This is incorrect. Integer factorization is NP-intermediate. Very much “in NP”.
https://en.m.wikipedia.org/wiki/NP-intermediate
Also, saying factorization lacks “complexity” because sieves exist misunderstands both concepts.
> In order to talk about complexity classes such as P, NP, and co-NP, the problem has to be stated as a decision problem.
> Decision problem (Integer factorization) — For every natural numbers n and k, does n have a factor smaller than k besides 1?
> It is known to be in both NP and co-NP
https://en.m.wikipedia.org/wiki/Integer_factorization#:~:tex....
Eg https://www.connellybarnes.com/documents/factoring.pdf
That is supported by the second wiki link you provide, which has "Unsolved problem in computer science: Can integer factorization be solved in polynomial time on a classical computer?" in a box at the side. https://en.m.wikipedia.org/w/index.php?title=Integer_factori...Integer factorization is unsolved and it’s decision problem is in NP.
IF’s decision problem’s complexity “may be in NP” because the question of whether P equalling NP is unknown.
Meaning IF is NP, but may well be P if P=NP. If P!=NP then IF is NP.