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%I #24 Nov 05 2019 13:52:55
%S 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,
%T 63,43,10,1,0,31,90,65,15,1,0,42,202,219,85,13,1,0,63,301,350,140,21,
%U 1,0,91,650,1058,618,154,17,1,0,123,965,1701,1050,266,28,1
%N Irregular triangle read by rows: T(n,k) is the number of primitive (period n) periodic palindromic structures using exactly k different symbols, 1 <= k <= n/2 + 1.
%C Permuting the symbols will not change the structure.
%C 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.
%D 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 Andrew Howroyd, <a href="/A285037/b285037.txt">Table of n, a(n) for n = 1..2600</a>
%F T(n, k) = Sum_{d | n} mu(n/d) * A285012(d, k).
%e Triangle starts:
%e 1
%e 0 1
%e 0 1
%e 0 2 1
%e 0 3 1
%e 0 4 5 1
%e 0 7 6 1
%e 0 10 18 7 1
%e 0 14 25 10 1
%e 0 21 63 43 10 1
%e 0 31 90 65 15 1
%e 0 42 202 219 85 13 1
%e 0 63 301 350 140 21 1
%e 0 91 650 1058 618 154 17 1
%e 0 123 965 1701 1050 266 28 1
%e 0 184 2016 4796 4064 1488 258 21 1
%e 0 255 3025 7770 6951 2646 462 36 1
%e 0 371 6220 21094 24914 12857 3222 410 26 1
%e 0 511 9330 34105 42525 22827 5880 750 45 1
%e ...
%e Example for n=6, k=2:
%e 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.
%o (PARI) \\ Ach is A304972
%o 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 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) )))}
%o { my(A=T(20)); for(n=1, matsize(A)[1], print(A[n,1..n\2+1])) } \\ _Andrew Howroyd_, Oct 01 2019
%o (PARI) \\ column sequence using above code.
%o ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ _Andrew Howroyd_, Oct 01 2019
%Y Columns 1..6 are: A063524, A056518, A056519, A056521, A056522, A056523.
%Y Partial row sums include A056513, A056514, A056515, A056516, A056517.
%Y Row sums are A285042.
%Y Cf. A284856, A284826, A284823, A285012, A304972.
%K nonn,tabf
%O 1,7
%A _Andrew Howroyd_, Apr 08 2017
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