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

1,2

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).

Links

Table of n, a(n) for n=1..60.

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).

Prog

(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++)

Crossrefs

Cf. A000217, A248983.

Sequence in context: A298738 A124776 A099285 * A188108 A166890 A197411

Adjacent sequences: A249261 A249262 A249263 * A249265 A249266 A249267

Keyword

nonn

Author

Derek Orr, Oct 23 2014

Status

approved