[seqfan] Re: Confused by Collatz-Fibonacci numbers

[Frank Adams-Watters]

I have edited it. Alonso, you were being confused by the use of F in the definition, which was intended to mean this sequence, not the Fibonacci numbers. I have changed it to a.

Tony's formula is correct. The example should help show why it is correct.

Franklin T. Adams-Watters

-----Original Message-----
From: Eric Schmidt <eric41293 at comcast.net>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Fri, Apr 3, 2015 12:45 am
Subject: [seqfan] Re: Confused by Collatz-Fibonacci numbers


On 4/2/2015 8:48 PM, Alonso Del Arte wrote:
> I'm still perplexed by the
Collatz-Fibonacci numbers, A174429. By Tony's
> formula from 2010, I get a(3) =
21. But by the formula in the title, I get
> a(3) = 60. Maybe I've made some
simple mistake of arithmetic or logic. Can
> someone take a look to see if they
don't or if they also see an
> inconsistency?

For me the formula in the title
agrees with the given terms, including 
a(3) = 21. I used the following Sage
code.

def collatz(n) :
     if n%2==0 : return n//2
     return 3*n+1

def a(n)
:
     if n==1 or n==2 : return 1
     return a(collatz(n)) +
a(collatz(collatz(n)))

I get the same results from Tony's formula.

-- 
Eric
Schmidt

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