 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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 Comments: ( Back to comment list | You must be logged in to post comments.) solution - spoiler | Comment 1 of 5

Consider the sequence a,a+1, a+2, a+3,… a+n-1.

The sum is S=a*n+ (n-1)*(n-2) /2=a*n+T(n-1),       T(k)=triangular number #k

To list all possible sequences   we need to find a and n such that both are positive
integers and satisfy the relation a=(2013- T(n-1))/n ; 0<n<2014 for a>0.
Clearly for every positive (a,n) there exist a longer series defined by (1-n,2a+n-1)
e.g. 670, 671, 672 (a=670  n=3) has a  counterpart series -669, -668, -667,  ….668,
669, 670, 671, 672 (a=-669,   n=2*670+2=1342).

Excel, manual or computer search outputs the following sequences, listed as (a,n):
( 2013,1  ); (1006,2 ); (670,3 ); (333,6 ); (178,11): (81,22 ); (45,33 ); (3,61 ); - each one of those eight strictly positive sequences is matched by a corresponding mixed sequence:

( -2012,4026  ); (-1005,2011 ):     ….
...( -2,126  );

sixteen sequences altogether.

Edited on January 7, 2013, 2:16 pm
 Posted by Ady TZIDON on 2013-01-07 11:16:10 Please log in:

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