[seqfan] Re: Is A049117 practically the same as A227946?

[eric41293 at comcast.net]

Due to the multiplicativity of phi, and phi(2^n) = 2^(n-1),
taking the odd part makes no difference as to the number of steps
needed to reach a power of two, and, if we have at least one iteration,
the power of two will be 1 if we take the odd part.
So, the only difference between
A227944(n) (number of steps of odd part of phi to reach 1) and
A049115(n) (number of steps of phi to reach power of 2)
is at powers of 2: A0227944(2^n) = 1 while A049115(2^n) = 0.
Also, since phi(n) < n, A049117(n-1) is the smallest number
which requires n steps to reach a power of 2.
>From this, we see that

A227946(n) = A49117(n-1), n >= 2.

----- Original Message ----- 
From: "Alonso Del Arte" <alonso.delarte at gmail.com> 
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu> 
Sent: Saturday, October 12, 2013 8:22:29 PM 
Subject: [seqfan] Is A049117 practically the same as A227946? 

Both have keyword:more, but A049117 has a few more terms than A227946. 

I tried out phi(14348907) = 9565938, the odd part of which is 4782969; then 
phi(4782969) = 3188646, odd part of which is 1594323; phi(1594323) = 
1062882... Then I realized 14348907 = 3^15 and obviously phi(3^n) = 2 * 
3^(n - 1). 

So unless there's a smaller number fitting the bill, 14348907 might be the 
next term of A227946, assuming we decide to not recycle A227946. 

Al 


-- 
Alonso del Arte 
Author at SmashWords.com<https://www.smashwords.com/profile/view/AlonsoDelarte> 
Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte> 

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