Timeline for How this expression leads to the given sequence
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10, 2013 at 23:55 | comment | added | Gottfried Helms | Carlo, see my accidental beautiful observation in my new answer... | |
| Dec 9, 2013 at 16:34 | comment | added | Carlo Beenakker | @GottfriedHelms --- my understanding is that $x$ is a variable, not an index; the function $a(n,x)$ is a polynomial of degree $n-1$ in the variable $x$; you are seeking a triangle of coefficients, such that the $n$-th row of the triangle has at the $m$-th position the coefficient of the term $x^m$ in $a(n,x)$. As I worked it out above, you see that this indeed gives the series mentioned by the OP. In particular, the fourth row has elements $1,13,18$, which are the coefficients of $a(4,x)=x+13x^2+18x^3$. | |
| Dec 9, 2013 at 14:54 | comment | added | Gottfried Helms | if x means the column-index, then all elements in the matrix $A$ in the first column are zero by $a(2,0)=0 \cdot a(1,0) = 0 $. Is that really meant? It would contradict the table-view in OEIS where the first column is 1. Are the indexes for S2 and A are thought differently (for S2 beginning at 0 and for A beginning at 1 ?) | |
| Jun 26, 2013 at 12:03 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |