A373178 Cardinality of the largest subset of {1,...,n} such that no five distinct elements of this subset multiply to a square.

1, 2, 3, 4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 10, 10, 10, 11, 11, 12, 12, 13, 13, 14, 15, 15, 16, 17, 18, 19, 19, 20, 20, 20, 21, 21, 21, 22, 23, 23, 24, 25, 26, 27, 28, 29, 30, 31, 31, 31, 31, 32, 33, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 42, 42, 43, 44, 45, 46, 46

Offset

1,2

Comments

a(n) >= A373114(n).

The limiting value of a(n)/n is unknown, but (if it exists), it is strictly less than 1, and at least A246849 ~ 0.828499... (see cited paper of Tao).

a(n+1)-a(n) is either 0 or 1 for any n.

If "five" is replaced by "one", "two", "three", "four", or "odd number of", one obtains A028391, A013928, A372306, A373119, A373114 respectively.

Links

Table of n, a(n) for n=1..70.

Terence Tao, On product representations of squares, arXiv:2405.11610 [math.NT], May 2024.

Example

a(8)=7, because the set {1,2,3,4,5,7,8} has no five distinct elements multiplying to a square, but {1,2,3,4,5,6,7,8} has 1*2*3*4*6 = 12^2.

Prog

(Python)
from math import isqrt
def is_square(n):
    return isqrt(n) ** 2 == n
def precompute_tuples(N):
    tuples = []
    for i in range(1, N + 1):
        for j in range(i + 1, N + 1):
            for k in range(j + 1, N + 1):
                for l in range(k + 1, N + 1):
                    for m in range(l + 1, N + 1):
                        if is_square(i * j * k * l * m):
                            tuples.append((i, j, k, l, m))
    return tuples
def valid_subset(A, tuples):
    set_A = set(A)
    for i, j, k, l, m in tuples:
        if i in set_A and j in set_A and k in set_A and l in set_A and m in set_A:
            return False
    return True
def largest_subset_size(N, tuples):
    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, tuples):
                return size
for N in range(1, 26):
    print(largest_subset_size(N, precompute_tuples(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 A373178(n):
    if n==1: return 1
    i = A373178(n-1)+1
    if sum(1 for p in combinations(range(1, n), 4) if is_square(n*prod(p))) > 0:
        a = [set(p) for p in combinations(range(1, n+1), 5) 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

Similar to A028391, A013928, A372306, A373119. Lower bounded by A373114.

Sequence in context: A125051 A064067 A202306 * A275579 A198292 A020892

Adjacent sequences: A373175 A373176 A373177 * A373179 A373180 A373181

Keyword

nonn

Author

Terence Tao, May 26 2024

Extensions

a(26)-a(38) from Michael S. Branicky, May 27 2024

a(39)-a(47) from Michael S. Branicky, May 30 2024

a(48)-a(70) from Martin Ehrenstein, May 31 2024

Status

approved