A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.
1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 6, 10, 10, 5, 1, 1, 8, 21, 20, 15, 6, 1, 1, 13, 39, 55, 35, 21, 7, 1, 1, 18, 92, 136, 120, 56, 28, 8, 1, 1, 30, 198, 430, 377, 231, 84, 36, 9, 1, 1, 46, 498, 1300, 1505, 888, 406, 120, 45, 10, 1
Offset: 0
Examples
Table begins with T[0,1]: 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 9 10 1 3 6 10 15 21 28 36 45 55 1 4 10 20 35 56 84 120 165 220 1 6 21 55 120 231 406 666 1035 1540 1 8 39 136 377 888 1855 3536 6273 10504 1 13 92 430 1505 4291 10528 23052 46185 86185 1 18 198 1300 5895 20646 60028 151848 344925 719290 1 30 498 4435 25395 107331 365260 1058058 2707245 6278140 1 46 1219 15084 110085 563786 2250311 7472984 21552969 55605670 1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022
Links
- C. G. Bower, Transforms (2)
Crossrefs
Programs
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Mathematica
b[n_, k_] := DivisorSum[n, EulerPhi[#]*k^(n/#) &] / n; c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2)]; T[0, ] = 1; T[n, k_] := (b[n, k] + c[n, k])/2; Table[T[n, k-n], {k, 1, 11}, {n, k-1, 0, -1}] // Flatten (* Robert A. Russell, Sep 21 2018 after Jean-François Alcover *)
Formula
T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d divides n} phi(d) * k^(n/d). - Robert A. Russell, Sep 21 2018
G.f. for column k: (kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. - Robert A. Russell, Sep 28 2018
T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i). (See A075195 formulas.) - Richard L. Ollerton, May 04 2021
Comments