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 Permutations count redux (Posted on 2015-10-09)
Consider all of the permutation of (1,2,3...n)

Each will have some number of local maxima, m.

For example if n=6 some permutations are
m=1 123465
m=2 143256
m=3 214356

Define f(n,m) as the number of permutations of (1,2,3...n) with m local maxima.

What may be ultimately sought is a formula for f(n,m) but here are some simpler considerations to prove:

Find a formula for f(n,2) in terms of values where m=1

Find a formula relating f(2a,a) and f(2a+1,2a).

Note: The problem of finding f(n,1) was investigated here

 No Solution Yet Submitted by Jer No Rating

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 first simpler consideration Comment 1 of 1
f(3,2) = 2, and thereafter there are 4 places for each new value that expands n to one higher, so f(n,2) thereafter if 2*4^(n-3). Placing this in terms of values for f(n,1), it's 2*4^(n-3) vs 2^(n-1). Dividing: 2*4^(n-3) / 2^(n-1) = 2^(2*n-5) / 2^(n-1) = 2^(n-4).  That is, f(n,2) = f(n,1) * 2^(n-4).

 Posted by Charlie on 2015-10-12 13:46:19

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