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j.c.
j.c.
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The sequence defined by $a_0=a_1 =1$ and $$ a_n = \frac{1}{n-1}\sum_{i=0}^{n-1}a_i^2, \quad n > 1 $$ fails to be integer for the first time at $a_{44}$. Why??

You can verify the statement by computing the sequence mod 43 (see more commentary here (day 5, problem 3)). That's not a very satisfying answer though. Is there a good reason for this behavior?

The sequence defined by $a_0=a_1 =1$ and $$ a_n = \frac{1}{n-1}\sum_{i=0}^{n-1}a_i^2, \quad n > 1 $$ fails to be integer for the first time at $a_{44}$. Why??

You can verify the statement by computing the sequence mod 43 (see more commentary here). That's not a very satisfying answer though. Is there a good reason for this behavior?

The sequence defined by $a_0=a_1 =1$ and $$ a_n = \frac{1}{n-1}\sum_{i=0}^{n-1}a_i^2, \quad n > 1 $$ fails to be integer for the first time at $a_{44}$. Why??

You can verify the statement by computing the sequence mod 43 (see more commentary here (day 5, problem 3)). That's not a very satisfying answer though. Is there a good reason for this behavior?

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Yufei Zhao
Yufei Zhao
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Simple recurrence that fails to be integer for the first time at the 44th term

The sequence defined by $a_0=a_1 =1$ and $$ a_n = \frac{1}{n-1}\sum_{i=0}^{n-1}a_i^2, \quad n > 1 $$ fails to be integer for the first time at $a_{44}$. Why??

You can verify the statement by computing the sequence mod 43 (see more commentary here). That's not a very satisfying answer though. Is there a good reason for this behavior?