A385328 -id:A385328 - cp's OEIS frontend

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

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

A386305 Numbers of people such that the first person is freed in the variant of the Josephus problem in which one person is skipped, then one is eliminated, then two people are skipped and one eliminated, then three people are skipped and so on.

Original entry on oeis.org

1, 2, 3, 18, 22, 171, 195, 234, 1262, 2136, 6040, 42545, 353067, 1332099, 1447753, 2789475, 3635021, 7857445, 9224024, 17128159, 27666710, 29279638

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Aug 20 2025

This sequence can also be described in terms of "AP dealing", in which one deals a deck of N cards into a new deck by moving one card to the bottom, dealing out the next card on top of the new deck, moving two cards to the bottom, etc. This sequence consists of all the deck sizes such that the top card remains the same after AP dealing.

Numbers k such that A291317(k) = 1.

			Suppose there are 5 people in a circle. We start with skipping one person and eliminating the next (person number 2). The leftover people are 3,4,5,1 in order. Then we skip two people and eliminate person number 5. The leftover people are 1,3,4 in order. Then we skip three people and person number 1 is eliminated. The leftover people are 3,4 in order. Then we skip four people and eliminate person number 3. Person 4 is freed. As person 1 is not freed, 5 is NOT in this sequence.

Cf. A081614, A291317, A385328, A386312, A386643.

def F(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        i = (i + c) % len(J)
        q = J.pop(i)
        c = c + 1
    return J[0]
print([n for n in range(1, 100000) if F(n) == 1])

a(20)-a(22) from Jinyuan Wang, Aug 31 2025

cp's OEIS Frontend

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

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