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I know the dimension of a Koch snowflake (log4/log3), but what numbers do I have to put in to obtain the dimension of a Sierpinski fractal?

Deschele Schilder
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    As stated on Wikipedia (https://en.wikipedia.org/wiki/Sierpinski_triangle#Properties), $\frac{\log 3}{\log 2}$. – Qiaochu Yuan May 01 '16 at 17:58
  • Do you mean the Sierpinski carpet or the Sierpinski triangle? – Arthur May 01 '16 at 17:59
  • The triangle is the one I mean (I didn´t know there was a carpet too). – Deschele Schilder May 01 '16 at 18:08
  • It's basically the same thing, only instead of removing the middle fourth of a triangle, and then repeat on the three remaining small triangles, you remove the middle ninth of a square, and repeat on the eight remaining small squares. – Arthur May 01 '16 at 18:11

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