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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. 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,7 LINKS Washington Bomfim, Rows 1..25, flattened Harold S. Grant, On a Formula for Circular Permutations, Mathematics Magazine, Vol. 23, No. 3 (Jan. - Feb., 1950), pp. 133-136. Hiroshi Kajimoto and Mai Osabe, Circular and Necklace Permutations, Bulletin of the Faculty of Education, Nagasaki University. Natural Sciences 2006; v.74, 1-14. S. Karim, J. Sawada, Z. Alamgirz, and S. M. Husnine, Generating bracelets with fixed content, (2011). S. Karim, J. Sawada, Z. Alamgirz, and S. M. Husnine, Generating bracelets with fixed content, Theoretical Computer Science, Volume 475, 4 March 2013, Pages 103-112. Frank Ruskey, Combinatorial Generation Algorithm Algorithm 4.24, p. 95. 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! * ...). EXAMPLE 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) => . PROG (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)); GCD = gcd(P, P); 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!; 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 CROSSREFS Cf. A000041 (row lengths), A213939 (another version with partitions found in a different order), A005654, A005656, A141783, A000010. Sequence in context: A238402 A016558 A154395 * A152159 A090341 A251092 Adjacent sequences:  A214606 A214607 A214608 * A214610 A214611 A214612 KEYWORD nonn,tabf AUTHOR Washington Bomfim, Jul 22 2012 STATUS approved

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Last modified September 20 05:51 EDT 2019. Contains 327212 sequences. (Running on oeis4.)