A293595 - OEIS

A293595

Triangle read by rows: T(n,k) = number of compositions of n into k parts such that no two cyclically adjacent parts are equal.

1, 1, 0, 1, 2, 0, 1, 2, 0, 0, 1, 4, 0, 0, 0, 1, 4, 6, 2, 0, 0, 1, 6, 6, 4, 0, 0, 0, 1, 6, 12, 10, 0, 0, 0, 0, 1, 8, 18, 16, 10, 2, 0, 0, 0, 1, 8, 24, 40, 20, 6, 0, 0, 0, 0, 1, 10, 30, 52, 50, 18, 0, 0, 0, 0, 0, 1, 10, 42, 84, 90, 50, 14, 2, 0, 0, 0, 0

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OFFSET

1,5

COMMENTS

Compositions of length 1 are included.

See theorem 4 in Hadjicostas reference for generating function.

LINKS

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

P. Hadjicostas, Cyclic, dihedral and symmetric Carlitz compositions of a positive integer, Journal of Integer Sequences, 20 (2017), Article 17.8.5.

FORMULA

G.f.: (Sum_{j>=1} x^(2*j)*y^2/(1+x^j*y)) + (Sum_{j>=1} x^j*y/(1+x^j*y)^2) / (1 - Sum_{j>=1} x^j*y/(1+x^j*y)).

EXAMPLE

Triangle begins:

1, 0;

1, 2, 0;

1, 2, 0, 0;

1, 4, 0, 0, 0;

1, 4, 6, 2, 0, 0;

1, 6, 6, 4, 0, 0, 0;

1, 6, 12, 10, 0, 0, 0, 0;

1, 8, 18, 16, 10, 2, 0, 0, 0;

1, 8, 24, 40, 20, 6, 0, 0, 0, 0;

...

Case n=6:

The included compositions are:

k=1: 6; => T(6,1) = 1

k=2: 15, 24, 42, 51; => T(6,2) = 4

k=3: 123, 132, 213, 231, 312, 321; => T(6,3) = 6

k=4: 1212, 2121; => T(6,4) = 2

MATHEMATICA

max = 10; gf = Sum[x^(2*j)*y^2/(1 + x^j*y), {j, 1, max}] + Sum[x^j*y/(1 + x^j*y)^2, {j, 1, max}]/(1 - Sum[ x^j*y/(1 + x^j*y), {j, 1, max}]) + O[x]^(max+1) + O[y]^(max+1) // Normal // Expand;

T[n_, k_] := SeriesCoefficient[gf, {x, 0, n}, {y, 0, k}];

Table[T[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 19 2018 *)

PROG

(PARI)

gf(n, y) = {my(A=sum(j=1, n, x^(2*j)*y^2/(1+x^j*y) + O(x*x^n)),

B=sum(j=1, n, x^j*y/(1+x^j*y)^2 + O(x*x^n)),

C=sum(j=1, n, x^j*y/(1+x^j*y) + O(x*x^n)));

A + B/(1-C)}

for(n=1, 10, my(p=polcoeff(gf(n, y), n)); for(k=1, n, print1(polcoeff(p, k), ", ")); print)

CROSSREFS

Row sums are in A212322.

Cf. A106351, A106369.