A249264 Sequence of distinct least nonnegative numbers such that the average of the first n terms is a triangular number.

0, 2, 1, 9, 3, 21, 6, 38, 10, 60, 15, 87, 112, 28, 148, 36, 189, 45, 235, 55, 286, 66, 342, 78, 403, 91, 469, 105, 540, 120, 616, 136, 697, 153, 783, 171, 874, 190, 970, 210, 1071, 231, 1177, 253, 1288, 276, 1404, 300, 1525, 325, 1651, 351, 1782, 378, 1918, 406, 2059, 435, 2205, 465

Offset: 1

Author

Derek Orr, Oct 23 2014

Keywords

• nonn

Comments

Similar to A248983 except a(1) = 0, the zeroth triangular number.

Note that the sum of the first 12 terms is 252. Also, one can show that (252+sum_{i=7..n}(A000566(n)+A000217(n)))/(2*n) = n*(n+1)/2. So, for n > 6, if a(2*k-1) = A000566(k) and a(2*k-2) = A000217(k) for all 6 < k <= n, then a(2*n) = n*(n+1)/2.

Similarly, for n > 6, if a(2*k-1) = A000566(k) and a(2*k) = A000217(k) for all 6 < k <= n, then a(2*n+1) = A000566(n).

Crossrefs

Cf. A000217, A248983.

Programs

PARI

v=[]; n=0; while(n<5000, num=(vecsum(v)+n); if(num%(#v+1)==0&&vecsearch(vecsort(v),n)==0,for(i=0,n,if(i*(i+1)/2>(num/(#v+1)), break); if(i*(i+1)/2==(num/(#v+1)), print1(n, ", "); v=concat(v, n); n=0; break))); n++)

Formula

Empirical g.f.: x^2*(66*x^16-66*x^15-153*x^14+153*x^13+91*x^12-91*x^11-3*x^2-x-2) / ((x-1)^3*(x+1)^3). - Colin Barker, Oct 24 2014

Conjectured: For n > 6, a(2*n-1) = A000566(n) and a(2*n) = A000217(n).