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 Consecutive in 2013 (Posted on 2013-01-07)
Determine all possible sequences of consecutive integers whose sum is precisely 2013.

 No Solution Yet Submitted by K Sengupta No Rating

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 Two computer methods. | Comment 2 of 5 |

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 A-Z
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 self-cancelling sequence from -(a-1) to (a-1) 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 A-Z
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 self-cancelling prepending of the sequences beginning in the negative values.

 Posted by Charlie on 2013-01-07 12:40:24

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