Number of 3 digit numbers in AP or GP How many three digit numbers have the property that their digits taken from left to right form an arithmetic or geometric progression?
Please check all the cases.
 A: Let us count. The list of geometric progressions is short, though not as short as I first believed!  There are the obvious  $1, 2, 4$, and $2, 4, 8$, and $1, 3, 9$, and their reverses. Then there is the less obvious $4,6,9$ and its reverse, which I had missed.  And of course $a, a, a$ where $a$ is any of $1$ to $9$. These last also happen to be arithmetic progressions.  What about "common ratio" $0$? We will go along with Wikipedia's definition and not allow that. 
Now let's count the arithmetic progressions. There are the $9$ with common difference $0$. 
There are $7$ increasing ones with common difference $1$, and $8$ decreasing ones, since $0$ can be the final digit in that case.
There are $5$ increasing ones with common difference $2$, and $6$ decreasing ones.
There are $3$ increasing ones with common difference $3$, and $4$ decreasing ones.
There is $1$ increasing one with common difference $4$, and there are $2$ decreasing ones.
If we decide to forget about the sequences $a, a, a$ we get a count of $44$, for we listed $6$ geometric progressions and $36$ arithmetic progressions. Adding in the sequences $a, a, a$, which we definitely should, since they are indeed in both categories, gives us $53$.
For whatever it is worth, Wikipedia does not allow common ratio $0$.  If we accept that, the correct count is $53$.  
A: 111 123 135 147 159 210 222 234 246 
258 321 333 345 357 369 420 432 444 
456 468 531 543 555 567 579 630 642 
654 666 678 741 753 765 777 789 840 
852 864 876 888 951 963 975 987 999 
should be all the arithmetic progressions, that's 45 right there, so 42 can't be right. 
124 139 248 421 469 842 931 964 so I get 53. What did I miss? 
Are we counting 100 200 ... 900 as geometric progressions with constant ratio zero? 
EDIT: For what it's worth, the number of 3-term geometric progressions with entries from $\{{1,2,\dots,n\}}$ is $(6/\pi^2)n\log n+O(n)$. There's a proof in my paper, Trifectas in geometric progression, Austral. Math. Soc. Gaz. 35 (2008) 189–194, available online at
http://www.austms.org.au/Publ/Gazette/2008/Jul08/TechPaperMyerson.pdf. In the paper, I leave counting the 3-term arithmetic progressions as an exercise. 
