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n 1 2 3 4 5 6 7 8 9. A n 1 2 7 42 429 7436 218348 10850216 911835460. = 2 3 7. = 3 11 13. = 2 2 11 13 2. = 2 2 13 2 17 19. = 2 3 13 17 2 19 2. = 2 2 5 17 2 19 3 23. 1 0 0 –1 1 0 –1 0 1. n 1 2 3 4 5 6 7 8 9. A n 1

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## 1 0 0 –1 1 0 –1 0 1

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**n**1 2 3 4 5 6 7 8 9 An 1 2 7 42 429 7436 218348 10850216 911835460 = 2 3 7 = 3 11 13 = 22 11 132 = 22 132 17 19 = 23 13 172 192 = 22 5 172 193 23 1 0 0 –1 1 0 –1 0 1**n**1 2 3 4 5 6 7 8 9 An 1 2 7 42 429 7436 218348 10850216 911835460 very suspicious 1 0 0 –1 1 0 –1 0 1 = 2 3 7 = 3 11 13 = 22 11 132 = 22 132 17 19 = 23 13 172 192 = 22 5 172 193 23**n**1 2 3 4 5 6 7 8 9 An 1 2 7 42 429 7436 218348 10850216 911835460 There is exactly one 1 in the first row 1 0 0 –1 1 0 –1 0 1**n**1 2 3 4 5 6 7 8 9 An 1 1+1 2+3+2 7+14+14+7 42+105+… There is exactly one 1 in the first row 1 0 0 –1 1 0 –1 0 1**1**1 1 2 3 2 7 14 14 7 42 105 135 105 42 429 1287 2002 2002 1287 429 1 0 0 –1 1 0 –1 0 1**There is exactly one 1 in the first row**n 1 2 3 4 5 6 7 8 9 An 1 1+1 2+3+2 7+14+14+7 42+105+… 1 0 0 –1 1 0 –1 0 1**There is exactly one 1 in the first row**n 1 2 3 4 5 6 7 8 9 An 1 1+1 2+3+2 7+14+14+7 42+105+… 1 0 0 –1 1 0 –1 0 1**1**1 1 2 3 2 7 14 14 7 42 105 135 105 42 429 1287 2002 2002 1287 429 1 0 0 –1 1 0 –1 0 1 + + +**1**1 1 2 3 2 7 14 14 7 42 105 135 105 42 429 1287 2002 2002 1287 429 1 0 0 –1 1 0 –1 0 1 + + +**1**1 2/2 1 2 2/3 3 3/2 2 7 2/4 14 14 4/2 7 42 2/5 105 135 105 5/2 42 429 2/6 1287 2002 2002 1287 6/2 429 1 0 0 –1 1 0 –1 0 1**1**1 2/2 1 2 2/3 3 3/2 2 7 2/4 14 5/5 14 4/2 7 42 2/5 105 7/9 135 9/7 105 5/2 42 429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429 1 0 0 –1 1 0 –1 0 1**2/2**2/33/2 2/45/54/2 2/57/99/75/2 2/69/1416/1614/96/2 1 0 0 –1 1 0 –1 0 1**Numerators:**1+1 1+11+2 1+12+31+3 1+13+43+61+4 1+14+56+104+101+5 1 0 0 –1 1 0 –1 0 1**1+1**1+11+2 1+12+31+3 1+13+43+61+4 1+14+56+104+101+5 Numerators: 1 0 0 –1 1 0 –1 0 1 Conjecture 1:**Conjecture 1:**1 0 0 –1 1 0 –1 0 1 Conjecture 2 (corollary of Conjecture 1):**1**0 0 –1 1 0 –1 0 1 Richard Stanley M.I.T.**1**0 0 –1 1 0 –1 0 1 Richard Stanley M.I.T. Andrews’ Theorem: the number of descending plane partitions of size n is George Andrews, Penn State**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 0 0 –1 1 0 –1 0 1**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 0 0 –1 1 0 –1 0 1 What is a descending plane partition?

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