = 0; GCD = gcdP(t);
    fordiv(GCD, d, x_P_div_d(t,d); S+= eulerphi(d) * F(N/d, t)); return(S) }
    OneOdd(t) = {K = 0; for(J = 2, t, if(P[J] % 2 == 1, K++)); return(K==1)}
    TwoOdd(t) = {K = 0; for(J = 2, t, if(P[J] % 2 == 1, K++)); return(K==2)}
    x_floor_P_div_2(t) = for(J = 2, t, x[J] = floor(P[J]/2));
    all_even_parts(t) = { for(J = 2, t, if(P[J] % 2 == 1, return(0) ) ); return(1) }
    x_P_div_2(t) = for(J = 2, t, x[J] = P[J]/2);
    \\
    A(t) = {S = Sum(t); L++;
    if((N%2==1) && OneOdd(t), x_floor_P_div_2(t);
       print(L," ",(S + N * F((N-1)/2, t))/(2*N)); return() );
    if(N%2==1, print(L," ", S/(2*N)); return() );
    if(all_even_parts(t), x_P_div_2(t);
       print(L," ",(S + N * F(N/2, t))/(2*N)); return() );
    if((N%2==0) &&  TwoOdd(t), x_floor_P_div_2(t);
       print(L," ",(S + N * F((N-2)/2, t))/(2*N)); return() );
    print(L," ", S/(2*N)) }
                                                \\ F. Ruskey algorithm 4.24
    Part(n, k, t) = { P[t] = k;
    if(n == k, A(t), for(j = 1, min(k, n-k), Part(n-k, j, t+1) ) )}
    for(n = 1, max_n, N = n; Part(2*n, n, 1) );  \\ b-file format

Formula

With S = Sum_( d | GCD of the parts of P ) { phi(d) * F(n/d, P/d) },
| (S+n*F((n-1)/2, [P/2]))/(2*n) if odd n, and only 1 odd part in P,
| S/(2*n) if odd n, and other P,
| (S + n * F(n/2, P/2)) / (2*n) if P has all even parts,
a(n)=| (S + n * F((n-2)/2, [P/2])) / (2*n)
| if even n, and P has exactly two odd parts,
| S/(2*n) if even n, and other P.
Where P is a partition of n, P/d is a vector of all the parts of P divided by d. Each element of vector [P/2] is equal to floor(p/2), p one part of P, and F(x,Y) = x! / (Y_1! * Y_2! * ...).
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