For an integer $n$  3 and $a$ permutation  = (p1, p2, …, pn) of {1, 2, …, n}, we say p1 is $a$ landmark point if 2 $\le$ l $\le$ $n$ – 1 and (pl –1 – pl)(pl + 1 – pl) > 0. For example, for $n$ = 7, the permutation (2, 7, 6, 4, 5, 1, 3) has four landmark points: p2 = 7, p4 = 4, p5 = 5 and p6 = 1. For $a$ given $n$  3, let L(n) denote the number of permutation of {1, 2, …, n} with exactly only landmark point. Find the maximum $n$ $\le$ 3 for which L(n) is $a$ perfect square.

Step 1: For the permutations of set {1, 2, 3, …, n}, the landmark point should be 1 or $n$ to satisfy given conditions.

Step 2: Ist IInd $r$ th nth

Step 3: If $n$ is at (similarly for 1) rth position, there is only one permutation of the remaining numbers for each selection.

Step 4: So, number of selections =  $n$ 1Cr 1

Step 5: = 2n – 1 – 2

Step 6: => Total number of selections = 2(2n–1– 2)

Step 7: L(n) = 4(2n–2 – 1)

Step 8: Now for L(n) to be $a$ perfect square $n$ should be equal to 3.

Step 9:  => n=3