The natural progression of (even perfect numbers)*3

[Dubslow]

Originally Posted by 10metreh

2σ(a)-a

Is there an OEIS entry for this sequence (starting with f(0)=3)?

Or, as a corollary, the secondary sequence g(n) where g(n) is the first index of f(n) which is divisible by the nth Mersenne prime?

Such that we know at least the first 873 terms of f(n); meanwhile g(1) = 0, g(2) = 2, g(3) = 29, with g(5) unknown as evidenced by this thread?

(Of course Mp | f(n) does not guarantee that the nth term loses the Mp driver, but it does mean that the sequence diverges from the f sequence.)

[10metreh]

Originally Posted by Dubslow

Is there an OEIS entry for this sequence (starting with f(0)=3)?

Yes: oeis.org/A146556

Or, as a corollary, the secondary sequence g(n) where g(n) is the first index of f(n) which is divisible by the nth Mersenne prime?

Such that we know at least the first 873 terms of f(n); meanwhile g(1) = 0, g(2) = 2, g(3) = 29, with g(5) unknown as evidenced by this thread?

Seems not - actually g(3) = 30, and g(4) = 492. The only OEIS match for these terms is oeis.org/A143414 which is clearly unrelated.

[Dubslow]

Originally Posted by 10metreh

Yes: oeis.org/A146556

Looks like the "Formula" is wrong? The "+ a(n-1)" at the beginning isn't right.

Originally Posted by 10metreh

Seems not - actually g(3) = 30, and g(4) = 492. The only OEIS match for these terms is oeis.org/A143414 which is clearly unrelated.

Oops, yes I had an off-by-one in my local recreation.

[10metreh]

Originally Posted by Dubslow

Looks like the "Formula" is wrong? The "+ a(n-1)" at the beginning isn't right.

It is correct - look carefully at the brackets.

[garambois]

Originally Posted by Dubslow

Originally Posted by 10metreh View Post
2σ(a)-a

Is there an OEIS entry for these sequence (starting with f(0)=3) ?

And what's about odd perfect numbers, if they exist ?



Starting with f(0)=2, we have : 2, 4, 10, 26, 58, 122, 250, 686 ...
There is no OEIS entry for this sequence.

Just for fun, I calculated the first 348 terms of this sequence. Curiously, we do not find the prime number 3 in the decomposition into prime numbers of these terms.

You can see here :

http://www.aliquotes.com/parfait_2_z.txt

[Dubslow]

Originally Posted by garambois

Curiously, we do not find the prime number 3 in the decomposition into prime numbers of these terms.

Really? I'd even say that's more than a bit curious. Has anyone looked at this sequence before?

[LaurV]

As we start with a number n which is either 1 or 2 (mod 3) and 0 (mod 2), it means this is either 2 or 4 (mod 6). If its sigma is 0 (mod 3), then we have the next term t=2*sigma-n is either t=2*0-1=2 (mod 3) or t=2*0-2=1 (mod 3). So the only combinations that would result in 0 (mod 3) are either (A) n=4 (mod 6) and sigma(n)=2 (mod 3) or (B) n=2 (mod 6) and sigma(n)=1 (mod 3). We can expand some formulae for sigma and see if we can get this, considering that if n=1 (mod 3) than its prime factors which are 2 (mod 3) can be grouped two-by-two in those products for sigma, etc.