Timeline for Number of walks on integer lattice with self-edge at zero
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2, 2016 at 1:49 | vote | accept | Raisin Bread | ||
| Jun 26, 2016 at 15:59 | comment | added | Johann Cigler | @MaxAlekseyev: Yes I know this. But I want to know if there is a simple direct argument in terms of walks. | |
| Jun 26, 2016 at 14:16 | comment | added | Max Alekseyev | @JohannCigler: Let $f(x)$ be the g.f. for A098615. One can easily verify that $\frac{f(x)-f(-x)}{2}=\frac{x}{1-5x^2}=x+5x^3+5^2x^5+5^3x^7+\dots$. | |
| Jun 24, 2016 at 15:32 | answer | added | Max Alekseyev | timeline score: 4 | |
| Jun 24, 2016 at 10:37 | history | edited | Raisin Bread | CC BY-SA 3.0 |
added 8 characters in body
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| Jun 24, 2016 at 8:19 | comment | added | Johann Cigler | Let $a(n)$ denote the number of walks from $0$ to $0$ in $n$ steps. Is there a simple reason why $a(2n+1)=5^n$? | |
| Jun 24, 2016 at 2:11 | history | asked | Raisin Bread | CC BY-SA 3.0 |