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f(n)=3^f(n-1)-2^f(n-2)
#3
I must note that - although trivial - if the sequence goes above 1 the sequence explodes to oo. This implies that solving for 1 must indeed give my desired constant.

The value of my constant is 0,516553200406323...

This value reminds me of eulers constant ?

tommy1729
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Messages In This Thread
f(n)=3^f(n-1)-2^f(n-2) - by tommy1729 - 02/18/2013, 01:50 PM
RE: f(n)=3^f(n-1)-2^f(n-2) - by tommy1729 - 02/18/2013, 11:08 PM
RE: f(n)=3^f(n-1)-2^f(n-2) - by tommy1729 - 02/26/2013, 11:30 PM



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