a(m) =a(m-1)+1/(1+1/a(m-1))

[Leroy Quet]

Sent to sci.math as well (but they usually have not much to say....):

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Let a(1) = 1 ;

Let, for m = any positive integer,
                       1
a(m+1) = a(m) + ---------------
                  1 + 1 /a(m)

In linear-mode:

a(m+1) = a(m) + 1 /(1 + 1 /a(m)).

This sequence begins:

1, 3/2, 21/10, 861/310,...

The numerators follow the recursion, for m >= 2,

n(m) = n(m-1) (n(m-1) + 2*d(m-1))

And the denominators...

d(m) = d(m-1) (n(m-1) + d(m-1))

The numerators in terms of only previous numerators:

n(m+1) = n(m)^2 + n(m)^3 /(2n(m-1)^2) - n(m)n(m-1)^2/2

And the denominators...

d(m+1) = d(m)^2 + d(m)^3 /d(m-1)^2 - d(m)d(m-1)^2

So, are there closed forms for the a's, the n's, or the d's?

Of course, every n(m) divides n(m+1), and d(m) divides d(m+1).

Thanks,
Leroy Quet