A179317 -id:A179317 - cp's OEIS frontend

A179519 'AP(n,k)' triangle read by rows. AP(n,k) is the number of aperiodic k-palindromes of n.

1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 0, 1, 2, 1, 0, 1, 0, 3, 0, 3, 0, 0, 1, 0, 3, 2, 3, 2, 1, 0, 1, 0, 3, 0, 6, 0, 4, 0, 0, 1, 0, 4, 4, 5, 4, 4, 4, 1, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 0, 1, 0, 4, 4, 10, 8, 10, 8, 4, 4, 1, 0

Offset: 1

John P. McSorley, Jul 17 2010

A k-composition of n is an ordered collection of k positive integers (parts) which sum to n.
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 AP(n,k) denote the number of aperiodic k-palindromes of n.
This sequence is the 'AP(n,k)' triangle read by rows.
The g.f. of this triangular array follows easily from A. Howroyd's formula for this sequence and P. Deleham's g.f. for sequence A051159. If T(n,k) = A051159(n,k), then g.f. = Sum_{n,k>=1} AP(n,k)*x^n*y^k = Sum_{n,k>=1} Sum_{d|gcd(n,k)} mu(d)*T(n/d-1,k/d-1)*x^n*y^k. Letting m = n/d and s = k/d, we get g.f. = Sum_{d>=1} mu(d)*Sum_{m,s>=1} T(m-1,s-1)*(x^d)^m*(y^d)^s. But P. Deleham's formula for sequence A051159 implies Sum_{m,s>=1} T(m-1,s-1)*x^m*y^s = x*y*(1+x+x*y)/(1-x^2-x^2*y^2). Thus, Sum_{n,k>=1} AP(n,k)*x^n*y^k = Sum_{d>=1} mu(d)*f(x^d,y^d), where f(x,y) = x*y*(1+x+x*y)/(1-x^2-x^2*y^2). - Petros Hadjicostas, Nov 04 2017

			The triangle begins
  1
  1,0
  1,0,0
  1,0,1,0
  1,0,2,0,0
  1,0,1,2,1,0
  1,0,3,0,3,0,0
  1,0,3,2,3,2,1,0
  1,0,3,0,6,0,4,0,0
  1,0,4,4,5,4,4,4,1,0
For example, row 8 is 1,0,3,2,3,2,1,0.
We have AP(8,3)=3 because there are 3 aperiodic 3-palindromes of 8, namely: 161, 242, and 323.
We have AP(8,4)=2 because there are 2 aperiodic 4-palindromes of 8, namely: 3113 and 1331.

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

If we count the aperiodic k-palindromes of n up to cyclic equivalence, APE(n, k), we get sequence A179317.

The row sums of this triangle give sequence A179781. - John P. McSorley, Jul 26 2010

T[n_, k_] := 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, Oct 30 2017, after Andrew Howroyd *)

\\ 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) = 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

T(n,k) = Sum_{d|gcd(n,k)} mu(d) * A051159(n/d-1, k/d-1). - Andrew Howroyd, Oct 07 2017

G.f.: Sum_{n>=1} mu(n)*f(x^n,y^n), where f(x,y) = x*y*(1+x+x*y)/(1-x^2-x^2*y^2). - Petros Hadjicostas, Nov 04 2017

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

A056513 Number of primitive (period n) periodic palindromic structures using a maximum of two different symbols.

Original entry on oeis.org

1, 1, 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: 0

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.

Number of Lyndon compositions (aperiodic necklaces of positive integers) summing to n that can be rotated to form a palindrome. - Gus Wiseman, Sep 16 2018

			From _Gus Wiseman_, Sep 16 2018: (Start)
The sequence of palindromic Lyndon compositions begins:
  (1)  (2)  (3)  (4)    (5)    (6)      (7)
                 (112)  (113)  (114)    (115)
                        (122)  (1122)   (133)
                               (11112)  (223)
                                        (11113)
                                        (11212)
                                        (11122)
(End)

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 = 0..1000

Row sums of A179317.

Cf. A008965, A025065, A056476, A059966, A242414, A285037, A317085, A317086, A317087, A318731.

(* b = A164090, c = A045674 *)
b[n_] := (1/4)*(7 - (-1)^n)*2^((1/4)*(2*n + (-1)^n - 1));
c[0] = 1;
c[n_] := c[n] = If[EvenQ[n], 2^(n/2 - 1) + c[n/2], 2^((n - 1)/2)];
a56503[n_] := If[OddQ[n], b[n]/2, (1/2)*(b[n] + c[n/2])];
a[n_] := DivisorSum[n, MoebiusMu[#] a56503[n/#]&];
Array[a, 45] (* Jean-François Alcover, Jun 29 2018, after Andrew Howroyd *)

a(n) = {if(n < 1, n==0, sumdiv(n, d, moebius(d)*(2 + d%2)*(2^(n/d\2)))/(4 - n%2))} \\ Andrew Howroyd, Sep 26 2019

seq(n) = Vec(1 + (1/2)*sum(k=1, n, moebius(k)*x^k*(2 + 3*x^k)/(1 - 2*x^(2*k)) - moebius(2*k)*x^(2*k)*(1 + x^(2*k))/(1 - 2*x^(4*k)) + O(x*x^n))) \\ Andrew Howroyd, Sep 27 2019

a(n) = Sum_{d|n} mu(d)*A056503(n/d) for n > 0.
a(n) = Sum_{k=1..2} A285037(n, k). - Andrew Howroyd, Apr 08 2017
G.f.: 1 + (1/2)*Sum_{k>=1} mu(k)*x^k*(2 + 3*x^k)/(1 - 2*x^(2*k)) - mu(2*k)*x^(2*k)*(1 + x^(2*k))/(1 - 2*x^(4*k)). - Andrew Howroyd, Sep 27 2019

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

a(0)=1 prepended by Andrew Howroyd, Sep 27 2019

A179181 'PE(n,k)' triangle read by rows. PE(n,k) is the number of k-palindromes of n up to cyclic equivalence.

Original entry on oeis.org

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

Offset: 1

John P. McSorley, Jun 30 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-palindrome of n is a k-composition of n which is a palindrome.
Let PE(n,k) denote the number of k-palindromes of n up to cyclic equivalence.
This sequence is the 'PE(n,k)' triangle read by rows.

			The triangle begins
  1
  1,1
  1,0,1
  1,1,1,1
  1,0,2,0,1
  1,1,2,1,1,1
  1,0,3,0,3,0,1
  1,1,3,2,3,2,1,1
  1,0,4,0,6,0,4,0,1
  1,1,4,2,6,4,4,2,1,1
For example, row 8 is 1,1,3,2,3,2,1,1.
We have PE(8,3)=3 because there are 3 3-palindromes of 8, namely: 161, 242, and 323, and none are cyclically equivalent to the others.
We have PE(8,4)=2 because there are 3 4-palindromes of 8, namely: 3113, 1331, and 2222, but 3113 and 1331 are cyclically equivalent.

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

Andrew Howroyd, Table of n, a(n) for n = 1..1275

If we ignore cyclic equivalence then we have sequence A051159.
The row sums of the 'PE(n, k)' triangle give sequence A056503.
Cf. A179317, A179519.

T[n_, k_] := (3 - (-1)^k)/4*Sum[MoebiusMu[d]*QBinomial[n/d - 1, k/d - 1, -1], {d, Divisors@ GCD[n, k]}]; Table[DivisorSum[GCD[n, k], T[n/#, k/#] &], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Oct 31 2021, after Jean-François Alcover at A179317 *)

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

PE(n, k) = Sum_{d | gcd(n,k)} A179317(n/d, k/d). - Andrew Howroyd, Oct 07 2017

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

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A179519 'AP(n,k)' triangle read by rows. AP(n,k) is the number of aperiodic k-palindromes of n.

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A056513 Number of primitive (period n) periodic palindromic structures using a maximum of two different symbols.

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A179181 'PE(n,k)' triangle read by rows. PE(n,k) is the number of k-palindromes of n up to cyclic equivalence.

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Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions