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A299927 Number of permutations of length n that avoid the patterns 213 and 312 and have k double ascents, read by rows. 1

%I

%S 1,1,2,3,1,4,3,1,5,6,4,1,6,10,10,5,1,7,15,20,15,6,1,8,21,35,35,21,7,1,

%T 9,28,56,70,56,28,8,1,10,36,84,126,126,84,36,9,1,11,45,120,210,252,

%U 210,120,45,10,1,12,55,165,330,462,462,330,165,55,11,1

%N Number of permutations of length n that avoid the patterns 213 and 312 and have k double ascents, read by rows.

%C In a permutation avoiding 213 and 312, all digits before n are increasing and all digits after n are decreasing. If k=0, either n is the first digit or the second digit of the permutation. If k >= 1, there are binomial(n-1, k+1) ways to choose k+1 digits before n; these digits together with n account for k double ascents.

%C For n >= 1, the sum of row n is 2^(n-1).

%H Michael De Vlieger, <a href="/A299927/b299927.txt">Table of n, a(n) for n = 0..11176</a> (rows 0 <= n <= 150, flattened).

%H M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, <a href="https://arxiv.org/abs/1812.07112">Distributions of Statistics over Pattern-Avoiding Permutations</a>, arXiv preprint arXiv:1812.07112 [math.CO], 2018.

%H Paul M. Rakotomamonjy, Sandrataniaina R. Andriantsoa, Arthur Randrianarivony, <a href="https://arxiv.org/abs/1910.13809">Crossings over permutations avoiding some pairs of three length-patterns</a>, arXiv:1910.13809 [math.CO], 2019.

%F If k=0 and n>0, a(n,k)=n.

%F If k >= 1, a(n,k) = binomial(n-1,k+1).

%e a(5,0)=5. This counts the permutations 15432, 25431, 35421, 45321, and 54321.

%e a(5,1)=6. This counts the permutations 12543, 13542, 14532, 23541, 24531, and 34521.

%e Triangle begins:

%e 1;

%e 1;

%e 2;

%e 3, 1;

%e 4, 3, 1;

%e 5, 6, 4, 1;

%e 6, 10, 10, 5, 1;

%e 7, 15, 20, 15, 6, 1;

%e 8, 21, 35, 35, 21, 7, 1;

%e 9, 28, 56, 70, 56, 28, 8, 1;

%e 10, 36, 84, 126, 126, 84, 36, 9, 1;

%e 11, 45, 120, 210, 252, 210, 120, 45, 10, 1;

%e 12, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1;

%p f:= proc(n, k)

%p if n = 0 and k = 0 then return 1:

%p elif k = 0 then return n:

%p else return binomial(n-1, k+1):

%p fi: end:

%p f(0, 0), f(1, 0), seq(seq(f(i, j), j = 0 .. i-2), i = 2 .. 12)

%t Table[Which[And[n > 0, k == 0], n, k >= 1, Binomial[n - 1, k + 1], True, 1], {n, 0, 12}, {k, 0, If[n < 2, 0, n - 2]}] // Flatten (* _Michael De Vlieger_, Feb 07 2019 *)

%K easy,nonn,tabf

%O 0,3

%A _Lara Pudwell_, Dec 15 2018

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