A381054 a(n) is the least k such that floor(sqrt(n*k/d(n*k))) - floor(sqrt(d(n*k))) = 1, where d(k) is the largest divisor of k which is <= sqrt(k).

5, 4, 4, 2, 1, 2, 1, 1, 2, 1, 4, 1, 4, 1, 1, 5, 9, 1, 9, 3, 1, 2, 9, 3, 2, 2, 2, 3, 16, 2, 16, 3, 2, 5, 2, 2, 25, 5, 2, 2, 25, 2, 25, 1, 1, 5, 25, 2, 2, 1, 3, 1, 36, 1, 1, 2, 3, 8, 36, 1, 36, 8, 1, 3, 1, 1, 49, 3, 3, 1, 49, 1, 49, 13, 1, 3, 1, 1, 49, 1

Offset: 1

Author

Hassan Baloui, Apr 14 2025

Keywords

• nonn
• look

Comments

In the case of semiprime numbers n=p*q, the sequence a(n) choses the first two multiples of the prime factors p and q such that there is exactly one square between these multiples (say k*p, m*q) 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 chose the first two multiples say (k_i*d_i, m_i*n/d_i) which are exactly one square apart, then a(n) is the smallest product k_i*m_i and k_i*d_i is the largest divisor of k_i*m_i*n which is <= sqrt(k_i*m_i*n), and i runs over half the number of divisors of n (since only pairs are considered).

a(n) can also factor numbers as follow :(to be proven)

One pair of the roots of the five quadratic equations X^2-(2*A000196(a(n)*n)+{1,2,3,4 or 5})*X+a(n)*n =0 is a pair of positive integers say (X_1,X_2) then gcd(X_1,n) and gcd(X_2,n) are nontrivial divisors of n.

A382286 is the first sequence of an infinite sequence of sequences and the current sequence a(n) is the second one. These sequences can be defined as:

For each positive integer m, the corresponding sequence evaluated at a positive integer n is the least k such that floor(sqrt(n*k/d(n*k))) - floor(sqrt(d(n*k))) = m, where d(k) is the largest divisor of k which is <= sqrt(k).

If we denote for each m the corresponding sequence a_m(n), it is conjectured that (a_m(n)=(A000196(n)-m)^2 iff n is a prime > m^2+1) the cases m=0:A382286,1 are answered affirmatively

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)=5 since C(k) > 1 for k=1 to 4 and C(5)=1
a(2)=4 since C(2*k) > 1 for k=1 to 3 and C(2*4)=1
a(3)=4 since C(3*k) > 1 for k=1 to 3 and C(3*4)=1
a(13*113)=7 because the first multiples of the prime factors which are exactly one square apart are 7*13 and 113.
a(13*114)=3 because from the pairs of conjugate divisors which are (1,1482), (2,741), (3,494), (6,247), (13,114), (19,78),(26,57) and (38,39) the first pair that have multiples exactly one square apart is (26,57) and these multiples are (3*26,57) and 57 is the largest divisor of 3*13*114 <= sqrt(3*13*114)
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000196, A033676, A033677, A048760, A382286, A383115.

Programs

Maple

d:= proc(n) max(select(t -> t^2 <= n, numtheory:-divisors(n))) end proc:
f:= proc(n) local k,t;
     for k from 1 do
       t:= d(n*k);
       if floor(sqrt(n*k/t)) - floor(sqrt(t)) = 1 then return k fi
     od
end proc:
map(f, [$1..100]); # Robert Israel, Jun 04 2025

PARI

d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
a(n) = my(k=1,dnk=d(n*k)); while (sqrtint(n*k/dnk) - sqrtint(dnk) != 1, k++;dnk=d(n*k)); k;

Formula

Conjecture: a(n) < (A000196(n)-1)^2 iff n>=9 is a composite number.

Conjecture: a(n) = (A000196(n)-1)^2 iff n>=9 is a prime number.