OFFSET
1,7
COMMENTS
Permuting the symbols will not change the structure.
Equivalently, the number of n-bead aperiodic necklaces (Lyndon words) with exactly k symbols, up to permutation of the symbols, which when turned over are unchanged. When comparing with the turned over necklace a rotation is allowed but a permutation of the symbols is not.
REFERENCES
M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..2600
FORMULA
T(n, k) = Sum_{d | n} mu(n/d) * A285012(d, k).
EXAMPLE
Triangle starts:
1
0 1
0 1
0 2 1
0 3 1
0 4 5 1
0 7 6 1
0 10 18 7 1
0 14 25 10 1
0 21 63 43 10 1
0 31 90 65 15 1
0 42 202 219 85 13 1
0 63 301 350 140 21 1
0 91 650 1058 618 154 17 1
0 123 965 1701 1050 266 28 1
0 184 2016 4796 4064 1488 258 21 1
0 255 3025 7770 6951 2646 462 36 1
0 371 6220 21094 24914 12857 3222 410 26 1
0 511 9330 34105 42525 22827 5880 750 45 1
...
Example for n=6, k=2:
There are 6 inequivalent solutions to A285012(6,2) which are 001100, 010010, 000100, 001010, 001110, 010101. Of these, 010010 and 010101 have a period less than 6, so T(6,2) = 6-2 = 4.
PROG
(PARI) \\ Ach is A304972
Ach(n, k=n) = {my(M=matrix(n, k, n, k, n>=k)); for(n=3, n, for(k=2, k, M[n, k]=k*M[n-2, k] + M[n-2, k-1] + if(k>2, M[n-2, k-2]))); M}
T(n, k=n\2+1) = {my(A=Ach(n\2+1, k), S=matrix(n\2+1, k, n, k, stirling(n, k, 2))); Mat(vectorv(n, n, sumdiv(n, d, moebius(d)*(S[(n/d+1)\2, ] + S[n/d\2+1, ] + if((n-d)%2, A[(n/d+1)\2, ] + A[n/d\2+1, ]))/if(d%2, 2, 1) )))}
{ my(A=T(20)); for(n=1, matsize(A)[1], print(A[n, 1..n\2+1])) } \\ Andrew Howroyd, Oct 01 2019
(PARI) \\ column sequence using above code.
ColSeq(n, k=2) = { Vec(T(n, k)[, k]) } \\ Andrew Howroyd, Oct 01 2019
CROSSREFS
Row sums are A285042.
KEYWORD
nonn,tabf
AUTHOR
Andrew Howroyd, Apr 08 2017
STATUS
approved