In the following, a is the first term of the sequence and b is the last.
A program could just do the counting, advancing b by 1 and eventually advancing a to bring the total down as needed:
DEFDBL AZ
CLS
a = 0: b = 1: tot = 1
WHILE b <= 2013
b = b + 1
tot = tot + b
WHILE tot > 2013
tot = tot  a
a = a + 1
WEND
IF tot = 2013 THEN
PRINT a; b, a + 1; b
END IF
WEND
finding the following pairs of starting and ending values:
3 63 2 63
45 77 44 77
81 102 80 102
178 188 177 188
333 338 332 338
670 672 669 672
1006 1007 1005 1007
2013 2013 2012 2013
Two pairs are listed on each line, the second pair accounting for the selfcancelling sequence from (a1) to (a1) that can be prepended any strictly positive sequence.
This list contains 16 sequences, but if 2013 fails to qualify as a "sequence" there are still 15 sequences remaining, including that from 2012 to 2013.
Alternatively we could use a formula for the sum of a given sequence:
2012 = ((a+b)/2) * (b  a + 1)
Then varying b we could solve for a. The quadratic solves to:
a = (1 +/ sqrt(1 + 4*(b^2 + b 4026))) / 2
which is the basis for the following:
DEFDBL AZ
FOR b = 1 TO 2013
discr = 1 + 4 * (b * b + b  4026)
IF discr >= 0 THEN
a = (1 + SQR(discr)) / 2
approx = INT(a + .5)
IF ABS(a  approx) < .0000001 THEN
PRINT a; b
END IF
a = (1  SQR(discr)) / 2
approx = INT(a + .5)
IF ABS(a  approx) < .0000001 THEN
PRINT a; b
END IF
END IF
NEXT b
finding:
3 63
2 63
45 77
44 77
81 102
80 102
178 188
177 188
333 338
332 338
670 672
669 672
1006 1007
1005 1007
2013 2013
2012 2013
where the dual solutions of the quadratic account for the selfcancelling prepending of the sequences beginning in the negative values.

Posted by Charlie
on 20130107 12:40:24 