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

1,7

Comments

For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.

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.

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

Column k is inverse Moebius transform of column k of A285037. - Andrew Howroyd, Oct 02 2019

Example

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.

Prog

(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

Crossrefs

Columns 2..6 are A056508, A056509, A056510, A056511, A056512.

Partial row sums include A056503, A056504, A056505, A056506, A056507.

Row sums are A285013.

Cf. A284855, A284826, A284823, A284877, A285037, A304972.

Sequence in context: A139002 A373438 A251913 * A139378 A038500 A091840

Adjacent sequences: A285009 A285010 A285011 * A285013 A285014 A285015

Keyword

nonn,tabf

Author

Andrew Howroyd, Apr 07 2017

Status

approved