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%I #2 Mar 30 2012 17:36:44
%S 15,36,45,78,105,153,210,276,325,378,465,528,561,630,741,820,861,903,
%T 990,1128,1225,1275,1326,1378,1485,1653,1770,1830,1953,2016,2145,2346,
%U 2485,2556,2775,2850,3003,3240,3403,3486,3655,3828,3916,4095,4371,4560
%N Triangular numbers remaining after arranging positive terms of A000217 in a triangle and removing an infinite number of "^"-shaped layers as described below.
%C 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).
%e The original triangle of positive triangular numbers begins like this:
%e ........................1
%e ......................3...6
%e ....................10..15..21
%e ..................28..36..45..55
%e ................66..78..91..105.120
%e ..............136.153.171.190.210.231
%e ......................................
%e 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, ....
%Y Cf. A000217, A001844.
%K nonn
%O 1,1
%A _Rick L. Shepherd_, Oct 28 2007
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