A385328 The number of people in a variation of the Josephus problem when the first person is freed and the elimination process is to skip the number of people equaling the number of letters in consecutive numbers, then eliminate the next person.

1, 2, 5, 26, 50, 82, 857, 1114, 3340, 3733, 3777, 11023, 12960, 17992, 47253, 329414, 367572, 382265

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Jun 25 2025, Jun 25 2025, Jul 06 2025

Keywords

• nonn
• more
• word

Comments

This sequence uses the US spelling. [Specifically, A005589. - Michael S. Branicky, Jul 23 2025]

This sequence can be used in magic tricks with SpellUnderDown dealing pattern. The first three people are skipped, corresponding to three letters in O-N-E, and the next person is eliminated. Then, three people are skipped corresponding to three letters in T-W-O, and the next person is eliminated. Then, 5 people are skipped, corresponding to 5 letters in T-H-R-E-E.

The deck sizes in this sequence guarantee that after the dealing, the last card is the one that was initially on top.

A naive probabilistic argument predicts the probability that A380204(k) = 1 is 1/k and expects this sequence to be infinite and distributed roughly as A002387. - Michael S. Branicky, Jul 23 2025

Examples

			Suppose there are 5 people in a circle. After three people are skipped (for O-N-E), the person number 4 is eliminated. The leftover people are 5,1,2,3 in order. Then three people are skipped (for T-W-O), and person number 3 is eliminated. The leftover people are 5,1,2 in order. Then 5 people are skipped (for T-H-R-E-E), and person 2 is eliminated. The leftover people are 5,1 in order. Then 4 people are skipped (for F-O-U-R), and person number 5 is eliminated. Person 1 is freed. Thus, 5 is in this sequence.
		

Crossrefs

Cf. A000079, A005589, A380201, A380202, A380204, A380246, A380247.

Formula

{k | A380204(k) = 1}. - Michael S. Branicky, Jul 23 2025

Extensions

a(15)-a(18) from Michael S. Branicky, Jul 23 2025