OFFSET
1,2
COMMENTS
a(n) >= A000720(n).
a(n) ~ n/log n (Erdős-Sárközy-Sós). Best bounds currently are due to Pach-Vizer.
a(n+1)-a(n) is either 0 or 1 for any n. (Is equal to 1 when n+1 is prime.)
LINKS
P. Erdős, A. Sárközy, and V. T. Sós, On Product Representations of Powers, I, Europ. J. Combinatorics 16 (1995), 567-588.
P. Pach and M. Vizer, Improved Lower Bounds for Multiplicative Square-Free Sequences, The Electronic Journal of Combinatorics, Volume 30, Issue 4 (2023), P4.31.
EXAMPLE
a(7)=6, because the set {1,2,3,4,5,7} has no four distinct elements multiplying to a square, but {1,2,3,4,5,6,7} has 1*2*3*6 = 6^2.
PROG
(Python)
from math import isqrt
def is_square(n):
return isqrt(n) ** 2 == n
def valid_subset(A):
length = len(A)
for i in range(length):
for j in range(i + 1, length):
for k in range(j + 1, length):
for l in range(k + 1, length):
if is_square(A[i] * A[j] * A[k] * A[l]):
return False
return True
def largest_subset_size(N):
from itertools import combinations
for size in reversed(range(1, N + 1)):
for subset in combinations(range(1, N + 1), size):
if valid_subset(subset):
return size
for N in range(1, 23):
print(largest_subset_size(N))
(Python)
from math import prod
from functools import lru_cache
from itertools import combinations
from sympy.ntheory.primetest import is_square
@lru_cache(maxsize=None)
def A373119(n):
if n==1: return 1
i = A373119(n-1)+1
if sum(1 for p in combinations(range(1, n), 3) if is_square(n*prod(p))) > 0:
a = [set(p) for p in combinations(range(1, n+1), 4) if is_square(prod(p))]
for q in combinations(range(1, n), i-1):
t = set(q)|{n}
if not any(s<=t for s in a):
return i
else:
return i-1
else:
return i # Chai Wah Wu, May 30 2024
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Terence Tao, May 26 2024
EXTENSIONS
a(22)-a(37) from Michael S. Branicky, May 26 2024
a(38)-a(63) from Martin Ehrenstein, May 27 2024
STATUS
approved