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

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

1,1

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

Links

Table of n, a(n) for n=1..51.

Ron Knott Runsums

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

Example

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.

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 *)

Crossrefs

Cf. A001227, A094550, A110701, A110702.

Sequence in context: A328387 A006872 A027179 * A121250 A105549 A080065

Adjacent sequences: A110700 A110701 A110702 * A110704 A110705 A110706

Keyword

nonn

Author

Ron Knott, Aug 04 2005

Extensions

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

Status

approved