A119963 - cp's OEIS frontend

A119963 Triangle T(n,k), 0 <= k <= n, read by rows, with T(2n,2k) = T(2n+1,2k) = T(2n+1,2k+1) = T(2n+2,2k+1) = binomial(n,k).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 2, 3, 1, 1, 1, 1, 3, 3, 3, 3, 1, 1, 1, 1, 4, 3, 6, 3, 4, 1, 1, 1, 1, 4, 4, 6, 6, 4, 4, 1, 1, 1, 1, 5, 4, 10, 6, 10, 4, 5, 1, 1, 1, 1, 5, 5, 10, 10, 10, 10, 5, 5, 1, 1, 1, 1, 6, 5, 15, 10, 20, 10, 15, 5, 6, 1, 1, 1, 1, 6, 6, 15, 15, 20, 20, 15, 15, 6, 6, 1, 1, 1, 1, 7, 6, 21, 15, 35, 20, 35, 15, 21, 6, 7, 1, 1

Offset: 0

Philippe Deléham, Aug 02 2006

From John P. McSorley, Aug 24 2010: (Start)
A combinatorial interpretation of this triangle is as follows:
Ignore the first column of 1's of the above triangle and the call the (n,k) entry of the new triangle formed RE(n,k).
Hence row 8 of the 'RE(n,k)' triangle is 1 4 3 6 3 4 1 1.
Now see sequence A180171 for the definition of a k-reverse of n.
Briefly, a k-reverse of n is a k-composition of n which is cyclically equivalent to its reverse.
Sequence A180171 is the 'R(n,k)' triangle read by rows where R(n,k) is the total number of k-reverses of n.
Then RE(n,k) is the number of k-reverses of n up to cyclic equivalence.
In sequence A180171 we have R(8,3)=9 because there are 9 3-reverses of 8.
In cyclically equivalent classes: {116,611,161} {224,422,242}, and {233,323,332}; since there are 3 such classes we have RE(8,3)=3.
Similarly, in A180171, we have R(8,6)=21 because all 21 6-compositions of 8 are 6-reverses of 8, but they come in 4 cyclically equivalent classes (with representatives 111113, 111122, 111212, and 112112) hence RE(8,6)=4.
There is another (equivalent) interpretation for RE(n,k) involving k-subsets of Z_n, the integers modulo n, and the multiplier -1. See the McSorley/Schoen paper below for more details.
In this case it is convenient to count k-subsets up to dihedral equivalence, rather than cyclic equivalence.
The counts are the same. The row sums of the 'RE(n,k)' triangle give sequence A052955.
(End)
From Petros Hadjicostas, Oct 12 2017: (Start)
When 1 <= k <= n, each cyclically equivalence class of k-reverses of n is a "Sommerville symmetrical cyclic composition," which was introduced by Sommerville (1909). On pp. 301-304 of his paper, he proves that the number of such (equivalence classes of) compositions of n with length k is exactly T(n,k) = RE(n,k).
The equivalence class of a Sommerville symmetrical cyclic composition contains at least one palindromic composition (type I) or a composition that becomes a palindromic composition if we remove the first part (type II). A composition with only one part is a palindromic composition of both types. Hadjicostas and Zhang (2017) have proved that each equivalence class of k-reverses of n contains exactly two compositions that are either of type I or type II (except in the case when k | n and all the parts are the same).
For example, consider the case with n=8 and k=3, where RE(8,3)=3. As pointed above by J. P. McSorley, in cyclically equivalent classes we have {116,611,161} {224,422,242}, and {233,323,332}. The first class contains one composition of type I (161) and one of type II (611); the second class contains one composition of type I (242) and one of type II (422); and the last class contains one composition of type I (323) and one of type II (233).
When n = 6 and k = 4, the class of 4-reverses {1221, 2211, 2112, 1122} contains two compositions of type I (1221 and 2112).
If A is a set of positive integers and 1 <= k <= n, let RE_A(n,k) be the total number of Sommerville symmetrical cyclic compositions of n with length k and parts only in A (= number of cyclically equivalence classes of k-reverses of n with parts only in A). Then the g.f. of RE_A(n,k) is Sum_{n,k >= 1} RE_A(n,k) * x^n * y^k = (-1/2) + (1 + y * f_A(x))^2/(2 * (1 - y^2 * f_A(x^2)), where f_A(x) = Sum_{m in A} x^m. (For this sequence, A = all positive integers.)
Sequence A292200 contains the total number of Sommerville symmetrical cyclic compositions of n that are Carlitz (compositions that have length one, or have length >= 1 and adjacent parts of the composition on a circle are distinct).
(End)

			Triangle begins (with rows for n >= 0 and columns for k >= 0) as follows:
  1;
  1, 1;
  1, 1, 1;
  1, 1, 1, 1;
  1, 1, 2, 1, 1;
  1, 1, 2, 2, 1, 1;
  1, 1, 3, 2, 3, 1, 1;
  1, 1, 3, 3, 3, 3, 1, 1;
  1, 1, 4, 3, 6, 3, 4, 1, 1;
  1, 1, 4, 4, 6, 6, 4, 4, 1, 1;
  ...

John P. McSorley, Counting k-compositions of n with palindromic and related structures, preprint, 2010. [From John P. McSorley, Aug 24 2010]

Andrew Howroyd, Table of n, a(n) for n = 0..1274
P. Hadjicostas and L. Zhang, Sommerville's symmetrical cyclic compositions of a positive integer with parts avoiding multiples of an integer, Fibonacci Quarterly 55 (2017), 54-73.
John P. McSorley and Alan H. Schoen, Rhombic tilings of (n,k)-Ovals, (n, k, lambda)-Cyclic Difference Sets, and Related Topics, Discrete Math., 313 (2013), 129-154. - From _N. J. A. Sloane_, Nov 26 2012
D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.

Cf. A007318, A292200, A051159.

The row sums of the T(n,k) triangle give sequence A029744 whose terms are 1 more than the terms of sequence A052955 (row sums of RE(n,k) triangle). See sequence A029744 where there is a reference to necklaces relevant to the combinatorial interpretation and the McSorley and McSorley/Schoen papers given here. - John P. McSorley, Aug 31 2010

Table[Binomial[Floor[(n - Boole[OddQ@ k])/2], Floor[k/2]], {n, 0, 10}, {k, 0, n}] (* Michael De Vlieger, Oct 11 2017, after PARI by Andrew Howroyd *)

T(n,k) = binomial((n-k%2)\2, k\2); \\ Andrew Howroyd, Oct 08 2017

G.f.: Sum_{n,k >= 1} RE(n,k)*x^n*y^k = (1+x*y-x^2)*x*y/((1-x)*(1-x^2-x^2*y^2)). - Petros Hadjicostas, Oct 12 2017
G.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = (1+x*y)*(1+x)/(1-x^2-x^2*y^2) as above, but adding 1/(1-x) to include n,k = 0 terms. - Paul Sampson, Nov 22 2017
T(n, k) = binomial(floor(n/2) - (k mod 2) * (1 - (n mod 2)), floor(k/2)) for 0 <= k <= n. - Petros Hadjicostas, May 29 2019

Corrected by Philippe Deléham, Aug 20 2010

cp's OEIS Frontend

A119963 Triangle T(n,k), 0 <= k <= n, read by rows, with T(2n,2k) = T(2n+1,2k) = T(2n+1,2k+1) = T(2n+2,2k+1) = binomial(n,k).

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