[seqfan] Re: n = number of divisors of 2^n -1
Jack Brennen
jfb at brennen.net
Wed Apr 21 20:46:09 CEST 2010
An obvious superset of your sequence, and one that is almost certainly
not finite, would be the set of n such that the number of divisors of
2^n-1 is divisible by n. The sequence, and the quotients d(2^n-1)/n,
assuming that I didn't make any mistakes:
1 1
2 1
4 1
6 1
8 1
12 2
16 1
24 4
32 1
48 16
64 2
96 64
128 4
192 256
256 8
...
Your sequence would be the subset which have quotient 1.
It's quite likely that this superset sequence contains all 2^n and 6*2^n
for nonnegative n, but does it contain any other values?
Leroy Quet wrote:
> I get the sequence where a positive integer n is included if
> n = the number of divisors of 2^n - 1,
> begins:
> 1,2,4,6,8,16
>
> (Every term from a(5) = 8 on is a multiple of 4, due to Catalan's conjecture.)
>
> Is this sequence finite? Is it in the OEIS?
>
> Thanks,
> Leroy Quet
>
> [ ( [ ([( [ ( ([[o0Oo0Ooo0Oo(0)oO0ooO0oO0o]]) ) ] )]) ] ) ]
>
>
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>
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