The natural progression of (even perfect numbers)*3
[QUOTE=10metreh;484026]2σ(a)a[/QUOTE]
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.) 
[QUOTE=Dubslow;484083]Is there an OEIS entry for this sequence (starting with f(0)=3)?[/quote]
Yes: [URL]oeis.org/A146556[/URL] [quote]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?[/quote] Seems not  actually g(3) = 30, and g(4) = 492. The only OEIS match for these terms is [url]oeis.org/A143414[/url] which is clearly unrelated. 
[QUOTE=10metreh;484085]Yes: [URL]oeis.org/A146556[/URL]
[/quote] Looks like the "Formula" is wrong? The "+ a(n1)" at the beginning isn't right. [QUOTE=10metreh;484085] Seems not  actually g(3) = 30, and g(4) = 492. The only OEIS match for these terms is [url]oeis.org/A143414[/url] which is clearly unrelated.[/QUOTE] Oops, yes I had an offbyone in my local recreation. 
[QUOTE=Dubslow;484088]Looks like the "Formula" is wrong? The "+ a(n1)" at the beginning isn't right.[/QUOTE]
It is correct  look carefully at the brackets. 
[QUOTE=Dubslow;484083]
Originally Posted by 10metreh View Post 2σ(a)a Is there an OEIS entry for these sequence (starting with f(0)=3) ? [/QUOTE] And what's about odd perfect numbers, if they exist ? :smile: 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 : [URL="http://www.aliquotes.com/parfait_2_z.txt"]http://www.aliquotes.com/parfait_2_z.txt[/URL] 
[QUOTE=garambois;487274] Curiously, we do not find the prime number 3 in the decomposition into prime numbers of these terms.[/QUOTE]
Really? I'd even say that's more than a bit curious. Has anyone looked at this sequence before? 
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 [URL="https://en.wikipedia.org/wiki/Divisor_function#Formulas_at_prime_powers"]formulae for sigma[/URL] 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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