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 May 30th, 2017, 05:11 AM #11 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 I rewrite the post correctly : You can generate a sequence of highly composite numbers using Euler totient phi(n). You start by any odd number m(1) You compute phi((m(1)) m(2)=m(1)+phi((m(1)) You compute phi((m(2)) m(3)=m(2)+phi((m(2)) ...... m(k)=m(k-1)+phi((m(k-1)) Some odd numbers will give you an infinite sequence highly composite. Can you find the odd numbers giving a sequence with 1.000 consecutive composite numbers? Can you prove why is it working?
 May 30th, 2017, 05:15 AM #12 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Im not asking anyone for computing the problem. Im waiting for deep explanation and more than that finding another non trivial sequences giving consecutive composites. 3,6,9,12,.... is a trivial one for example.
 May 30th, 2017, 06:34 AM #13 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 I start with 3 I don't get 6 m(1) = 3 phi(m(1)) = 2 m(2) = 3 + 2 = 5 phi(m(2)) = 4 m(3) = 5 + 4 = 9 phi(9) = 6 m(4) = 9 + 6 = 15 phi(15) = 8 m(5) = 15 + 8 = 23 phi(23) = 22 m(6) = 23 + 22 = 45 am I doing it wrong?
May 30th, 2017, 06:57 AM   #14
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Quote:
 Originally Posted by agentredlum I start with 3 I don't get 6 m(1) = 3 phi(m(1)) = 2 m(2) = 3 + 2 = 5 phi(m(2)) = 4 m(3) = 5 + 4 = 9 phi(9) = 6 m(4) = 9 + 6 = 15 phi(15) = 8 m(5) = 15 + 8 = 23 phi(23) = 22 m(6) = 23 + 22 = 45 am I doing it wrong?
No. You are correct.
I did not say all the odd numbers.
I said some odd numbers.
Read me first and then try to find some of those numbers.

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