A382286 a(n) is the least k such that floor(sqrt(n*k/d(n*k))) = floor(sqrt(d(n*k))), where d(k) is the largest divisor of k which is <= sqrt(k).

1, 1, 1, 1, 4, 1, 4, 2, 1, 2, 9, 2, 9, 2, 2, 1, 16, 2, 16, 1, 2, 5, 16, 1, 1, 5, 3, 1, 25, 1, 25, 1, 3, 8, 1, 1, 36, 8, 3, 1, 36, 1, 36, 3, 2, 8, 36, 1, 1, 2, 6, 3, 49, 2, 2, 1, 6, 13, 49, 2, 49, 13, 2, 1, 2, 2, 64, 4, 6, 2, 64, 2, 64, 18, 2, 4, 2, 2, 64, 4, 1, 18, 81, 2, 4, 18, 9, 4, 81

Offset: 1

Author

Hassan Baloui, Mar 20 2025

Keywords

• nonn

Comments

In the case of semiprime numbers n=p*q, the sequence a(n) chooses the first two multiples of the prime factors p and q such that these multiples (say k*p, m*q) are between two consecutive squares, then a(n)=m*k.

In the case of any composite number n, for each pair of conjugate divisors (d_i, n/d_i) we choose the first two multiples say (k_i*d_i, m_i*n/d_i) which are between two consecutive squares. Then a(n) is the smallest product k_i*m_i where i runs over half the number of divisors of n (since only pairs are considered).

a(n) can also factor numbers as follow:

gcd(n,A000196(a(n)*n)+1-sqrt((A000196(a(n)*n)+1)^2-a(n)*n))

gcd(n,A000196(a(n)*n)+1+sqrt((A000196(a(n)*n)+1)^2-a(n)*n)) are proper divisors of n if sqrt(a(n)*n)-A000196(a(n)*n) < 1/2

Or

gcd(n,(2*A000196(a(n)*n)+1-sqrt((2*A000196(a(n)*n)+1)^2-4*a(n)*n))/2)

gcd(n,(2*A000196(a(n)*n)+1+sqrt((2*A000196(a(n)*n)+1)^2-4*a(n)*n))/2) are proper divisors of n if sqrt(a(n)*n)-A000196(a(n)*n) > 1/2

Examples

			Let C(k)=floor(sqrt(k/d(k)))-floor(sqrt(d(k))), where d(k) is the largest divisor of k which is <= sqrt(k):
a(1)=a(2)=a(3)=1 since  C(n)=0, for n=1,2,3.
a(4)=1 (and generally a(n^2)=1).
a(5)=4, since C(k*5) > 0 for k=1 to 3 and C(4*k)=0.
a(6)=1 since C(6)=0.
a(7)=4 since C(k*7) > 0 for k=1 to 3 and C(4*7)=0.
a(8)=2 since C(8)=1 and C(2*8)=0.
a(9)=1
a(11)=9 since C(k*11) > 0 for k=1 to 8 and C(9*11)=0.
		

Crossrefs

Cf. A000196, A033676, A033677, A048760.

Programs

PARI

d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
a(n) = my(k=1); while (sqrtint(n*k/d(n*k)) != sqrtint(d(n*k)), k++); k; \\ Michel Marcus, Mar 21 2025

Formula

a(n) < A048760(n) iff n is a composite number.

a(n) = A048760(n) iff n is a prime number.

a(p^(2*m+1)) = p*a(p)/A048760(p) for any pair (p,m) of a prime number p and a natural number m (except the case (2,2)).

a(n^2) = 1.

If we combines the first two properties we get floor(a(n)/A048760(n))=A010051(n),

And partial sum of floor(a(n)/A048760(n)) from 2 to N = A000720(N).