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+--- Thread: f(n)=3^f(n-1)-2^f(n-2) (/showthread.php?tid=772)
f(n)=3^f(n-1)-2^f(n-2) - tommy1729 - 02/18/2013
Consider f(n)=3^f(n-1)-2^f(n-2) with f(1)=0 and f(2)=a.
For a>>1.51... the sequence goes to oo.
For a<<1.51... the sequence goes to 0.
Can we express the constant x=1.51... ?
Is a sine type wave possible for f(2)= x ?
What happens with complex numbers ?
RE: f(n)=3^f(n-1)-2^f(n-2) - tommy1729 - 02/18/2013
Notice that 1.51... is NOT a real repelling fixpoint of x=3^x-2^x.
The only fixpoints of x=3^x-2^x on the extended real line are 0,1 and +oo.
The value 1 seems to suggest that f(2)=1.51... might make the sequence oscillate around 1.
Assuming the above it might work to compute 1.51... by starting at value 1 and doing the iterations backwards until we reach approximately 0. But of course nicer ways should be available.
I have not seen this sequence investigated before and it seems it might require new ideas, unless I overlooked something trivial.
I must add that this thread also deserves some credit from my student mick who some of you have already met on MSE.
Mick and I have given eachother permission to post eachothers (or common ) ideas concerning some math subjects including dynamical systems and tetration.
RE: f(n)=3^f(n-1)-2^f(n-2) - tommy1729 - 02/26/2013
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