# Tag Info

5

Here is an approach different from the one in the link of Will's answer: think $2$-adically. Since $(1+\sqrt{-2})^4 = 1 + (-8-4\sqrt{-2}) = 1 + \gamma$, with $|\gamma|_2 = 1/(4\sqrt{2}) < 1/2$, we can view $a_m$ as a $2$-adic analytic function by fixing $m \bmod 4$: picking $r \in \{0,1,2,3\}$ and writing $m = 4q+r$ where $q$ runs through the natural ...

4

EEDDIITT: you have all the solutions. Proved in Y. Bugeaud and T.N. Shorey, On the number of solutions of the generalized Ramanujan-Nagell equation, I. Jour. reine angew. Math. vol. 539 (2001) pages 55-74. Preprint is item number 92 at http://www.math.tifr.res.in/~shorey/ Statement found in N. Saradha and Anitha Srivanasan, Genaralized ...

4

In fact $x_{n+1}=a_n^2$, where $\{a_n\}$ are determined by the recurrence relation $$a_{n+1}=n a_n+a_{n-1}$$ with the initial conditions $a_0=1$, $a_1=0$. This can be exprerimentally discovered using the online encyclopedia of integer sequences, and then the proof can be easily carried out by induction. Proof: The only nontrivial induction step: ...

2

We start from the convolution $$1=\sum_{i=1}^j K_i \binom{n+j}{j-i}$$ then multiply both sides by $x^{j-1}$ and sum from $j=1$ to $\infty$. For the RHS we obtain $$\sum_{j=1}^\infty \sum_{i=1}^j K_i \binom{n+j}{j-i}x^{j-1}=\sum_{i=1}^\infty \sum_{j=i}^\infty K_i \binom{n+j}{j-i}x^{j-1}$$ where we have reversed the order of summation. The inner sum over $j$ ...

2

I think you misinterpret repeating pattern. We know that $a_n$ only depends on $a_{n-1}$ and $a_{n-2}$. Now, we start with some given $a_1$ and $a_2$ (modulo $10$, since only the last digit is of interest). There are only $10\cdot 10=100$ possible combinations for a pair $(a_{n},a_{n+1})$. Since the sequence $(a_n)_{n\in \mathbb N}$ is infinite, we know by ...

2

If $f(z)=\sum_{n=0}^\infty \frac{a_n}{n!}z^n$ is a solution to the differential equation, then $a_i$ is a solution to the related recurrence relation, and visa versa. You can just do the arithmetic. So if $b_n=a_1r_1^n+a_2r_2^n$ then $$f(z)=\sum_{n=0}^\infty \frac{b_n}{n!}z^n = a_1e^{r_1z} + a_2e^{r_2z}$$

1

The solution $y = ar^n$ is the solution to $y_{n + 1} = r y_n$, and not $\Delta y_n = r y_n$. In discrete calculus, $\Delta$ is the analogue (or at least one analogue) of $\text D \equiv \dfrac{d}{dx}$. But $y_{n + 1} = \text{E}\,y_n$, where $\text{E}$ is the shift operator, and that is not analogous to $\text D$. You are asking why we guess this solution. ...

1

An alternative of trying to 'guess' a solution is to derive it using the "generating function" approach. (To continue the ODE comparison, it's the equivalent of solving via Laplace transforms.) Suppose we start with the simple recurrence relation $a_{n}=ra_{n-1}$. Consider the formal power series in $x$ defined by $A(x)=\sum_{n=0} a_n x^n$. (By formal, I ...

Only top voted, non community-wiki answers of a minimum length are eligible