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A285012 Irregular triangle read by rows: T(n,k) is the number of periodic palindromic structures of length n using exactly k different symbols, 1 <= k <= n/2 + 1.

%I A285012 #19 Oct 02 2019 17:46:20
%S A285012 1,1,1,1,1,1,3,1,1,3,1,1,6,5,1,1,7,6,1,1,13,19,7,1,1,15,25,10,1,1,25,
%T A285012 64,43,10,1,1,31,90,65,15,1,1,50,208,220,85,13,1,1,63,301,350,140,21,
%U A285012 1,1,99,656,1059,618,154,17,1,1,127,966,1701,1050,266,28,1
%N A285012 Irregular triangle read by rows: T(n,k) is the number of periodic palindromic structures of length n using exactly k different symbols, 1 <= k <= n/2 + 1.
%C A285012 For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.
%C A285012 Equivalently, the number of necklaces, 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.
%D A285012 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]
%H A285012 Andrew Howroyd, <a href="/A285012/b285012.txt">Table of n, a(n) for n = 1..2600</a>
%F A285012 Column k is inverse Moebius transform of column k of A285037. - _Andrew Howroyd_, Oct 02 2019
%e A285012 Triangle starts:
%e A285012 1
%e A285012 1   1
%e A285012 1   1
%e A285012 1   3    1
%e A285012 1   3    1
%e A285012 1   6    5     1
%e A285012 1   7    6     1
%e A285012 1  13   19     7     1
%e A285012 1  15   25    10     1
%e A285012 1  25   64    43    10     1
%e A285012 1  31   90    65    15     1
%e A285012 1  50  208   220    85    13    1
%e A285012 1  63  301   350   140    21    1
%e A285012 1  99  656  1059   618   154   17   1
%e A285012 1 127  966  1701  1050   266   28   1
%e A285012 1 197 2035  4803  4065  1488  258  21  1
%e A285012 1 255 3025  7770  6951  2646  462  36  1
%e A285012 1 391 6250 21105 24915 12857 3222 410 26 1
%e A285012 1 511 9330 34105 42525 22827 5880 750 45 1
%e A285012 ...
%e A285012 Example for n=6, k=2:
%e A285012 Periodic symmetry means results are either in the form abccba or abcdcb.
%e A285012 There are 3 binary words in the form abccba that start with 0 and contain a 1 which are 001100, 010010, 011110. Of these, 011110 is equivalent to 001100 after rotation.
%e A285012 There are 7 binary words in the form abcdcb that start with 0 and contain a 1 which are 000100, 001010, 001110, 010001, 010101, 011011, 011111. Of these, 011111 is equivalent to 000100, 010001 is equivalent to 001010 and 011011 is equivalent to 010010 from the first set.
%e A285012 There are therefore a total of 7 + 3 - 4 = 6 equivalence classes so T(6,2) = 6.
%o A285012 (PARI) \\ Ach is A304972, Prim is A285037.
%o A285012 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}
%o A285012 Prim(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) )))}
%o A285012 T(n,k=n\2+1) = {my(A=Prim(n,k)); Mat(vectorv(n, n, sumdiv(n, d, A[d, ])))}
%o A285012 { my(A=T(20)); for(n=1, matsize(A)[1], print(A[n,1..n\2+1])) } \\ _Andrew Howroyd_, Oct 02 2019
%o A285012 (PARI) \\ column sequence using above code.
%o A285012 ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ _Andrew Howroyd_, Oct 02 2019
%Y A285012 Columns 2..6 are A056508, A056509, A056510, A056511, A056512.
%Y A285012 Partial row sums include A056503, A056504, A056505, A056506, A056507.
%Y A285012 Row sums are A285013.
%Y A285012 Cf. A284855, A284826, A284823, A284877, A285037, A304972.
%K A285012 nonn,tabf
%O A285012 1,7
%A A285012 _Andrew Howroyd_, Apr 07 2017