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May 30th, 2017, 04:11 AM   #11
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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?
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May 30th, 2017, 04:15 AM   #12
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I`m not asking anyone for computing the problem.
I`m 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.
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May 30th, 2017, 05:34 AM   #13
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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?
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May 30th, 2017, 05:57 AM   #14
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Quote:
Originally Posted by agentredlum View Post
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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