List questions from recurrence-relations

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Closed form of $\int_{0}^1 \operatorname{Li}^2_n(x) dx$ for positive integer $n$

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Periodicity of recursive sequence

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Even and odd Hermite polynomials

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random Pell sequence

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Solve recurrence using generating functions if possible.

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If $u_{n+2}-4u_{n+1}+4u_n=0$ then why $u_n=A2^n+Bn2^n$?

Published on 29 Aug 2019 - 8:46

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Proof by induction that given the recurrence $b_{n+1} = b_n-\frac{b_n^2}{2}$, $\frac{1}{n+1}\leq b_n\leq\frac{2}{n+2}$

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Simplify system of 3 equations of 3 recursive functions

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Solve the recurrence relation $a_n=a_{n/2} +n +1$

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Alternating Fibonacci-like sequences are periodic?

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Where does the identity $\tau(nm) = \sum_{d \mid (m, n)} \mu(d) \tau(n/d) \tau(m/d)$ come from?

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Master Theorem gives different results from Iteration solving Recurrence Relation

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$\begin{cases} T(1) = 1 \quad \quad \quad \\ T(n) = 4T(n-1) + 3 \end{cases}$ using method of summation factors?

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Odd convergents of $\sqrt 2$ and Pythagorean triangles with consecutive legs

Published on 25 Mar 2026 - 19:11

List Question

Proof that the Fibonacci sequence grows exponentially

Closed form of $\int_{0}^1 \operatorname{Li}^2_n(x) dx$ for positive integer $n$

Periodicity of recursive sequence

Even and odd Hermite polynomials

random Pell sequence

Solve recurrence using generating functions if possible.

If $u_{n+2}-4u_{n+1}+4u_n=0$ then why $u_n=A2^n+Bn2^n$?

Proof by induction that given the recurrence $b_{n+1} = b_n-\frac{b_n^2}{2}$, $\frac{1}{n+1}\leq b_n\leq\frac{2}{n+2}$

Simplify system of 3 equations of 3 recursive functions

Solve the recurrence relation $a_n=a_{n/2} +n +1$

Alternating Fibonacci-like sequences are periodic?

Where does the identity $\tau(nm) = \sum_{d \mid (m, n)} \mu(d) \tau(n/d) \tau(m/d)$ come from?

Master Theorem gives different results from Iteration solving Recurrence Relation

$\begin{cases} T(1) = 1 \quad \quad \quad \\ T(n) = 4T(n-1) + 3 \end{cases}$ using method of summation factors?

Odd convergents of $\sqrt 2$ and Pythagorean triangles with consecutive legs

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