cp's OEIS Frontend

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).

%I A382286 #52 Jun 19 2025 20:57:43
%S A382286 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,
%T A382286 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,
%U A382286 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
%N 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).
%C A382286 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.
%C A382286 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).
%C A382286 a(n) can also factor numbers as follow:
%C A382286 gcd(n,A000196(a(n)*n)+1-sqrt((A000196(a(n)*n)+1)^2-a(n)*n))
%C A382286 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
%C A382286 Or
%C A382286 gcd(n,(2*A000196(a(n)*n)+1-sqrt((2*A000196(a(n)*n)+1)^2-4*a(n)*n))/2)
%C A382286 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
%F A382286 a(n) < A048760(n) iff n is a composite number.
%F A382286 a(n) = A048760(n) iff n is a prime number.
%F A382286 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)).
%F A382286 a(n^2) = 1.
%F A382286 If we combines the first two properties we get floor(a(n)/A048760(n))=A010051(n),
%F A382286 And partial sum of floor(a(n)/A048760(n)) from 2 to N = A000720(N).
%e A382286 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):
%e A382286 a(1)=a(2)=a(3)=1 since  C(n)=0, for n=1,2,3.
%e A382286 a(4)=1 (and generally a(n^2)=1).
%e A382286 a(5)=4, since C(k*5) > 0 for k=1 to 3 and C(4*k)=0.
%e A382286 a(6)=1 since C(6)=0.
%e A382286 a(7)=4 since C(k*7) > 0 for k=1 to 3 and C(4*7)=0.
%e A382286 a(8)=2 since C(8)=1 and C(2*8)=0.
%e A382286 a(9)=1
%e A382286 a(11)=9 since C(k*11) > 0 for k=1 to 8 and C(9*11)=0.
%o A382286 (PARI) d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
%o A382286 a(n) = my(k=1); while (sqrtint(n*k/d(n*k)) != sqrtint(d(n*k)), k++); k; \\ _Michel Marcus_, Mar 21 2025
%Y A382286 Cf. A000196, A033676, A033677, A048760.
%K A382286 nonn
%O A382286 1,5
%A A382286 _Hassan Baloui_, Mar 20 2025