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From Pontus von Brömssen, Dec 11 2024: (Start)

T(n,k) = A378635(n-1,k-1) + 1 for 2 <= k <= n.

T(n,k) = A378635(n,(k mod n) + 1).

(End)

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1;

1, 2;

1, 3, 2;

1, 3, 2, 4;

1, 5, 2, 4, 3;

1, 4, 2, 6, 3, 5;

1, 6, 2, 5, 3, 7, 4;

1, 5, 2, 7, 3, 6, 4, 8;

1, 9, 2, 6, 3, 8, 4, 7, 5;

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

This card dealing is related to a variation of the Josephus problem, where the first person is eliminated and the second person is skipped. The card in row n and column k is x if and only if in the Josephus problem variation with n people, the person number x is the k-th person eliminated. Equivalently, every each row of the Josephus triangle ??? A378682 related to this variation is an inverse permutation of every the corresponding row of this triangle.

T(1,1) = 1, for n > 1, T(n,1) = 1 and T(n,2) = T(n-1,n-1) + 1. For n > 1 and k > 2, T(n,k) = T(n-1,k-2) + 1.

Cf. A005843, A006257, A152423, A225381, A321298, A378635, A378682.

Cf. A005843, A006257, A225381, A321298, A378635, A378682.

nonn,changed,tabl

Suppose there are four cards arranged in order 1,3,2,4. Card 1 is dealt, and card 3 goes under, then card 2 is dealt , and card 4 goes under. Now, the leftover deck is ordered 3,4. Card 3 is dealt, and card 4 goes under. Now , the leftover deck is card 4, which is dealt. The dealt cards are in order. Thus, the fourth row of the triangle is 1,3,2,4.

Triangle begins:

1;

1, 2;

1, 3, 2;

1, 3, 2, 4;

1, 5, 2, 4, 3;

1, 4, 2, 6, 3, 5;

1, 6, 2, 5, 3, 7, 4;

1, 5, 2, 7, 3, 6, 4, 8;

1, 9, 2, 6, 3, 8, 4, 7, 5;