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A394247
Maximum number of parts i with multiplicity i possible in a partition of n.
6
0, 1, 0, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
OFFSET
0,6
COMMENTS
This sequence was inspired by a puzzle, "IQ Tests" from the book Mathematical Puzzles and Curiosities. Consider a multiple-choice question: "How many correct answer choices are there: (a) 1, (b) 1, (c) 2, (d) 2?" Answer 2 can be considered correct as it appears twice.
REFERENCES
I. David, T. Khovanova, and Y. Shpilman, Mathematical Puzzles and Curiosities, World Scientific, 2026, p.5.
LINKS
FORMULA
If n = A000330(k), then a(n) = k. Otherwise, if A000330(k) + k < n <= A000330(k+1) + k+1, then a(n) = k.
a(n) = A328806(n) - 1. - Alois P. Heinz, May 08 2026
EXAMPLE
Consider the 7 partitions of 5: (1,1,1,1,1),(1,1,1,2),(1,2,2),(1,1,3),(2,3),(1,4), and (5). Only one partition (1,4) has one distinct number appearing the same number of times as itself. Only one partition (1,2,2) has two distinct numbers appearing the same number of times as themselves. The remaining partitions do not have parts appearing the same number of times as themselves. Thus, the maximum number of distinct parts appearing the same number of times as their values, implying that a(5)=2.
MAPLE
b:= proc(n, i, t) option remember; `if`(n=0, t, `if`(i<1, 0,
max(seq(b(n-i*j, i-1, t+`if`(i=j, 1, 0)), j=0..n/i))))
end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..107); # Alois P. Heinz, Mar 26 2026
KEYWORD
nonn
AUTHOR
Tanya Khovanova and PRIMES STEP senior group, Mar 13 2026
STATUS
approved