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Code Golf

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    \$\begingroup\$ The classic approach, but I think the parameters can be better tuned. Specifically, I think that f x=(x,x,x) at the cost of one n. in the last line gives the optimal score for this overall structure. \$\endgroup\$ Commented Feb 12, 2016 at 9:11
  • \$\begingroup\$ I don't know Haskell, so I could be off base here, but I'll point out that 4^(4^4) is less than 3^(4^5) \$\endgroup\$ Commented Feb 12, 2016 at 18:01
  • \$\begingroup\$ Pretty sure the 4th n. is going to be larger. 2^18 vs 3 * (2^16) unless I made a mistake calculating the original exponentiation: 2^(4^9) vs. 3^((4^8)*3) \$\endgroup\$ Commented Feb 12, 2016 at 18:20
  • \$\begingroup\$ No, @PeterTaylor is correct: 2^(4^9) = 16^(4^8) < 27^(4^8) = 3^(4^8 ⋅ 3). \$\endgroup\$ Commented Jun 22, 2017 at 23:53
  • \$\begingroup\$ (,) (or (,,)) can be used to save some bytes and improve the score by using more ns. \$\endgroup\$ Commented Aug 8, 2018 at 12:29