A381050 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-under-down dealing is used, then the resulting cards will be dealt in increasing order.

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

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Apr 14 2025

Keywords

• nonn
• tabl

Comments

Down-under-down dealing is a dealing pattern where the top card is dealt, the second card is placed at the bottom of the deck, then the third card is dealt. 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 person is eliminated, the second person is skipped, and the third person is eliminated. 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 A383076 is an inverse permutation of the corresponding row of this triangle.

The total number of moves for row n is A032766(n) = floor(3n/2).

The index of the largest number in row n is A381051(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 1,4,2,3. In round 1, card 1 is dealt, card 4 goes under, card 2 is dealt. Now the deck is ordered 3,4. In round 2, card 3 is dealt, card 4 goes under, then card 4 is dealt. The dealt cards are in order. Thus, the fourth row of the triangle is 1,4,2,3.
Table begins:
  1;
  1, 2;
  1, 3, 2;
  1, 4, 2, 3;
  1, 4, 2, 3, 5;
  1, 5, 2, 3, 6, 4;
  1, 7, 2, 3, 6, 4, 5;
  1, 6, 2, 3, 7, 4, 5, 8;
		

Crossrefs

Cf. A006257, A032766, A225381, A321298, A378635, A381050, A381051, A383076, A378674.

Programs

Mathematica

row[n_]:=Module[{ds,res,k,i=1,len},ds=CreateDataStructure["Queue",Range[n]];res=CreateDataStructure["FixedArray",n];While[(ds["Length"]>=2),res["SetPart",i++,ds["Pop"]];ds["Push",ds["Pop"]];If[ds["Length"]>1,res["SetPart",i++,ds["Pop"]];]];res["SetPart",n,ds["Pop"]];Flatten[PositionIndex[res["Elements"]]/@Range[n]]];
Array[row, 13, 1] // Flatten (* Shenghui Yang, May 11 2025 *)

Python

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        i = i%len(J)
        out.append(J.pop(i))
        i = (i + 1)%len(J)
        #i = i%len(J)
        if len(J) > 1:
            out.append(J.pop(i))
    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)]) # Michael S. Branicky, Apr 28 2025

Formula

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