A383847 Triangle T(n,k) read by rows, where row n is a permutation of the numbers 1 through n, such that if a deck of n cards is prepared in this order, and down-down-under dealing is used, then the resulting cards will be dealt in increasing order.

1, 1, 2, 1, 2, 3, 1, 2, 4, 3, 1, 2, 5, 3, 4, 1, 2, 5, 3, 4, 6, 1, 2, 6, 3, 4, 7, 5, 1, 2, 8, 3, 4, 7, 5, 6, 1, 2, 7, 3, 4, 8, 5, 6, 9, 1, 2, 8, 3, 4, 10, 5, 6, 9, 7, 1, 2, 11, 3, 4, 9, 5, 6, 10, 7, 8, 1, 2, 9, 3, 4, 10, 5, 6, 12, 7, 8, 11, 1, 2, 10, 3, 4, 13, 5, 6, 11, 7, 8, 12, 9

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, May 12 2025

Keywords

• nonn
• tabl

Comments

Down-down-under dealing is a dealing pattern where the top two cards are dealt, then the third card is placed at the bottom of the deck. This pattern repeats until all of the cards have been dealt.

This card dealing is related to a variation on the Josephus problem, where the first two people are eliminated, and the third person is skipped. The card in row n and column k is x if and only if in the corresponding Josephus problem with n people, the person number x is the k-th person eliminated. Equivalently, each row of Josephus triangle A383845 is an inverse permutation of the corresponding row of this triangle.

The total number of moves for row n is A001651(n) = floor((3*n-1)/2).

The index of the largest number in row n is A383846(n), corresponding to the index of the freed person in the corresponding Josephus problem.

Examples

			Consider a deck of four cards arranged in the order 1,2,4,3. In round 1, card 1 is dealt, card 2 is dealt, and card 4 goes under. Now the deck is ordered 3,4. In round 2, card 3 is dealt, then card 4 is dealt. The dealt cards are in order. Thus, the fourth row of the triangle is 1,2,4,3.
The table begins:
  1;
  1, 2;
  1, 2, 3;
  1, 2, 4, 3;
  1, 2, 5, 3, 4;
  1, 2, 5, 3, 4, 6;
  1, 2, 6, 3, 4, 7, 5;
  1, 2, 8, 3, 4, 7, 5, 6;
		

Crossrefs

Cf. A006257, A001651, A225381, A321298, A378635, A383845, A383846, A381050, A382528.

Programs

Python

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        out.append(J.pop(i))
        i = i%len(J)
        if len(J) > 1:
            out.append(J.pop(i))
        i = (i + 1)%len(J)
    out += [J[0]]
    return [out.index(j)+1 for j in list(range(1, n+1))]
print([e for n in range(1, 14) for e in row(n)])

Formula

T(n,3j+1) = 2j+1, for 3j+1 <= n. T(n,3j+2) = 2j+2, for 3j+2 <= n.