NB. diagonalization http://adventofcode.com/day/25

c=: ".>cutLF 0 : 0
20151125  18749137  17289845  30943339  10071777  33511524
31916031  21629792  16929656   7726640  15514188   4041754
16080970   8057251   1601130   7981243  11661866  16474243
24592653  32451966  21345942   9380097  10600672  31527494
   77061  17552253  28094349   6899651   9250759  31663883
33071741   6796745  25397450  24659492   1534922  27995004
)

f=: 33554393&| @ *&252533

NB. So the table is:
NB. f^:0 f^:2 f^:5 ...
NB. f^:1 f^:4 f^:8 ...
NB. f^:3 f^:7 f^:12 ...

NB. zero-based index: (r+1)*c+(r*(r+1)+c*(c+1))/2
NB. one-based index:  (c-1)*r+(r*(r-1)+c*(c-1))/2
NB.                   ((c+r)²-3*r-c)/2
i=: 4 : '(x*<:y)+<.-:(x*<:x)+y*<:y'
i=: <.@-:@(*:@+ - 3&*@[ + ])

NB. find value @ one-based row 2947, column 3029
echo n=:2947 i 3029 NB. index is 17850353
echo f^:n {.{.c     NB. value is 19980801 (August 1st, 1998?)

NB. (33554393&|@*&252533)^:(2947(<.&-:@(*:@+-3&*@[+]))3029)20151125
NB. Or with power-modulo optimization, blindingly fast:
NB. 33554393|20151125*(33554393&|@^&(2947(<.&-:@(*:@+-3&*@[+]))3029))252533

exit 0