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1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

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1 0 0 –1 1 0 –1 0 1

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  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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 + + +

  9. 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 + + +

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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:

  15. Conjecture 1: 1 0 0 –1 1 0 –1 0 1 Conjecture 2 (corollary of Conjecture 1):

  16. 1 0 0 –1 1 0 –1 0 1 Richard Stanley M.I.T.

  17. 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

  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 0 0 –1 1 0 –1 0 1

  19. 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?