OFFSET
1,2
COMMENTS
This Josephus problem is related to under-down-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 skips one number and the next two numbers are 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.
LINKS
Eric Huang, Tanya Khovanova, Timur Kilybayev, Ryan Li, Brandon Ni, Leone Seidel, Samarth Sharma, Nathan Sheffield, Vivek Varanasi, Alice Yin, Boya Yun, and William Zelevinsky, Card Dealing Math, arXiv:2509.11395 [math.NT], 2025. See p. 17.
EXAMPLE
Consider 4 people in a circle. Initially, person number 1 is skipped and persons number 2 and 3 are eliminated. The remaining people are now in order 4, 1. Then 4 is skipped and 1 is eliminated. Then, 4 is eliminated. Thus, the fourth row of the triangle is 2, 3, 1, 4, the order of elimination.
Triangle begins;
1;
2, 1;
2, 3, 1;
2, 3, 1, 4;
2, 3, 5, 1, 4;
2, 3, 5, 6, 4, 1;
2, 3, 5, 6, 1, 4, 7;
2, 3, 5, 6, 8, 1, 7, 4;
2, 3, 5, 6, 8, 9, 4, 7, 1;
...
PROG
(Python)
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
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Apr 28 2025
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Apr 14 2025
STATUS
approved