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
solution - spoiler
