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A374954 Positive integers k for which sqrt(k) < sqrt(p_1) + ... + sqrt(p_r), where p_1*...*p_r is the prime factorization of k.

%I A374954 #10 Aug 17 2024 23:27:05
%S A374954 4,6,8,9,10,12,14,15,16,18,20,22,24,28,32,36,40,48,64
%N A374954 Positive integers k for which sqrt(k) < sqrt(p_1) + ... + sqrt(p_r), where p_1*...*p_r is the prime factorization of k.
%C A374954 This sequence is finite. Proof: First, let's assume that p_1 = ... = p_r = p, i.e. k = p^r. Then sqrt(p^r) < r*sqrt(p) or p < r^(2/(r-1)) respectively must apply. This inequality is satisfied for p = 2 and 2 <= r <= 6 as well as for p = 3 and r = 2. k can therefore contain at most r = 6 prime factors and is not a prime. By examining the individual ways for the highest value of k as a function of r, we find k = 2*2*2*2*2*2 = 64 for r = 6, k = 2*2*2*2*3 = 48 for r = 5, 2*2*2*5 = 40 for r = 4, 2*2*7 = 28 for r = 3 and 2*11 = 22 for r = 2. Therefore, this sequence is finite and its terms lie between 4 and 64.
%e A374954 24 = 2*2*2*3 is in the sequence, because sqrt(24) < sqrt(2) + sqrt(2) + sqrt(2) + sqrt(3).
%p A374954 A374954:=proc(k)
%p A374954    local i,r,s,L;
%p A374954    if not isprime(k) then
%p A374954       L:=ifactors(k)[2];
%p A374954       r:=numelems(L);
%p A374954       s:=0;
%p A374954       for i to r do
%p A374954          s:=s+sqrt(L[i,1])*L[i,2]
%p A374954       od;
%p A374954       s:=evalf(s^2);
%p A374954       if k<s then
%p A374954          return k
%p A374954       fi;
%p A374954    fi;
%p A374954 end proc;
%p A374954 seq(A374954(k),k=4..64);
%Y A374954 Cf. A001414, A002808, A046343, A063538, A063539, A063762, A063763, A101550.
%K A374954 nonn,fini,full
%O A374954 1,1
%A A374954 _Felix Huber_, Jul 29 2024