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How can I define a sequence from a recurrence?

Posted 9 years ago

I have some trouble with defining a sequence from a recurrence in Wolfram Alpha.

If I type u(n+1) = (2^(u(n)))/(n+1), u(0) = 1, Wolfram seems to understand what I mean, and gives the correct first terms:

n | 0 | 1 | 2 | 3 | 4
u(n) | 1 | 2 | 2 | 1.33333 | 0.629961

However, when I type u(n+1) = (3^(u(n)))/(n+1), u(0) = 1 (the only difference is the 2 being replaced by a 3), then I get totally unexpected results:

n | 0 | 1 | 2 | 3 | 4
u(n) | 0.6 | 0.322197 | 0.20353 | 0.156321 | 0.131929

How can u(0) = 0.6 when I give u(0) = 1? I would have expected to get 1, 3^1/1=3, 3^3/2, etc. in the same way that the first sequence values were 1, 2^1/1 = 2, 2^2/2 = 2, 2^2/3 = 1.33333 etc.

What am I doing wrong, and how to explain the result in the second case?

Thanks!

POSTED BY: Jean F.
2 Replies
Posted 9 years ago

Thank you for your answer... I realize I might have given the wrong tags to my question, it is not a mathematica program, just the input that I give to Wolfram Alpha... ( Sequence with 2 and Sequence with 3 )

Could this overflow that you mention be the reason of the strange output in the second case? Or am I doing a stupid mistake somewhere?

POSTED BY: Jean F.
POSTED BY: Christian Neel
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