added 632 characters in body

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edited Jul 5 at 18:41

JavaScript (Node.js), 51 bytes

f=(n,i=6)=>n>1&&Math.min(i+f(n/--i),i>2?f(n,i):1/0)

JavaScript (Node.js), 50 bytes by Shaggy porting xnor

f=n=>n>1&&Math.min(...[2,3,4,5].map(x=>x-~f(n/x)))

f=(n,i=6)=>n>1&&Math.min(i+f(n/--i),i>2?f(n,i):1/0)

f=(n,i=6)=>n>1&&Math.min(i+f(n/--i),i>2?f(n,i):1/0)

f=n=>n>1&&Math.min(...[2,3,4,5].map(x=>x-~f(n/x)))

deleted 93 characters in body

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edited Jul 5 at 11:19

f=n=>n>1&&Mathf=(n,i=6)=>n>1&&Math.min(fi+f(n/2--i)+3,i>2?f(n/3)+4,f(n/4i)+5,f(n:1/5)+60)

f(n/6)+7 can be f(n/2/3)+3+4 so up to f(n/5)+6 is enough. Test case don't show it though.Try it online!

f=n=>n>1&&Math.min(f(n/2)+3,f(n/3)+4,f(n/4)+5,f(n/5)+6)

f(n/6)+7 can be f(n/2/3)+3+4 so up to f(n/5)+6 is enough. Test case don't show it though.

f=(n,i=6)=>n>1&&Math.min(i+f(n/--i),i>2?f(n,i):1/0)

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created Jul 5 at 11:08

f=n=>n>1&&Math.min(f(n/2)+3,f(n/3)+4,f(n/4)+5,f(n/5)+6)

f(n/6)+7 can be f(n/2/3)+3+4 so up to f(n/5)+6 is enough. Test case don't show it though.