A378635 -id:A378635 - cp's OEIS frontend

A381667 Triangle read by row: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people where two people are skipped each step.

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

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 03 2025

In this variation of the Josephus elimination process, the numbers 1 through n are arranged in a circle. A pointer starts at position 1. With each turn, the pointer skips two numbers and the next number is eliminated. The process repeats until no numbers remain. This sequence represents the triangle T(n, k), where n is the number of people in the circle, and T(n, k) is the elimination order of the k-th number in the circle.

This variation of the Josephus problem is related to under-under-down card dealing, where the cards of a deck are dealt by alternately cycling two cards from the top "under", and then dealing the next card "down". In particular, T(n,k) is the k-th card dealt in under-under-down dealing if the deck begins in order 1,2,3,...,n.

			Consider 4 people in a circle. Initially, people numbered 1 and 2 are skipped, and person 3 is eliminated. The remaining people are now in order 4, 1, 2. Then 4 and 1 are skipped, and person 2 is eliminated. The remaining people are in order 4, 1. Now, 4 and 1 are skipped, and 4 is eliminated. Person 1 is eliminated last. Thus, the fourth row of the triangle is 3, 2, 4, 1.
Triangle begins;
  1;
  1, 2;
  3, 1, 2;
  3, 2, 4, 1;
  3, 1, 5, 2, 4;
  3, 6, 4, 2, 5, 1;
  3, 6, 2, 7, 5, 1, 4;
  3, 6, 1, 5, 2, 8, 4, 7;

Cf. A006257, A054995, A225381, A321298, A378635, A380195, A008585, A381591 A381667.

def UUDES(n):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=2
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
sequence = []
for i in range(1,20):
    sequence += [str(v) for v in UUDES(i)]
print(", ".join(sequence))

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 0:
        i = (i + 2)%len(J)
        out.append(J.pop(i))
    return out
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Mar 24 2025

A382354 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 under-down-under dealing is used, then the resulting cards will be dealt in increasing order.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 22 2025

Under-down-under dealing is a dealing pattern where the top card is placed at the bottom of the deck; then the next card is 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 person is skipped, the second person is eliminated, and the third person is skipped. Then, the process repeats. 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 A382358(n) is an inverse permutation of the corresponding row of this triangle.
The total number of moves for row n is 3n-1.
The first column is A382356(n): the order of elimination of the first person in the Josephus problem.
The index of the largest number in row n is A382355(n), corresponding to the index of the freed person in the corresponding Josephus problem.
T(n,3j-1) = j, for 3j <= n.

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

Cf. A006257, A225381, A321298, A378635, A382355, A382356, A382358.

def T(n, A):
    return invPerm(J(n,A))
def J(n,A):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=A[i]
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
def UDU(n):
    return [1] + [2 for i in range(n)]
seq = []
for i in range(20):
    seq += T(i,UDU(i))
print(", ".join([str(v) for v in seq]))

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        i = (i + 1)%len(J)
        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)]) # Michael S. Branicky, Mar 27 2025

For any n, we have T(n,2) = 1. T(1,1) = 1; T(2,1) = 2. For n > 2, T(n,1) = T(n-1,n-2) + 1 and T(n,3) = T(n-1,n-1) + 1. For n > 3 and k > 3, T(n,k) = T(n-1,k-3) + 1.

A382355 A version of the Josephus problem: a(n) is the surviving integer under the skip-eliminate-skip version of the elimination process.

Original entry on oeis.org

1, 1, 1, 4, 3, 6, 3, 6, 9, 3, 6, 9, 12, 1, 4, 7, 10, 13, 16, 19, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 3, 6

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 22 2025

nonn

This variation of the Josephus problem is related to under-down-under card dealing.

			Consider 4 people in a circle in order 1,2,3,4. In the first round, person 1 is skipped, then person 2 is eliminated, then person 3 is skipped. Now people are in order 4,1,3. In the second round, person 4 is skipped, person 1 is eliminated, and person 3 is skipped. Now people are in order 4,3. In the third round person 3 is eliminated. Person 4 is freed. Thus, a(4) = 4.

Cf. A006257, A225381, A321298, A378635, A382354, A382356, A382358.

def J(n,A):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=A[i]
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
def survivor(n, A):
    return J(n, A)[n-1]
def UDU(n):
    return [1] + [2 for i in range(n)]
seq = []
for i in range(1,20):
    seq += [survivor(i, UDU(i))]
print(", ".join([str(v) for v in seq]))

def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        i = (i + 1)%len(J)
        q = J.pop(i)
        i = (i + 1)%len(J) # skip the third
    return J[0]
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Mar 24 2025

A382356 Elimination order of the first person in a variation of the Josephus problem, where there are n people total. During each round the first person is skipped, the second is eliminated and the third person is skipped. Then the process repeats.

Original entry on oeis.org

1, 2, 3, 2, 4, 4, 3, 5, 7, 4, 10, 9, 5, 14, 9, 6, 10, 15, 7, 18, 21, 8, 19, 14, 9, 15, 24, 10, 21, 28, 11, 23, 19, 12, 20, 26, 13, 31, 28, 14, 36, 24, 15, 25, 43, 16, 47, 39, 17, 44, 29, 18, 30, 44, 19, 40, 50, 20, 42, 34, 21, 35, 45, 22, 57, 47, 23, 55, 39, 24

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 22 2025

nonn

This elimination process is related to under-down-under card dealing.

a(3k-2) = k.

			Consider n people. Four people are in order 1,2,3,4. In the first round, the first person is skipped, the second person is eliminated, and the third person is skipped. Now people are ordered 4,1,3. In the second round, person 4 is skipped, and person 1 is eliminated. Thus, person 1 is eliminated in the second round and a(4) = 2.

Cf. A006257, A225381, A321298, A378635, A382354, A382355, A382358.

def T(n, A):
    return invPerm(J(n,A))
def J(n,A):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=A[i]
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
def firstPersonElimOrder(n, A):
    return T(n, A)[0]
def UDU(n):
    return [1] + [2 for i in range(n)]
seq = []
for i in range(1,88):
    seq += [firstPersonElimOrder(i, UDU(i))]
print(", ".join([str(v) for v in seq]))

def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 0:
        i = (i + 1)%len(J)
        q = J.pop(i)
        if q == 1: return c
        i = (i + 1)%len(J) # skip the third
        c = c+1
print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Mar 24 2025

A382358 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people, where in each round, the first person is skipped, the second eliminated and the third is skipped.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 22 2025

This variation of the Josephus problem is related to under-down-under card dealing.
The n-th row has n elements.
In this variation of the Josephus elimination process, the numbers 1 through n are arranged in a circle. A pointer starts at position 1. With each turn, the pointer skip one number, eliminates the next, and skips one more number. The process repeats until no numbers remain. This sequence represents the triangle T(n,k), where n is the number of people in the circle, and T(n,k) is the elimination order of the k-th number in the circle.

			Consider 4 people in a circle. In the first round, person 1 is skipped, person 2 is eliminated, and person 3 is skipped. The remaining people are now in order 4,1,3. In the second round, person 4 is skipped, person 1 is eliminated, and person 3 is skipped. The remaining people are in order 4, 3. In the third round, person 3 is eliminated, and in the last round, person 4 is eliminated. Thus, the order of elimination is 2, 1, 3, 4, and this is the fourth row of the triangle.
Triangle begins;
  1;
  2, 1;
  2, 3, 1;
  2, 1, 3, 4;
  2, 5, 4, 1, 3;
  2, 5, 3, 1, 4, 6;
  2, 5, 1, 6, 4, 7, 3;
  2, 5, 8, 4, 1, 7, 3, 6;
  ...

Cf. A006257, A225381, A321298, A378635, A382354, A382355, A382356.

def J(n,A):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=A[i]
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
def UDU(n):
    return [1] + [2 for i in range(n)]
seq = []
for i in range(1,20):
    seq += J(i, UDU(i))
print(", ".join([str(v) for v in seq]))

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        i = (i + 1)%len(J)
        out.append(J.pop(i))
        i = (i + 1)%len(J)
    return out + [J[0]]
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Mar 24 2025

A382528 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 under-down-down dealing is used, then the resulting cards will be dealt in increasing order.

Original entry on oeis.org

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

Offset: 1

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

Under-down-down dealing is a dealing pattern where the top card is placed at the bottom of the deck, then the next two cards are 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 one person is skipped, and the next two are 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 ??? is an inverse permutation of the corresponding row of this triangle.
The total number of moves for row n is A007494(n) = ceiling(3n/2).
The first column is A381048, the order of elimination of the first person in the Josephus problem.
The index of the largest number in row n is A337191(n), corresponding to the index of the freed person in the corresponding Josephus problem.

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

Cf. A006257, A007494, A225381, A321298, A378635, A337191, A381048, A381049.

def T(n, A):
    return invPerm(J(n,A))
def J(n,A):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=A[i]
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
def DDU(n):
    return [0] + [(i)%2 for i in range(n)]
def DUD(n):
    return DDU(n+1)[1:]
def UDD(n):
    return DUD(n+1)[1:]
seq = []
for i in range(1,10):
    seq += T(i, UDD(i))
print(", ".join([str(v) for v in seq]))

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        i = (i + 1)%len(J)
        out.append(J.pop(i))
        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

T(n,k) = T(n-2,k-3) + 2. T(1,1) = 1. For the first 3 columns, we have T(n,1) = T(n-2,n-2) + 2, T(n,2) = 1, and T(n,3) = 2.

It follows that T(n,3k+2) = 2k+1, T(n,3k) = 2k.

A378674 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.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Dec 03 2024

Down-under dealing is a dealing pattern where the top card is dealt then the next card is put on the bottom of the deck. Then, this pattern repeats until all cards are dealt.
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, each row of the Josephus triangle A378682 related to this variation is an inverse permutation of the corresponding row of this triangle.
The total number of moves for row n is 2n-1.
The first column is the order of elimination of the first person in the corresponding variation of the Josephus problem and consists of all ones.
The index of the largest number in row n is A152423(n), corresponding to the index of the freed person in this variation of the Josephus problem.
T(n,2j-1) = j, for 2j-1 <= n.
Sequence A378635 represents a similar triangle for under-down dealing.

			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;

Cf. A006257, A152423, A225381, A321298, A378635, A378682, A381050.

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"]];]; res["SetPart", n, ds["Pop"]]; Flatten[PositionIndex[res["Elements"]] /@ Range[n]]]; Array[row, 14, 1] // Flatten (* Shenghui Yang, May 16 2025 *)

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.
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)

A383076 Triangle T(n,k) read by rows: where T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people, where the elimination pattern is eliminate-skip-eliminate.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Apr 15 2025

This Josephus problem is related to down-under-down card dealing.
The n-th row has n elements.
In this variation of the Josephus elimination process, the numbers 1 through n are arranged in a circle. A pointer starts at position 1. With each turn, the pointer eliminates the first number, skips the second, then eliminates the third. The process repeats until no numbers remain. This sequence represents the triangle T(n, k), where n is the number of people in the circle, and T(n, k) is the elimination order of the k-th number in the circle.

			Consider 4 people in a circle. Initially, person number 1 is eliminated, person number 2 is skipped, and person number 3 is eliminated. The remaining people are now in order 4, 2. Then, 4 is eliminated, and 2 is left. Thus, the fourth row of the triangle is 1, 3, 4, 2, the order of elimination.
Triangle begins;
1;
1, 2;
1, 3, 2;
1, 3, 4, 2;
1, 3, 4, 2, 5;
1, 3, 4, 6, 2, 5;
1, 3, 4, 6, 7, 5, 2;
1, 3, 4, 6, 7, 2, 5, 8;

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

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)
        if len(J) > 1:
            out.append(J.pop(i))
    out += [J[0]]
    return out
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Apr 28 2025

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.

Original entry on oeis.org

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

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

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.

			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;

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

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)])

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

A386639 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 the AP dealing is used, then the resulting cards will be dealt in increasing order.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 2, 1, 4, 3, 3, 1, 4, 5, 2, 4, 1, 6, 3, 2, 5, 5, 1, 3, 4, 2, 6, 7, 3, 1, 7, 5, 2, 6, 8, 4, 7, 1, 8, 6, 2, 9, 4, 5, 3, 9, 1, 8, 5, 2, 4, 7, 6, 3, 10, 5, 1, 6, 4, 2, 10, 11, 7, 3, 8, 9, 7, 1, 4, 9, 2, 11, 10, 8, 3, 6, 5, 12, 4, 1, 13, 11, 2, 10, 6, 7, 3, 5, 12, 9, 8, 10, 1, 7, 6, 2, 12, 8, 5, 3, 14

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Jul 27 2025

The AP dealing is a dealing pattern where x cards are placed at the bottom of the deck, and then the next card is dealt. The number of cards x placed at the bottom changes with every dealt card according to the arithmetic progression 1, 2, 3, and so on. This pattern repeats until all of the cards have been dealt.
This card dealing can equivalently be seen as a variation on the Josephus problem, where one person is skipped, then the next person is eliminated, then two people are skipped and one person is eliminated, then three people are skipped, and so on. T(n,k) is the order of elimination of the k-th person in the Josephus problem. Equivalently, each row of T is the inverse permutation of the corresponding row of the Josephus triangle A386641, i.e., A386641(n,T(n,k)) = k.
The total number of moves for row n is A000096.
The first column is A386643(n), the order of elimination of the first person in the Josephus problem.
The index of the largest number in row n is A291317(n), corresponding to the index of the freed person in the corresponding Josephus problem.

			Consider a deck of four cards arranged in the order 2,1,4,3. We put one card under and deal the next card, which is card number 1. Now the deck is ordered 4,3,2. We place 2 cards under and deal the next one, which is card number 2. Now the deck is 4,3. Again, placing 3 cards under and dealing the next, we will deal card number 3, leaving card number 4 to be dealt last. The dealt cards are in order. Thus, the fourth row of the triangle is 2,1,4,3.
The triangle begins as follows:
  1;
  2, 1;
  3, 1, 2;
  2, 1, 4, 3;
  3, 1, 4, 5, 2;
  4, 1, 6, 3, 2, 5;
  5, 1, 3, 4, 2, 6, 7;
  3, 1, 7, 5, 2, 6, 8, 4;
  7, 1, 8, 6, 2, 9, 4, 5, 3;

Cf. A000096, A291317, A386641, A386643.

Cf. A378635 (classical elimination process).

T(n,A000096(k)) = k, for A000096(k) <= n.

cp's OEIS Frontend

A381667 Triangle read by row: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people where two people are skipped each step.

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A382354 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 under-down-under dealing is used, then the resulting cards will be dealt in increasing order.

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A382355 A version of the Josephus problem: a(n) is the surviving integer under the skip-eliminate-skip version of the elimination process.

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A382356 Elimination order of the first person in a variation of the Josephus problem, where there are n people total. During each round the first person is skipped, the second is eliminated and the third person is skipped. Then the process repeats.

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A382358 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people, where in each round, the first person is skipped, the second eliminated and the third is skipped.

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A382528 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 under-down-down dealing is used, then the resulting cards will be dealt in increasing order.

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A378674 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.

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A383076 Triangle T(n,k) read by rows: where T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people, where the elimination pattern is eliminate-skip-eliminate.

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