cp's OEIS Frontend

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).

%I A381054 #43 Jun 05 2025 02:04:20
%S A381054 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,
%T A381054 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,
%U A381054 1,1,49,3,3,1,49,1,49,13,1,3,1,1,49,1
%N 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).
%C A381054 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.
%C A381054 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).
%C A381054 a(n) can also factor numbers as follow :(to be proven)
%C A381054 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.
%C A381054 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:
%C A381054 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).
%C A381054 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
%H A381054 Robert Israel, <a href="/A381054/b381054.txt">Table of n, a(n) for n = 1..10000</a>
%F A381054 Conjecture: a(n) < (A000196(n)-1)^2 iff n>=9 is a composite number.
%F A381054 Conjecture: a(n) = (A000196(n)-1)^2 iff n>=9 is a prime number.
%e A381054 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 A381054 a(1)=5 since C(k) > 1 for k=1 to 4 and C(5)=1
%e A381054 a(2)=4 since C(2*k) > 1 for k=1 to 3 and C(2*4)=1
%e A381054 a(3)=4 since C(3*k) > 1 for k=1 to 3 and C(3*4)=1
%e A381054 a(13*113)=7 because the first multiples of the prime factors which are exactly one square apart are 7*13 and 113.
%e A381054 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)
%p A381054 d:= proc(n) max(select(t -> t^2 <= n, numtheory:-divisors(n))) end proc:
%p A381054 f:= proc(n) local k,t;
%p A381054      for k from 1 do
%p A381054        t:= d(n*k);
%p A381054        if floor(sqrt(n*k/t)) - floor(sqrt(t)) = 1 then return k fi
%p A381054      od
%p A381054 end proc:
%p A381054 map(f, [$1..100]); # _Robert Israel_, Jun 04 2025
%o A381054 (PARI) d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
%o A381054 a(n) = my(k=1,dnk=d(n*k)); while (sqrtint(n*k/dnk) - sqrtint(dnk) != 1, k++;dnk=d(n*k)); k;
%Y A381054 Cf. A000196, A033676, A033677, A048760, A382286, A383115.
%K A381054 nonn,look
%O A381054 1,1
%A A381054 _Hassan Baloui_, Apr 14 2025