[seqfan] Re: Observation on the divisors of 136
Brad Klee
bradklee at gmail.com
Sat Dec 30 11:03:35 CET 2017
Hi Jeremy and Seqfans,
After: http://list.seqfan.eu/pipermail/seqfan/2017-December/018222.html
If I understand correctly, 136 belongs to a probably finite ( cf. A019434,
A092506 ) class of factorizations with a(n) = 2^k*p, p prime, p=2^(k+1)+1 .
This begs the question: are there any examples with factor 2^k and two
additional prime factors?
Hypothesize that: a=2^k*p1*p2, p1<p2 . We require p1=2^(k+1)+1 and
p2=p1*2^(k+1)+1, which yields condition:
p2 - p1 = (p1-1)*2^(k+1) = ( p1 - 1 )^2
p2 = p1^2 - p1 + 1
Then check values p1 = 5, 17, 257, 65537 and find that none of the
corresponding p2 are prime. It seems your proposed sequence will just be
powers of two interspersed with a few extra values related to Fermat Primes
or the primes of A092506,
2^(2^n - 1)*(2^(2^n) + 1) : 10, 136, 32896, 2147516416 .
As for adding 1's rather than 0's, after a = 3, consider a = 3 * 7, but
binary decomposition of 21 includes contradictory zero digit values. This
precludes any odd values appearing in the sequence other than 3. Not sure
though, have I missed something or mistaken the definitions?
Best wishes finding interesting new sequences in 2018, 2019, 2020, etc...
Cheers,
Brad
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