OFFSET
19,3
COMMENTS
This sequence was inspired by the following puzzle from the book Mathematical Puzzles and Curiosities. If you choose the answer to this question at random, what is the probability you will be correct? A:25%, B:50%, C:60%, D:25%.
If we look at parts as percentages, the sequence counts partition such that there exists part p, such that if you choose a part uniformly at random, the part p is chosen with probability p%.
REFERENCES
I. David, T. Khovanova, and Y. Shpilman, Mathematical Puzzles and Curiosities, World Scientific, 2026.
LINKS
Sean A. Irvine, Table of n, a(n) for n = 19..100
EXAMPLE
For n = 19, the partition (10,1,1,1,1,1,1,1,1,1) has 10 parts. The part equaling 10 appears 10% of the time. Thus, this partition is counted by the sequence, implying that a(19) >= 1. For n < 19, no such partition exists. For n = 19. This is the only such partition.
For n = 21, there are two partitions with the given property: (10,2,2,1,1,1,1,1,1,1,1,1) and (10,3,1,1,1,1,1,1,1,1). Hence, a(21) = 2.
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Tanya Khovanova and PRIMES STEP senior group, May 15 2026
STATUS
approved
