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URL: https://oeis.org/A392081

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A392081
Number of distinct orders of derangements of n items.
1
1, 0, 1, 1, 2, 2, 4, 4, 5, 7, 9, 11, 12, 16, 19, 22, 26, 29, 36, 38, 46, 51, 61, 64, 75, 83, 96, 106, 117, 132, 145, 162, 179, 199, 220, 240, 259, 287, 319, 344, 376, 407, 445, 482, 525, 568, 618, 663, 714, 773, 836, 901, 966, 1039, 1110, 1195, 1282, 1382, 1476
OFFSET
0,5
COMMENTS
By rough experimentation, for n <= 175, a(n) is approximately equal to (1/16)*n*e^(sqrt(n)*Pi/4). This is similar in form to that of Hardy and Ramanujan's approximation for the integer partition function.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..175 (first 101 terms from Sean Lipton)
FORMULA
a(n) = |{lcm(P) for P in [partitions of n with no part equal to 1]}|.
EXAMPLE
a(9) = 7 since there are 7 possible orders:
lcm(9) = 9
lcm(2,7) = 14
lcm(3,6) = lcm(2,2,2,3) = 6
lcm(4,5) = 20
lcm(2,2,5) = 10
lcm(2,3,4) = 12
lcm(3,3,3) = 3
MATHEMATICA
a[0]=1; a[n_]:=CountDistinct[ LCM@@@Select[IntegerPartitions[n], ContainsNone[#, {1}]&]]; Array[a, 55, 0] (* James C. McMahon, Feb 25 2026 *)
PROG
(Python)
from functools import cache
from math import lcm
@cache
def orders(n):
if n == 1:
return set()
output = {n}
for i in range(2, n//2+1):
for o in orders(n-i):
output.add(lcm(i, o))
return output
for i in range(1, 50):
print(f"{i}: {len(orders(i))}")
(PARI) a(n) = if(n < 2, n==0, my(S=[]); forpart(p=n, S=setunion(S, [lcm(Vec(p))]), [2, n]); #S); \\ Andrew Howroyd, Feb 17 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Sean Lipton, Feb 17 2026
STATUS
approved