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  • \$\begingroup\$ You should be able to save a byte on your fibonacci function g with g=n=>n<3||g(n-1)+g(n-2) \$\endgroup\$ Commented Apr 6, 2024 at 12:07
  • \$\begingroup\$ @noodleman Yeah originally I had a raw g(n) somewhere but I think the only one left is in the stroke-width adjustment version. \$\endgroup\$ Commented Apr 6, 2024 at 12:40
  • \$\begingroup\$ [e=g(n)*6+1,e*4-2,e+g(n-2)*6+1] saves a few bytes \$\endgroup\$ Commented Apr 6, 2024 at 20:38
  • \$\begingroup\$ Also doing the initial call to h with s=-3 saves a few more bytes \$\endgroup\$ Commented Apr 6, 2024 at 20:41
  • \$\begingroup\$ -5 more by replacing g with g=n=>i=n<3?6:g(n-1)+g(n-2), removing all *6s, replacing *24 with *4, and adding a /6 in the map call in h (code) \$\endgroup\$ Commented Apr 6, 2024 at 23:06