A361648 Number of permutations p of [n] such that p(i), p(i+2), p(i+4),... form an up-down sequence for i in {1,2}.
1, 1, 2, 3, 6, 20, 80, 350, 1750, 10080, 64512, 450912, 3438204, 28471872, 253913088, 2424193200, 24687555750, 267199961600, 3062092267520, 37037541651968, 471565937953396, 6304419553216512, 88298062293762048, 1292879475255280640, 19753693667117055100
Offset: 0
Keywords
Examples
a(0) = 1: (), the empty permutation. a(1) = 1: 1. a(2) = 2: 12, 21. a(3) = 3: 123, 132, 213. a(4) = 6: 1234, 1243, 1324, 2134, 2143, 3142. a(5) = 20: 12453, 12534, 12543, 13452, 13542, 14352, 21453, 21534, 21543, 23451, 23541, 24351, 31452, 31524, 31542, 32451, 32541, 41523, 41532, 42531. a(6) = 80: 124635, 125634, 125643, 126453, ..., 526413, 526431, 536412, 536421.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..485
Programs
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Maple
b:= proc(u, o) option remember; `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u)) end: a:= n-> (h-> b(h, 0)*b(n-h, 0)*binomial(n, h))(iquo(n, 2)): seq(a(n), n=0..30);
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Mathematica
b[u_, o_] := b[u, o] = If[u+o == 0, 1, Sum[b[o-1+j, u-j], {j, 1, u}]]; a[n_] := With[{h = Quotient[n, 2]}, b[h, 0] b[n-h, 0] Binomial[n, h]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 26 2023, after Alois P. Heinz *)
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Python
from math import comb from itertools import accumulate def A361648(n): if n<=1: return 1 blist = (0,1) for _ in range((m:=n>>1)-1): blist = tuple(accumulate(reversed(blist),initial=0)) return blist[-1]*sum(blist)*comb(n,m) if n&1 else blist[-1]**2*comb(n,m) # Chai Wah Wu, Apr 16 2023
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