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%I #19 May 26 2017 22:32:57
%S 3,15,27,30,42,75,90,105,135,147,165,243,252,270,273,315,363,375,378,
%T 420,462,495,507,612,660,675,693,735,750,780,810,855,858,867,945,1050,
%U 1083,1155,1170,1215,1287,1323,1365,1470,1485,1518,1587,1785,1815,1875,1950
%N 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.
%C 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
%C 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.
%C 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
%C 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
%H Ron Knott <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/">Runsums</a>
%H T. Verhoeff, <a href="https://cs.uwaterloo.ca/journals/JIS/trapzoid.html">Rectangular and Trapezoidal Arrangements</a>, J. Integer Sequences, Vol. 2, 1999, #99.1.6.
%e 3 = 1+2 = 3, so 3 is a term.
%e 15 = 4+5+6 = 7+8 so 15 is a term.
%e a(6) = 75 because 75 = 3+4+5+6+7+8+9+10+11+12 = 13+14+15+16+17.
%t 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 *)
%Y Cf. A001227, A094550, A110701, A110702.
%K nonn
%O 1,1
%A _Ron Knott_, Aug 04 2005
%E Initial 3 added by _A. D. Skovgaard_, May 22 2017