A381114 -id:A381114 - cp's OEIS frontend

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.

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

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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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