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.

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 22, 24, 28, 32, 36, 40, 48, 64

Offset: 1

Author

Felix Huber, Jul 29 2024

Keywords

• nonn
• fini
• full

Comments

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.

Examples

			24 = 2*2*2*3 is in the sequence, because sqrt(24) < sqrt(2) + sqrt(2) + sqrt(2) + sqrt(3).
		

Crossrefs

Cf. A001414, A002808, A046343, A063538, A063539, A063762, A063763, A101550.

Programs

Maple

A374954:=proc(k)
   local i,r,s,L;
   if not isprime(k) then
      L:=ifactors(k)[2];
      r:=numelems(L);
      s:=0;
      for i to r do
         s:=s+sqrt(L[i,1])*L[i,2]
      od;
      s:=evalf(s^2);
      if kA374954(k),k=4..64);