Let ( n k ) be a sequence of natural numbers starting at n 1 = 2 and growing as follows n k ≤ n k + 1 ≤ n k 2 + 2As far as I know this sort of dynamical system may have no closed-form solution, but even if it has one, the = replaced in it by ≤ would include more sequences than the above recurrence allows.

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Let   (      n    k    ) be a sequence of natural numbers starting at       n    1    =  2 and growing as follows      n    k    ≤      n          k      +      1        ≤      n    k    2    +  2As far as I know this sort of dynamical system may have no closed-form solution, but even if it has one, the   = replaced in it by   ≤ would include more sequences than the above recurrence allows.

Hugo Bruce · Accepted Answer

For the sequence       u    k   such that       u    1    =      n    1   and       u          k      +      1        =      u    k    2    +  2, we can write            log      ⁡              u                  k          +          1                            2              k        +        1              =            log      ⁡              u              k                    2          k        +      1          2              k        +        1              log  ⁡  (  1  +      2          u      k      2        )hence we get            log      ⁡              u              n                    2          n        =            log      ⁡              n        1              2    +      ∑          k      =      1              n      −      1            1          2              k        +        1              log  ⁡  (  1  +      2          u      k      2        )As the series is clearly convergent,                     log        ⁡                  u                  n                            2              n              →    C  , with       C    =                  log        ⁡                  n          1                    2        +          ∑              k        =        1                ∞              1              2                  k          +          1                      log    ⁡    (    1    +          2              u        k        2              )  and   C can be well approximated.

bs1tuaz · Accepted Answer

&quot;Obviously&quot;,       n    j   is maximal if       n          k      +      1        =      n    k    2    +  2 for   k  =  1  ,  …  ,  j  −  1. In this case the sequence goes   2  ,  6  ,  38  ,  1446  ,  … and grows roughly   ∼      c            2      k       for suitable   c. For example,       n    k    ≤            3                2      k        −  2 follws quickly by induction, and       3   can be lowered to   ≈  1.576 except for small indices.

Let ( n k </msub> ) be a sequence of natural numbers starting at

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