A308678 Kuba-Panholzer Table 2 pattern 231, 132 for Stirling permutation k = 2.

0, 4, 0, 5, 26, 135, 685, 3453, 17379, 87503, 441101, 2226900, 11260144, 57023761, 289204167, 1468757683, 7468800180, 38024694279, 193801106977, 988750469868, 5049220893012, 25807115270500, 132009416652052, 675765729570936, 3461711974830844, 17744712728166057

Offset

0,2

Links

Table of n, a(n) for n=0..25.

Markus Kuba, Alois Panholzer, Stirling permutations containing a single pattern of length three, Australasian Journal of Combinatorics (2019) Vol. 74, No. 2, 216-239.

Formula

For k = 2, a(n) = Sum_{j = 0..(n - 1)} (binomial(n - 1, j) * (k^2 * binomial(n + j (k - 1) + k - 4, n - j - 4) + (k - 1) * binomial(n + j (k - 1) + k - 3, n - j - 2) + binomial(n + j (k - 1) + k - 2, n - j - 1) - (2 k - 1) * binomial(n + j (k - 1) - 3, n - j - 2) - binomial(n + j (k - 1) - 2, n - j - 1)))

Mathematica

With[{k = 2}, Table[Sum[Binomial[n - 1, j] (k^2*Binomial[n + j (k - 1) + k - 4, n - j - 4] + (k - 1) Binomial[n + j (k - 1) + k - 3, n - j - 2] + Binomial[n + j (k - 1) + k - 2, n - j - 1] - (2 k - 1) Binomial[n + j (k - 1) - 3, n - j - 2] - Binomial[n + j (k - 1) - 2, n - j - 1]), {j, 0, n - 1}], {n, 0, 25}]]

Crossrefs

Cf. A001700, A003517.

Sequence in context: A108174 A134530 A351571 * A184365 A126813 A056141

Adjacent sequences: A308675 A308676 A308677 * A308679 A308680 A308681

Keyword

nonn,easy

Author

Michael De Vlieger, Jun 16 2019

Status

approved