JavaScript (Node.js), 47 bytes

• -1 by Arnauld.

f=x=>x<6?x:Math.min(f(x+1>>1)+2,f(x-=x/3<<1)+3)

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Python 2, 49 bytes

f=lambda x:x*(x<6)or min(f(x-x/2)+2,f(x-x/3*2)+3)

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I don't really ensure this function is correct. But it at least pass all testcases given.

After I ask this solution on Math SE, caduk had provided a nice prove to it. You may read the prove by caduk on Math SE.

Anyway, the formula is

$$ f(x)= \begin{cases} n & n<6 \\ \min \left\{f\left(n-\left\lfloor\frac{n}{2}\right\rfloor\right)+2, f\left(n-2\left\lfloor\frac{n}{3}\right\rfloor\right)+3 \right\} & \text{otherwise} \end{cases} $$

JavaScript (Node.js), 68 bytes

f=(n,m,i=n>>1)=>i?Math.min(f(n-i,i)-~(i!=m),f(n,m,i-1),f(n-1,m)+1):n

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Charcoal, 51 38 bytes

⊞υ⟦⁰⟧ＦＮ⊞υＥ⁺²ι⌊Ｅ⮌υ⌊Ｅμ⁺ξ⁺∨⁼π⊕ν⊕ν¬⁼πκＩ⌊⊟υ

Try it online! Link is to verbose version of code. Explanation: Uses dynamic programming on the basic recurrence relation, which may be equivalent to @l4m2's approach, but I can't read it.

⊞υ⟦⁰⟧

Start with 0 keystrokes to type 0 as.

ＦＮ

Loop over all numbers up to n.

⊞υＥ⁺²ι⌊Ｅ⮌υ⌊Ｅμ⁺ξ⁺∨⁼π⊕ν⊕ν¬⁼πκ

Loop over all clipboard sizes up to n and calculate the minimum number of keystrokes needed for the current number while finishing with that clipboard size. All previous entries are considered; 1 is added to the entry if it has a different clipboard size, as that means another ^C is needed, plus the number of extra as is added, unless the clipboard is of exactly the right size, in which case 1 is added, as in that case only a ^V is needed. (The values where more than one a is needed are unnecessary but it's golfier to calculate and discard them.)

Ｉ⌊⊟υ

Output the minimum number of keystrokes needed to output n in unary.

A port of @tsh's formula using dynamic programming takes 33 bytes:

Ｆ⊕Ｎ⊞υ⎇‹ι⁶ι⌊Ｅ²⁺⁺²κ§υ⁻ι×⊕κ÷ι⁺²κＩ⌊⊟υ

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Swift 5.9, 48 bytes

let f={x in x<6 ?x:2+min(f(x-x/2),1+f(x-x/3*2))}

A direct port of @tsh's Python 2 answer, because I don't know how else to do this concisely.

Python3, 327 bytes

R=range
def f(n):
 q,s,r=[(0,0)],[],[]
 for a,c in q:
  if a==n:r+=[c];continue
  if r and c>=min(r):continue
  if a<3:q+=[(a+1,c+1)]
  else:
   U=[(a+1,c+1)]
   for i in R(3,min(a,n-a)+1):
    for j in R(1,(n-a)//i+1):U+=[(a+i*j,c+1+j)]
   for A,B in U:
    if all(k<A or K>B for k,K in s):q+=[(A,B)];s+=[(A,B)]
 return min(r)

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Ruby, 42 bytes

f=->x{x<6?x:2-[-f[x-x/2],~f[x-x/3*2]].max}

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Not very original, basically a port of tsh's answer.

Scala 3, 64 bytes

A direct port of @tsh's Python 2 answer.

def f(x:Int):Int=if(x<6)x else 2+Math.min(f(x-x/2),1+f(x-x/3*2))

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