Kevin Ryde: For your step function f(n) which Andrew said, the superseeker proposes f(n) = A220114(2*n)+1.  Looks about right.  Might double-check "nontrivial divisor" condition here though.

Enric Reverter i Bigas: I don't know any function for this sequence with input n. I've found a nice one that works till n = 20 but fails for n >20, so if you think is not convenient tu publish it on the basis a(n) = f(a(n-1)), the only possibility is to eliminate this draft. If I find in a future a function as  required, probably I shall revert to this matter.

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Andrew Howroyd: Is there a sequence that gives the largest integer m such that m = ((2x * n) /(x+1)) - x, with x a positive nontrivial divisor of m? (input n instead of a(n)). Similarly one for the smallest?

Enric Reverter i Bigas: Answer to Andrew Howroyd
For the largest integer, I don’t know any. For the smallest, Mathematica has found this one which works up to a(100):
a(n) = m = 6 + (2^(-2+n/2))*( 3 + 2√2 + ((-1)^(1+n))*(-3+2√2))
Best regards

Andrew Howroyd: The point I was trying to make, is that it would be preferable, in my opinion, to submit the sequence for f(n) prior to the one that is a(n)=f(a(n-1)).

Enric Reverter i Bigas: Yes, of course, I would like it too.

N. J. A. Sloane: Perhaps you should start by submitting the sequence that Howroyd suggests, and then we can come back to this on.

Michel Marcus: with x a positive nontrivial divisor of a(n) : should rather be:  with x a positive nontrivial divisor of m ?

Michel Marcus: replaced p with m because p sounds like prime