A056513 -id:A056513 - cp's OEIS frontend

A285037 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.

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

Offset: 1

Andrew Howroyd, Apr 08 2017

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.

			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.

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]

Andrew Howroyd, Table of n, a(n) for n = 1..2600

Columns 1..6 are: A063524, A056518, A056519, A056521, A056522, A056523.
Partial row sums include A056513, A056514, A056515, A056516, A056517.
Row sums are A285042.
Cf. A284856, A284826, A284823, A285012, A304972.

\\ 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

\\ column sequence using above code.
ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ Andrew Howroyd, Oct 01 2019

T(n, k) = Sum_{d | n} mu(n/d) * A285012(d, k).

A056519 Number of primitive (period n) periodic palindromic structures using exactly three different symbols.

Original entry on oeis.org

0, 0, 0, 1, 1, 5, 6, 18, 25, 63, 90, 202, 301, 650, 965, 2016, 3025, 6220, 9330, 18970, 28495, 57650, 86526, 174210, 261624, 525693, 788945, 1582462, 2375101, 4759482, 7141686, 14300976, 21457735, 42951825, 64439003, 128946888, 193448101, 387037370, 580606145, 1161485370

Offset: 1

Marks R. Nester

nonn

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

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]

Andrew Howroyd, Table of n, a(n) for n = 1..200

Column 3 of A285037.

Cf. A056482, A056509, A056513, A056514.

a(n) = A056514(n) - A056513(n).

Moebius transform of A056509. - Andrew Howroyd, Oct 01 2019

a(17)-a(35) from Andrew Howroyd, Apr 08 2017

Terms a(36) and beyond from Andrew Howroyd, Oct 01 2019

A179317 'APE(n,k)' triangle read by rows. APE(n,k) is the number of aperiodic k-palindromes of n up to cyclic equivalence.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 3, 0, 3, 0, 0, 1, 0, 3, 1, 3, 1, 1, 0, 1, 0, 3, 0, 6, 0, 4, 0, 0, 1, 0, 4, 2, 5, 2, 4, 2, 1, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 0, 1, 0, 4, 2, 10, 4, 10, 4, 4, 2, 1, 0

Offset: 1

John P. McSorley, Jul 10 2010

A k-composition of n is an ordered collection of k positive integers (parts) which sum to n.
Two k-compositions of n are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
A k-composition is aperiodic (primitive) if its period is k, or if it is not the concatenation of a smaller composition.
A k-palindrome of n is a k-composition of n which is a palindrome.
Let APE(n,k) denote the number of aperiodic k-palindromes of n up to cyclic equivalence.
This sequence is the 'APE(n,k)' triangle read by rows.
The only possibility for two distinct aperiodic palindromes to be cyclically equivalent is with an even number of terms and with a rotation by half the number of terms. For example, 123321 is cyclically equivalent to 321123. - Andrew Howroyd, Oct 07 2017

			The triangle begins
1
1,0
1,0,0
1,0,1,0
1,0,2,0,0
1,0,1,1,1,0
1,0,3,0,3,0,0
1,0,3,1,3,1,1,0
1,0,3,0,6,0,4,0,0
1,0,4,2,5,2,4,2,1,0
For example, row 8 is 1,0,3,1,3,1,1,0.
We have APE(8,3)=3 because there are 3 aperiodic 3-palindromes of 8, namely: 161, 242, and 323, and none are cyclically equivalent to the others.
We have APE(8,4)=1 because there are 2 aperiodic 4-palindromes of 8, namely: 3113 and 1331, but they are cyclically equivalent.

John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (Rows n=1..50 of triangle, flattened)

The row sums of the 'APE(n, k)' triangle give sequence A056513.

If cyclic equivalence is ignored, we get sequence A179519. - John P. McSorley, Jul 26 2010

T[n_, k_] := (3-(-1)^k)/4*Sum[MoebiusMu[d]*QBinomial[n/d - 1, k/d - 1, -1], {d, Divisors[GCD[n, k]]}];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 24 2019 *)

\\ here p(n, k)=A051159(n-1, k-1) is number of k-palindromes of n.
p(n, k) = if(n%2==1&&k%2==0, 0, binomial((n-1)\2, (k-1)\2));
T(n, k) = if(k%2,1,1/2) * sumdiv(gcd(n,k), d, moebius(d) * p(n/d, k/d));
for(n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print) \\ Andrew Howroyd, Oct 07 2017

APE(n,k) = (3-(-1)^k)/4 * A179519(n,k). - Andrew Howroyd, Oct 07 2017

Terms a(56) and beyond from Andrew Howroyd, Oct 07 2017

A056518 Number of primitive (period n) periodic palindromic structures using exactly two different symbols.

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 7, 10, 14, 21, 31, 42, 63, 91, 123, 184, 255, 371, 511, 750, 1015, 1519, 2047, 3030, 4092, 6111, 8176, 12222, 16383, 24486, 32767, 49024, 65503, 98175, 131061, 196308, 262143, 392959, 524223, 785910, 1048575, 1572256, 2097151, 3144702, 4194162

Offset: 1

Marks R. Nester

nonn

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

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]

Column 2 of A285037.

Cf. A056481.

A056513(n)-A000007(n-1).

a(17)-a(45) from Andrew Howroyd, Apr 08 2017

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A285037 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.

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A056519 Number of primitive (period n) periodic palindromic structures using exactly three different symbols.

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A179317 'APE(n,k)' triangle read by rows. APE(n,k) is the number of aperiodic k-palindromes of n up to cyclic equivalence.

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A056518 Number of primitive (period n) periodic palindromic structures using exactly two different symbols.

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