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.
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, 64, 43, 10, 1, 1, 31, 90, 65, 15, 1, 1, 50, 208, 220, 85, 13, 1, 1, 63, 301, 350, 140, 21, 1, 1, 99, 656, 1059, 618, 154, 17, 1, 1, 127, 966, 1701, 1050, 266, 28, 1
Offset: 1
Examples
Triangle starts: 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 64 43 10 1 1 31 90 65 15 1 1 50 208 220 85 13 1 1 63 301 350 140 21 1 1 99 656 1059 618 154 17 1 1 127 966 1701 1050 266 28 1 1 197 2035 4803 4065 1488 258 21 1 1 255 3025 7770 6951 2646 462 36 1 1 391 6250 21105 24915 12857 3222 410 26 1 1 511 9330 34105 42525 22827 5880 750 45 1 ... Example for n=6, k=2: Periodic symmetry means results are either in the form abccba or abcdcb. 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. 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. There are therefore a total of 7 + 3 - 4 = 6 equivalence classes so T(6,2) = 6.
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
Crossrefs
Programs
-
PARI
\\ Ach is A304972, Prim is A285037. 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} 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) )))} T(n,k=n\2+1) = {my(A=Prim(n,k)); Mat(vectorv(n, n, sumdiv(n, d, A[d, ])))} { my(A=T(20)); for(n=1, matsize(A)[1], print(A[n,1..n\2+1])) } \\ Andrew Howroyd, Oct 02 2019
-
PARI
\\ column sequence using above code. ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ Andrew Howroyd, Oct 02 2019
Formula
Column k is inverse Moebius transform of column k of A285037. - Andrew Howroyd, Oct 02 2019
Comments