1. n 1 2 3 4 5 6 7 8 9 An 42
429
7436
218348
= 2  3  7
= 3  11  13
= 22  11  132
= 22  132  17  19
= 23  13  172  192
= 22  5  172  193  23 1 –1

2. n 1 2 3 4 5 6 7 8 9 An 42
429
7436
218348
very suspicious 1 –1
= 2  3  7
= 3  11  13
= 22  11  132
= 22  132  17  19
= 23  13  172  192
= 22  5  172  193  23

3. There is exactly one 1 in the first row2 3 4 5 6 7 8 9 An 42
429
7436
218348
There is exactly one 1 in the first row 1 –1

4. There is exactly one 1 in the first row2 3 4 5 6 7 8 9 An
1+1
2+3+2 …
There is exactly one 1 in the first row 1 –1

5. 1 1 –1

6. There is exactly one 1 in the first row2 3 4 5 6 7 8 9 An
1+1
2+3+2 … 1 –1

7. There is exactly one 1 in the first row2 3 4 5 6 7 8 9 An
1+1
2+3+2 … 1 –1

8. 1 1 –1 + + +

9. 1 1 –1 + + +

10. 1
1 2/2 1
2 2/ /2 2
7 2/ /2 7
42 2/ /2 42
429 2/ /2 429 1 –1

11. 1
1 2/2 1
2 2/ /2 2
7 2/ / /2 7
42 2/ / / /2 42
429 2/ / / / /2 429 1 –1

12. 2/2
2/ /2
2/ / /2
2/ / / /2
2/ / / / /2 1 –1

13. 1+1 1+2 1+1 2+3 1+3 1+1 3+4 3+6 1+4 1+1 4+5 6+10 4+10 1+5 Numerators: 1 –1

14. Numerators: 1+1 1+1 1+2 1+1 2+3 1+3 1+1 3+4 3+6 1+4
Numerators: 1 –1
Conjecture 1:

15. Conjecture 2 (corollary of Conjecture 1):–1
Conjecture 2 (corollary of Conjecture 1):

16. 1 –1
Richard Stanley M.I.T.

17. George Andrews, Penn State1 –1
Richard Stanley M.I.T.
Andrews’ Theorem: the number of descending plane partitions of size n is
George Andrews, Penn State

18. All you have to do is find a 1-to-1 correspondence between n by n alternating sign matrices and descending plane partitions of size n, and conjecture 2 will be proven! 1 –1

19. What is a descending plane partition?
All you have to do is find a 1-to-1 correspondence between n by n alternating sign matrices and descending plane partitions of size n, and conjecture 2 will be proven! 1 –1
What is a descending plane partition?