A110703 Numbers S with two neighboring run sums (sum of positive integer runs) S = a+(a+1)+..+b=(b+1)+(b+2)...+c, 0

3, 15, 27, 30, 42, 75, 90, 105, 135, 147, 165, 243, 252, 270, 273, 315, 363, 375, 378, 420, 462, 495, 507, 612, 660, 675, 693, 735, 750, 780, 810, 855, 858, 867, 945, 1050, 1083, 1155, 1170, 1215, 1287, 1323, 1365, 1470, 1485, 1518, 1587, 1785, 1815, 1875, 1950

Offset: 1

Author

Ron Knott, Aug 04 2005

Keywords

• nonn

Comments

In other words, numbers n such that a list of consecutive numbers can be split into two parts in which their sums both equal n. - A. D. Skovgaard, May 22 2017

If the two runs overlap in one number, the runs are Friends and their sums are A110701. The sums are the difference of two triangular numbers A000217.

The subsequence where there is more than one possible splitting begins 105, 945, 1365, 2457, 2625, 3990, 5145, 8505, ... - Jean-François Alcover, May 22 2017

a(n) seems to always be divisible by 3.- A. D. Skovgaard, May 22 2017. This is true. Sequence lists values of n = t(t+1)/2 - k(k+1)/2 = m(m+1)/2 - t(t+1)/2 with k < t < m. Since any triangular number must be of the form 3w or 3w+1, then there are two possibilities for n = 3w - k(k+1)/2 = m(m+1)/2 - 3w or n = 3w + 1 - k(k+1)/2 = m(m+1)/2 - 3w - 1. For first case, if k(k+1)/2 = 3u+1, there is no solution for m. Similarly for second case, if k(k+1)/2 = 3u, there is no solution for m. So always n must be divisible by 3. - Altug Alkan, May 22 2017

Examples

			3 = 1+2 = 3, so 3 is a term.
15 = 4+5+6 = 7+8 so 15 is a term.
a(6) = 75 because 75 = 3+4+5+6+7+8+9+10+11+12 = 13+14+15+16+17.
		

Links

• Ron Knott Runsums
• T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Crossrefs

Cf. A001227, A094550, A110701, A110702.

Programs

Mathematica

Select[Range[1000], False =!= Reduce[# == Sum[k, {k, x, y}] == Sum[k, {k, y + 1, z}] && z >= y >= x > 0, {x, y, z}, Integers] &] (* Giovanni Resta, May 22 2017 *)

Extensions

Initial 3 added by A. D. Skovgaard, May 22 2017