Determine all possible sequences of consecutive integers whose sum is precisely 2013.

TABLE
  x     y
4026    1    2013
2013    2    1006.5
1342    3     671
671    6     335.5
366    11     183
183    22      91.5
122    33      61
66    61      33
61    66      30.5
33    122      16.5
22    183      11
11    366       5.5
6    671       3
3    1342       1.5
2    2013       1
1    4026       0.5

2013 is half of 4026.

x and y are divisor pairs of 4026.

y is the number of consecutive integers which will total 2013 with the FIRST term being:
a) if 2013/y is an integer:
     2013/y -(y-1)/2
b) if 2013/y terminates with  ".5":
     2013/y-(y/2-0.5)

For y= 2 and 3 we have 1006 + 1007 and 670 + 671 + 672 respectively.
For y= 4026 the series is -2012 - 2011 - ....    +2012 +2013.

Posted by brianjn on 2013-01-07 19:01:11