[seqfan] Re: Observation on the divisors of 136
davidwwilson at comcast.net
Mon Jan 15 02:57:13 CET 2018
If n >= 1 is odd, and divisible by all of its odd prefixes, then n has one or two 1's.
Since n >= 1, n has at least one 1.
Suppose n has more than two 1's.
Then n has an odd proper prefix.
Therefore, let m be the longest odd proper prefix of n.
We can show that m > 1 and n = km + 1.
Hence m does not divide n.
Hence n is not divisible by all its odd prefixes, a contradiction.
So n has one or two 1's, QED.
If n >= 1 is divisible by all of its prefixes, then n has one or two 1's.
(Note: The numbers divisible by all their binary prefixes are given by A175381. Gardiner's numbers are a subset, divisible by all and only their binary prefixes.)
If n is divisible by all its prefixes, it is divisible by all its odd prefixes.
Let m be equal to n with its trailing 0's removed.
m is odd and divisible by all its odd prefixes.
By Lemma 1, m has one or two 1's.
But m and n have the same number of 1's, so n has one or two 1's, QED.
If n >= 1 is divisible by all of its prefixes, then one of:
(a) n = 2^k where 0 <= k.
(b) n = (2^j + 1)*2^k where 0 <= k < j.
Proof left to the reader, uses Lemma 2.
If the divisors of n are precisely the prefixes of n, then n is of one of the forms:
(a) n = 2^k, where k >= 0
(b) n = (2^(k + 1) + 1)*2^k, where k >= 0 and 2^(k + 1) + 1 is prime.
Proof left to the reader, uses Lemma 3.
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