Timeline for The sequence $G(n,k)=G(n-2,k)+G(n,k-2)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2019 at 7:49 | comment | added | Alexey Ustinov | @joro Yes, because in this case the answer is the sum of 2 trangles. First one has 0, 0, 0, ... and 0, A, 2A, 3A,... on it's sides, second one has 0, B, 2B,... and 0, 0, 0,... respectively. In both cases you'll get Pascal triangle multiplied by A or by B and shifted. For n and k not divisable by A and B one should correct boundary conditions using additional Pascal triangles. | |
| Nov 5, 2019 at 7:38 | comment | added | joro | Is the following generalization solvable: for natural A,B define $G(n,k)=G(n-A,k)+G(n,k-B)$? | |
| Nov 4, 2019 at 16:25 | comment | added | Alexey Ustinov | @joro Sorry, probably it was not clear. In this case Pascal triangle fills the first coordinate quarter: $C(n,0)=C(0,k)=1,$ $C(n,k)=C(n-1,k)+C(n,k-1).$ | |
| Nov 4, 2019 at 15:56 | comment | added | Alexey Ustinov | @joro No, everything is like in Pascal triangle $(n,k \ge 0).$ | |
| Nov 4, 2019 at 15:52 | comment | added | joro | Thank you. Do you require $k \le n$? | |
| Nov 4, 2019 at 15:03 | history | answered | Alexey Ustinov | CC BY-SA 4.0 |