A321791 Table read by descending antidiagonals: T(n,k) is the number of unoriented cycles (bracelets) of length n using up to k available colors.
1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 6, 1, 0, 1, 6, 15, 20, 21, 8, 1, 0, 1, 7, 21, 35, 55, 39, 13, 1, 0, 1, 8, 28, 56, 120, 136, 92, 18, 1, 0, 1, 9, 36, 84, 231, 377, 430, 198, 30, 1, 0
Offset: 0
Examples
Table begins with T(0,0): 1 1 1 1 1 1 1 1 1 1 1 ... 0 1 2 3 4 5 6 7 8 9 10 ... 0 1 3 6 10 15 21 28 36 45 55 ... 0 1 4 10 20 35 56 84 120 165 220 ... 0 1 6 21 55 120 231 406 666 1035 1540 ... 0 1 8 39 136 377 888 1855 3536 6273 10504 ... 0 1 13 92 430 1505 4291 10528 23052 46185 86185 ... 0 1 18 198 1300 5895 20646 60028 151848 344925 719290 ... 0 1 30 498 4435 25395 107331 365260 1058058 2707245 6278140 ... 0 1 46 1219 15084 110085 563786 2250311 7472984 21552969 55605670 ... 0 1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022 ... For T(3,3)=10, the unoriented cycles are 9 achiral (AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, CCC) and 1 chiral pair (ABC-ACB).
Crossrefs
Programs
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Mathematica
Table[If[k>0, DivisorSum[k, EulerPhi[#](n-k)^(k/#)&]/(2k) + ((n-k)^Floor[(k+1)/2]+(n-k)^Ceiling[(k+1)/2])/4, 1], {n, 0, 12}, {k, 0, n}] // Flatten
Formula
T(n,k) = [n==0] + [n>0] * (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d|n} phi(d) * k^(n/d).
Linear recurrence for row n: T(n,k) = Sum_{j=0..n} -binomial(j-n-1,j+1) * T(n,k-1-j) for k >= n + 1.
O.g.f. for column k >= 0: Sum_{n>=0} T(n,k)*x^n = 3/4 + (1 + k*x)^2/(4*(1 - k*x^2)) - (1/2) * Sum_{d >= 1} (phi(d)/d) * log(1 - k*x^d). - Petros Hadjicostas, Feb 07 2021