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A318354
Triangle read by rows: T(n,k) is the number of permutations p of {1..n} such that p(1)=k and p(i+1) < p(i) iff a strict majority of {1..n} \ {p(1)..p(i)} are < p(i).
1
1, 1, 1, 2, 1, 2, 5, 3, 3, 5, 16, 11, 8, 11, 16, 62, 46, 35, 35, 46, 62, 286, 224, 178, 143, 178, 224, 286, 1519, 1233, 1009, 831, 831, 1009, 1233, 1519, 9184, 7665, 6432, 5423, 4592, 5423, 6432, 7665, 9184, 62000, 52816, 45151, 38719, 33296, 33296, 38719, 45151, 52816, 62000
OFFSET
1,4
COMMENTS
If you shuffle n cards numbered 1 to n and then turn them over one at a time, guessing whether the next will be larger than the previous by the (optimal) rule that you guess "larger" unless there are more cards remaining smaller than the one just revealed, T(n,k) is the number of arrangements such that the first card revealed is k and you guess correctly every time.
LINKS
The Riddler Express, O. Roeder, solution to 2018-Aug 17 problem by K. Hudson
FORMULA
For k <= n/2 + 1: T(n+1,k) = Sum_{i=k..n} T(n,i);
For k >= n/2 + 1: T(n+1,k) = Sum_{i=1..k-1} T(n,i).
T(n+1,k+1) = f(n,k), where f(n,k) is the auxiliary function defined in the formula for A144188.
EXAMPLE
Suppose you are playing with four cards and you initially turn over a "2". You guess "larger" because there are two larger cards, 3 and 4, remaining, and only 1 smaller card, 1, remaining. You continue playing in this way, guessing larger unless there are (strictly) more smaller cards remaining. You guess correctly every time if the order of the cards was 2,3,4,1; 2,4,3,1; or 2,4,1,3. Thus T(4,2) = 3.
The triangle begins:
1
1 1
2 1 2
5 3 3 5
16 11 8 11 16
62 46 35 35 46 62
CROSSREFS
T(n+1,1) = A144188(n).
Sequence in context: A345278 A212431 A346517 * A380079 A348373 A106480
KEYWORD
nonn,tabl
AUTHOR
Glen Whitney, Aug 24 2018
STATUS
approved