A134509 Triangular numbers remaining after arranging positive terms of A000217 in a triangle and removing an infinite number of "^"-shaped layers as described below.

15, 36, 45, 78, 105, 153, 210, 276, 325, 378, 465, 528, 561, 630, 741, 820, 861, 903, 990, 1128, 1225, 1275, 1326, 1378, 1485, 1653, 1770, 1830, 1953, 2016, 2145, 2346, 2485, 2556, 2775, 2850, 3003, 3240, 3403, 3486, 3655, 3828, 3916, 4095, 4371, 4560

Offset: 1

Author

Rick L. Shepherd, Oct 28 2007

Keywords

• nonn

Comments

Let "s" be any sequence (finite or infinite) and "b" be any set of real numbers. Define an operation 'triangular removal', TriRem(s,b), that produces a subsequence from s as follows: Arrange the terms s(i) by rows into a triangle, which can be viewed as a (possibly infinite) set of nested "^"-shaped layers. Count each layer from the outside as layer 1, 2, 3, .... During the following removal process, these layer numbers are considered fixed: For each positive integer n in b, remove layer n if it exists. TriRem(s,b) is the sequence of remaining terms read by rows. The current sequence, A134509, is TriRem(A000217-{0},A000217). A complementary operation 'triangular retention', TriRet(s,b), can be defined similarly that instead retains the layers specified by b. The index of an original term s(i) at the apex of a removed/retained "^"-layer is a centered square number (A001844).

Examples

			The original triangle of positive triangular numbers begins like this:
........................1
......................3...6
....................10..15..21
..................28..36..45..55
................66..78..91..105.120
..............136.153.171.190.210.231
......................................
The upside-down "V" with 1 at the top is layer 1, with 15 at the top is layer 2, with 91 at the top is layer 3, etc. Because 1 and 3 are elements of b=A000217, layers 1 and 3 are among those completely removed. The remaining terms by row begin the infinite subsequence: 15, 36, 45, 78, 105, ....
		

Crossrefs

Cf. A000217, A001844.