A380248 -id:A380248 - cp's OEIS frontend

A380201 Triangle T(n,k) read by rows, where row n is a permutation of numbers 1 through n, such that if a deck of n cards is prepared in this order, and SpellUnder-Down dealing is used, then the resulting cards are put down in increasing order.

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

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 16 2025

In Spell Under-Down dealing, we spell the positive integers starting from O-N-E, moving 1 card from the top of the deck underneath the deck for each letter, followed by dealing or "putting down" the top card. So we start by putting 3 cards under for O-N-E, then we deal a card. Then we put 3 cards under for T-W-O, then we deal a card. Then we put 5 cards under for T-H-R-E-E, and subsequently deal a card. This dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: UUUDUUUDUUUUUD, where each "U" corresponds to putting a card “under” and each "D" corresponds to dealing a card “down”.
This card dealing can be thought of as a generalized version of the Josephus problem. In this version of the Josephus problem, we spell the positive integers in increasing order, each time skipping past 1 person for each letter and executing the next person. The card in row n and column k is x if and only if in the corresponding Josephus problem with n people, the person numbered x is the k-th person eliminated.
  Equivalently, each row of the corresponding Josephus triangle A380247 is an inverse permutation of the corresponding row of this triangle. The first column is A380246, the order of elimination of the first person in the corresponding Josephus problem. The index of the largest number in row n is A380204(n), corresponding to the index of the freed person in the corresponding Josephus problem. The number of card moves if we start with n cards is A380202 = A067278(n) + n.

			Triangle begins:
  1;
  2, 1;
  1, 3, 2;
  2, 4, 3, 1;
  5, 3, 2, 1, 4;
  4, 2, 5, 1, 3, 6;
  2, 3, 4, 1, 6, 5, 7;
  5, 6, 8, 1, 7, 4, 3, 2;
  ...
For n = 4 suppose there are four cards arranged in order 2, 4, 3, 1. Three cards go under for each letter in O-N-E, then 1 is dealt. Now the deck is ordered 2,4,3. Three cards go under for each letter in T-W-O, then card 2 is dealt. Now the leftover deck is ordered 4,3. Five cards go under for each letter in T-H-R-E-E, then card 3 is dealt. Finally, card 4 is dealt. The dealt cards are in numerical order. Thus, the fourth row of the triangle is 2, 4, 3, 1.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248.

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(",","").replace("-", ""))
def nthRow(n):
    l = []
    for i in range(0,n):
        l.append(0)
    zp = 0
    for j in range(1,n+1):
        zc = 0
        while zc <= spell(j):
            if l[zp] == 0:
                zc += 1
            zp += 1
            zp = zp % n
        l[zp-1] = str(j)
    return l
l = []
for i in range(1,20):
    l += nthRow(i)
print(", ".join(l))

A380204 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process.

Original entry on oeis.org

1, 1, 2, 2, 1, 6, 7, 3, 5, 3, 5, 6, 10, 9, 2, 13, 3, 16, 10, 2, 15, 6, 15, 6, 21, 1, 7, 23, 26, 6, 20, 12, 27, 29, 7, 2, 36, 11, 6, 7, 32, 6, 32, 43, 10, 31, 7, 5, 42, 1, 17, 48, 7, 31, 53, 25, 42, 43, 29, 39, 51, 25, 43, 7, 26, 59, 15, 10, 60, 69, 13, 57, 54, 66, 57, 30, 9, 35, 64, 9, 65, 1, 15, 3, 79, 47, 86, 7

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 16 2025

Arrange n people numbered 1, 2, 3, ..., n in a circle, increasing clockwise. Starting with the person numbered 1, spell the letters of O-N-E, moving one person clockwise for each letter. Once you are done, eliminate the next person. Then, spell the letters of T-W-O; in other words, skip three people and eliminate the next person. Following this, spell the letters of T-H-R-E-E; in other words, skip five people and eliminate the next person. Continue until one person remains. The number of this person is a(n).

			Consider n = 4 people. The first person eliminated is number 4. This leaves the remaining people in the order 1, 2, 3. The second person eliminated is number 1; the people left are in the order 2, 3. The next person eliminated is numbered 3, leaving only the person numbered 2. Thus a(4) = 2.

Michael S. Branicky, Table of n, a(n) for n = 1..20000

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380246, A380247, A380248.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        i = (i + f(c))%len(J)
        q = J.pop(i)
        c = c+1
    return J[0]
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Jan 26 2025

Terms a(22) and beyond corrected by Michael S. Branicky, Feb 15 2025

A380246 Elimination order of the first person in a variation of the Josephus problem, where the number of skipped people correspond to the number of letters in consecutive numbers, called SpellUnder-Down.

Original entry on oeis.org

1, 2, 1, 2, 5, 4, 2, 5, 6, 4, 6, 10, 3, 12, 6, 8, 15, 4, 13, 19, 14, 17, 5, 22, 18, 26, 6, 20, 13, 17, 19, 23, 7, 25, 21, 31, 22, 32, 8, 31, 38, 20, 29, 9, 27, 18, 43, 10, 15, 50, 37, 20, 16, 41, 11, 21, 39, 36, 34, 32, 29, 12, 36, 50, 27, 53, 35, 19, 45, 67, 13, 20, 70, 59, 74, 26, 21, 40, 65, 14, 49, 82, 33, 43, 28, 34, 53, 15

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 17 2025

Arrange n people numbered 1,2,3,...,n in a circle, increasing clockwise. Starting with the person numbered 1, spell the letters of O-N-E, moving one person clockwise for each letter. Once you are done, eliminate the next person. Then, spell the letters of T-W-O; in other words, skip three people and eliminate the next person. Following this, spell the letters of T-H-R-E-E; in other words, skip five people and eliminate the next person. Continue until one person remains. a(n) is the order of elimination of the first person.

			Consider n = 4 people. The first person eliminated is number 4. This leaves the remaining people in order 1, 2, 3. The second person eliminated is number 1. Thus, person number 1 is eliminated in the second round, implying that a(4) = 2.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248.

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(chr(44), "").replace("-", ""))
def nthRow(n):
    l = []
    for i in range(0,n):
        l.append(0)
    zp = 0
    for j in range(1,n+1):
        zc = 0
        while zc <= spell(j):
            if l[zp] == 0:
                zc += 1
            zp += 1
            zp = zp % n
        l[zp-1] = str(j)
    return l
l = []
for i in range(1,89):
    l += [nthRow(i)[0]]
print(l)

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 0:
        i = (i + f(c))%len(J)
        q = J.pop(i)
        if q == 1: return c
        c = c+1
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Feb 15 2025

A380202 Number of card moves to deal n cards using the SpellUnder-Down dealing.

Original entry on oeis.org

4, 8, 14, 19, 24, 28, 34, 40, 45, 49, 56, 63, 72, 81, 89, 97, 107, 116, 125, 136, 146, 156, 168, 179, 190, 200, 212, 224, 235, 246, 256, 266, 278, 289, 300, 310, 322, 334, 345, 355, 364, 373, 384, 394, 404, 413, 424, 435, 445, 455, 464, 473, 484, 494, 504, 513, 524, 535, 545, 555, 564, 573, 584, 594, 604, 613, 624, 635, 645

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 16 2025

In SpellUnder-Down dealing, we spell the number of the next card, putting a card under for each letter in the number, then we deal the next card. So we start with putting 3 cards under, for O-N-E, then deal, then 3 under for T-W-O, then deal, then 5 under for T-H-R-E-E, then deal. The dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: UUUDUUUDUUUUUD.

			The dealing pattern to deal three cards is UUUDUUUDUUUUUD. It contains 14 letters, thus, a(3) = 14.

Cf. A005589, A006257, A067278, A225381, A321298, A378635, A380201, A380204, A380246, A380247, A380248.

a(n) = A067278(n) + n.

A380247 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 the number of people skipped correspond to the number of letters in the next number in English alphabet.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 17 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. Then three people are skipped because number O-N-E has three letters, then the next person is eliminated. Next, three people are skipped because T-W-O has three letters, and the next person is eliminated. Then, five people are skipped because T-H-R-E-E has five letters, and so on. This 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 person in the circle.

In rows 4 and after, the first number is 4. In rows 8 and after, the second number is 8. In rows 14 and after, the third number is 14. In the limit the numbers form sequence A380202.

			Triangle begins:
  1;
  2, 1;
  1, 3, 2;
  4, 1, 3, 2;
  4, 3, 2, 5, 1;
  4, 2, 5, 1, 3, 6;
  4, 1, 2, 3, 6, 5, 7;
  ...
For n = 4 suppose four people are arranged in a circle corresponding to the fourth row of the triangle. Three people are skipped for each letter in O-N-E; then the 4th person is eliminated. This means the row starts with 4. The next three people are skipped, and the person eliminated is number 1. Thus, the next element in the row is 1. Then, 5 people are skipped, and the next person eliminated is number 3. Similarly, the last person eliminated is number 2. Thus, the fourth row of this triangle is 4, 1, 3, 2.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380248.

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(chr(44), "").replace("-", ""))
def inverse_permutation(p):
    inv = [0] * len(p)
    for i, x in enumerate(p):
        inv[x-1] = i +1
    return inv
def nthRow(n):
    l = []
    for i in range(0,n):
        l.append(0)
    zp = 0
    for j in range(1,n+1):
        zc = 0
        while zc <= spell(j):
            if l[zp] == 0:
                zc += 1
            zp += 1
            zp = zp % n
        l[zp-1] = j
    return l
l = []
for i in range(1,15):
    l += inverse_permutation(nthRow(i))
print(l)

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def row(n):
    c, i, J = 1, 0, list(range(1, n+1))
    out = []
    while len(J) > 1:
        i = (i + f(c))%len(J)
        q = J.pop(i)
        out.append(q)
        c = c+1
    out.append(J[0])
    return out
print([e for n in range(1, 15) for e in row(n)]) # Michael S. Branicky, Feb 15 2025

A381114 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in a variation of the Josephus elimination process for n people, where the number of people skipped is equal to the number of letters in the previous number's English name.

Original entry on oeis.org

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

Offset: 1

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

In this variation of the Josephus elimination process, people numbered 1 through n are arranged in a circle. A pointer starts at person 1. Person 1 is eliminated, then three people are skipped because the number O-N-E has three letters. Then, the next person is eliminated. Next, three people are skipped because T-W-O has three letters, and the next person is eliminated. Then, five people are skipped because T-H-R-E-E has five letters, and so on. This repeats until no people remain. This sequence represents the triangle T(n, k), the order of elimination for the person numbered k in a circle of n people.
Every entry of the first column is 1.
In rows 5 and after, the second number is 5. In rows 9 and after, the third number is 9. In rows 15 and after, the fourth number is 15. In the limit, the numbers in a row form sequence A381128(k).

			Triangle begins:
 1;
 1, 2;
 1, 3, 2;
 1, 2, 4, 3;
 1, 5, 2, 4, 3;
 1, 5, 4, 3, 6, 2;
 1, 5, 3, 6, 2, 4, 7;
 ...
For n = 4, suppose four people are arranged in a circle. The first person is eliminated, then three people are skipped because O-N-E has three letters. The leftover people are now in order 2,3,4. Then, the next person (numbered 2) is eliminated. The leftover people are now ordered 3,4. The next three people are skipped, so the person eliminated is number 4. Then, 5 people are skipped, and the next person eliminated is number 3. Thus, the fourth row of the triangle is 1,2,4,3.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381127, A381128, A381129.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def row(n):
    c, i, J, out = 1, 0, list(range(1, n+1)), []
    while len(J) > 1:
        q = J.pop(i)
        out.append(q)
        i = (i + f(c))%len(J)
        c = c+1
    return out + [J[0]]
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Feb 16 2025

A381127 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-SpellUnder dealing is used, then the resulting cards will be dealt in increasing order.

Original entry on oeis.org

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

Offset: 1

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

In Down-SpellUnder dealing, we deal the first card, then spell the positive integers starting from O-N-E, moving a card from the top of the deck underneath the deck for each letter in the English spelling of the number, followed by dealing or "putting down" the top card. So we start by dealing the first card, then putting 3 cards under because O-N-E has three letters, then we deal the next card. Then we put 3 cards under because T-W-O has three letters, then we deal a card. Then we put 5 cards under for T-H-R-E-E, and so on. This dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: DUUUDUUUDUUUUUD, where each "U" corresponds to putting a card “under” and each "D" corresponds to dealing a card “down”.
This card dealing can be thought of as a generalized version of the Josephus problem. In this version of the Josephus problem, we start by executing the first person, then spell the positive integers in increasing order, each time skipping past 1 person for each letter and executing the next person. The card in row n and column k is x if and only if in the corresponding Josephus problem with n people, the person numbered x is the k-th person eliminated.
Equivalently, each row of the corresponding Josephus triangle A381114 is an inverse permutation of the corresponding row of this triangle. The first column contains only ones, since person number 1 always dies first in the corresponding Josephus problem. The index of the largest number in row n is A381129(n), corresponding to the index of the freed person in the corresponding Josephus problem. The number of card moves that we need to take if we start with n cards is A381128(n).

			Triangle begins:
 1;
 1, 2;
 1, 3, 2;
 1, 2, 4, 3;
 1, 3, 5, 4, 2;
 1, 6, 4, 3, 2, 5;
 1, 5, 3, 6, 2, 4, 7;
 1, 3, 4, 5, 2, 7, 6, 8;
  ...
For n = 4, suppose there are four cards arranged in the order 1,2,4,3. Card 1 is dealt, and then three cards go under the deck because O-N-E has three letters. Now, the deck is ordered 2,4,3. Card 2 is dealt, and three cards go under because T-W-O has three letters. Now, the leftover deck is ordered 3,4. Card 3 is dealt, and five cards go under because T-H-R-E-E has five letters. Then card 4 is dealt. The dealt cards are in numerical order. Thus, the fourth row of the triangle is 1, 2, 4, 3.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381128, A381129.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def row(n):
    c, i, J, out = 1, 0, list(range(1, n+1)), []
    while len(J) > 1:
        q = J.pop(i)
        out.append(q)
        i = (i + f(c))%len(J)
        c = c+1
    out.append(J[0])
    return [out.index(j)+1 for j in list(range(1, n+1))]
print([e for n in range(1, 15) for e in row(n)]) # Michael S. Branicky, Feb 16 2025

A381128 The number of card moves required to deal n cards using Down-SpellUnder dealing.

Original entry on oeis.org

1, 5, 9, 15, 20, 25, 29, 35, 41, 46, 50, 57, 64, 73, 82, 90, 98, 108, 117, 126, 133, 143, 153, 165, 176, 187, 197, 209, 221, 232, 239, 249, 259, 271, 282, 293, 303, 315, 327, 338, 344, 353, 362, 373, 383, 393, 402, 413, 424, 434, 440, 449, 458, 469, 479, 489, 498, 509, 520, 530, 536, 545, 554, 565, 575, 585, 594, 605

Offset: 1

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

In Down-SpellUnder dealing, after dealing the i-th card we move a card from the top of the deck to the bottom for each letter in the English spelling of i. Then we deal the next card and proceed likewise. So we start by dealing card 1, then putting 3 cards under because O-N-E has three letters. Then, we deal the next card, and put three cards under 3 because T-W-O has three letters. We then deal again, put 5 under for T-H-R-E-E, and so on. This dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: DUUUDUUUDUUUUUD, where D means 'deal', and U means 'under'.

			The dealing pattern to deal four cards is DUUUDUUUDUUUUUD. It contains 15 letters, so a(4) = 15.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381127, A381129.

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(",", "").replace("-", ""))
moves_so_far = 0
l = []
for i in range(1, 41):
    moves_so_far += 1
    l += [str(moves_so_far)]
    moves_so_far += spell(i)
print(", ".join(l))

a(n) = A067278(n-1) + n.

A381129 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process, called Down-SpellUnder.

Original entry on oeis.org

1, 2, 2, 3, 3, 2, 7, 8, 4, 6, 4, 6, 7, 11, 10, 3, 14, 4, 17, 11, 3, 16, 7, 16, 7, 22, 2, 8, 24, 27, 7, 21, 13, 28, 30, 8, 3, 37, 12, 7, 8, 33, 7, 33, 44, 11, 32, 8, 6, 43, 2, 18, 49, 8, 32, 54, 26, 43, 44, 30, 40, 52, 26, 44, 8, 27, 60, 16, 11, 61, 70, 14, 58, 55

Offset: 1

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

Arrange n people numbered 1, 2, 3, ..., n in a circle, increasing clockwise. Execute the person numbered 1, then spell the letters of O-N-E, moving one person clockwise for each letter. Once you are done, eliminate the next person. Then, spell the letters of T-W-O; in other words, skip three people and eliminate the next person. Following this, spell the letters of T-H-R-E-E; in other words, skip five people and eliminate the next person. Continue until one person remains. The number of this person is a(n).

			Consider n = 4 people. The first person eliminated is number 1. This leaves the remaining people in the order 2, 3, 4. The second person eliminated is number 2; the people left are in the order 3, 4. The next person eliminated is numbered 4, leaving only the person numbered 3. Thus, a(4) = 3.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381127, A381128.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        q = J.pop(i)
        i = (i + f(c))%len(J)
        c = c+1
    return J[0]
print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Feb 16 2025

a(22) and beyond from Michael S. Branicky, Feb 16 2025

A385513 The numbers of people in the "SpellUnder-Down" variant of the Josephus problem such that the last person is freed.

Original entry on oeis.org

1, 6, 7, 105, 181, 215, 821, 1907, 3176, 23388, 55058

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Jul 01 2025

In SpellUnder-Down dealing, we spell the number of the next card, putting a card under for each letter in the number, then we deal the next card. So we start with putting 3 cards under, for O-N-E, then deal, then 3 under for T-W-O, then deal, then 5 under for T-H-R-E-E, then deal. The dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: UUUDUUUDUUUUUD. In the corresponding Josephus problem, we skip the next person for each under dealing, and eliminate the next person for each down dealing.
This sequence can be used in magic tricks with the SpellUnder-Down dealing pattern. The deck sizes in this sequence guarantee that after the dealing, the last card dealt is the one that was initially on the bottom.
The classical Josephus problem corresponds to under-down dealing. In this case, the last person is freed when the number of people is a power of 2 minus 1.
A naive probabilistic argument predicts the probability that A380204(k) = k is 1/k and expects this sequence to be infinite and distributed roughly as A002387. - Michael S. Branicky, Jul 24 2025

			Suppose there are 5 people in a circle. We start with skipping three people for O-N-E. After three people are skipped, the person number 4 is eliminated. The leftover people are 5,1,2,3 in order. Then we skip three people for T-W-O. The person number 3 eliminated, and the leftover people are 5,1,2 in order. Then we skip 5 people for T-H-R-E-E, and person number 2 is eliminated, and the leftover people are 5,1 in order. Then we skip four people for F-O-U-R. person number 5 is eliminated. Person 1 is freed. As person 1 is not last, 5 is NOT in this sequence.

Cf. A005589, A006257, A182459, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A385328.

{k | A380204(k) = k}. - Michael S. Branicky, Jul 24 2025

a(10)-a(11) from Michael S. Branicky, Jul 24 2025

cp's OEIS Frontend

A380201 Triangle T(n,k) read by rows, where row n is a permutation of numbers 1 through n, such that if a deck of n cards is prepared in this order, and SpellUnder-Down dealing is used, then the resulting cards are put down in increasing order.

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A380204 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process.

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A380246 Elimination order of the first person in a variation of the Josephus problem, where the number of skipped people correspond to the number of letters in consecutive numbers, called SpellUnder-Down.

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A380202 Number of card moves to deal n cards using the SpellUnder-Down dealing.

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A380247 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 the number of people skipped correspond to the number of letters in the next number in English alphabet.

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A381114 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in a variation of the Josephus elimination process for n people, where the number of people skipped is equal to the number of letters in the previous number's English name.

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A381127 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-SpellUnder dealing is used, then the resulting cards will be dealt in increasing order.

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A381128 The number of card moves required to deal n cards using Down-SpellUnder dealing.

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A381129 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process, called Down-SpellUnder.

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