A395725 - OEIS

%I #7 May 11 2026 02:34:24

%S 1,1,1,2,2,2,6,6,6,5,23,23,23,20,14,103,103,103,92,70,42,513,513,513,

%T 466,373,252,132,2762,2762,2762,2537,2097,1520,924,429,15793,15793,

%U 15793,14619,12348,9362,6206,3432,1430,94776,94776,94776,88226,75674,59168,41470,25352,12870,4862

%N Triangle read by rows: T(n,k) is the number of permutations of [n] avoiding the pattern 1324 with the maximum element n in position k, 1 <= k <= n.

%C This triangle refines A061552 by the position of the maximum element n inside each 1324-avoiding permutation of [n]. Row sums recover A061552(n); the rightmost column equals A000108(n-1), and A391315(n) = A061552(n) - T(n,n) recovers the existing entry as the partial row sum Sum_{k=1..n-1} T(n,k).

%C Machine-verified theorem (Coq, proven): appending the pair [n; v] to a prefix bounded below n preserves 1324-avoidance if and only if the prefix avoids 132, independently of v. The next-to-rightmost diagonal is therefore A000984(n-2).

%C Machine-verified theorem (Coq, proven): a permutation in Av_n(1324) starts with 1 iff its tail (the last n-1 entries) avoids 213. Combined with the standard Catalan convolution identity, this yields a closed form for the count of 1324-avoiders that simultaneously pin the smallest element to a fixed position j and the largest to a fixed position k, for j = 1 (and dually k = n by reverse-complement invariance of 1324).

%H Sean A. Irvine, <a href="/A395725/b395725.txt">Table of n, a(n) for n = 1..120</a> (rows 1..15 flattened)

%H Charles C. Norton, <a href="https://github.com/CharlesCNorton/1324-Joint-Positions">1324-Joint-Positions Coq formalization</a>

%F T(n,n) = A000108(n-1).

%F T(n,n-1) = (n-1)*A000108(n-2) = A000984(n-2).

%F T(n,k) = A061552(n-1) for 1 <= k <= 3.

%F Sum_{k=1..n} T(n,k) = A061552(n).

%F A391315(n) = A061552(n) - A000108(n-1) = Sum_{k=1..n-1} T(n,k).

%e Triangle begins:

%e       n=1:   1;

%e       n=2:   1,   1;

%e       n=3:   2,   2,   2;

%e       n=4:   6,   6,   6,   5;

%e       n=5:  23,  23,  23,  20,  14;

%e       n=6: 103, 103, 103,  92,  70,  42;

%e       n=7: 513, 513, 513, 466, 373, 252, 132;

%e T(4,4) = 5: among the 23 permutations of [4] avoiding 1324, five have 4 at the last position, namely (1,2,3,4), (1,3,2,4), (2,1,3,4), (2,3,1,4), and (3,2,1,4); they biject with the 5 = A000108(3) permutations of [3] avoiding 132 by deleting the trailing 4.

%t avoids1324[p_] := !AnyTrue[Subsets[Range[Length[p]], {4}], p[[#[[1]]]] < p[[#[[3]]]] < p[[#[[2]]]] < p[[#[[4]]]] &]; T[n_, k_] := Count[Permutations[Range[n]], p_ /; Position[p, n][[1, 1]] == k && avoids1324[p]]; Table[T[n, k], {n, 1, 9}, {k, 1, n}] // Flatten

%o (Python)

%o from itertools import permutations, combinations

%o def avoids1324(p):

%o     return not any(p[i]<p[k]<p[j]<p[l] for i,j,k,l in combinations(range(len(p)),4))

%o def T(n,k): return sum(1 for p in permutations(range(1,n+1)) if p[k-1]==n and avoids1324(p))

%o print([T(n,k) for n in range(1,10) for k in range(1,n+1)])

%Y Cf. A000108, A000984, A061552, A391315.

%K nonn,tabl

%O 1,4

%A _Charles Cornell Norton_, May 04 2026

%E More terms from _Sean A. Irvine_, May 11 2026