20180402, 23:56  #1 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
The natural progression of (even perfect numbers)*3
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.) 
20180403, 00:12  #2  
Nov 2008
2×3^{3}×43 Posts 
Quote:
Quote:
Last fiddled with by 10metreh on 20180403 at 00:12 

20180403, 00:26  #3  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Quote:
Quote:


20180403, 12:25  #4 
Nov 2008
2·3^{3}·43 Posts 

20180509, 12:37  #5  
"Garambois JeanLuc"
Oct 2011
France
1000101100_{2} Posts 
Quote:
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 

20180515, 12:43  #6 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 

20180515, 15:59  #7 
Romulan Interpreter
Jun 2011
Thailand
9,377 Posts 
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*sigman is either t=2*01=2 (mod 3) or t=2*02=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 twobytwo in those products for sigma, etc.

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