%I #45 Oct 31 2018 05:49:50
%S 0,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153,171,190,210,231,
%T 253,276,300,325,351,378,406,435,465,496,528,561,595,630,666,703,741,
%U 780,820,861,903,946,990,1035,1081,1128,1176,1225,1275
%N Triangular numbers (A000217) omitting the term 1.
%C The full triangle of the inverse Akiyama-Tanigawa transform applied to (-1)^n*A062510(n)=3*(-1)^n*A001045(n) yielding a(n) is
%C 0, 3, 6, 10, 15, 21, 28, 36, ...
%C -3, -6, -12, -20, -30, -42, -56, ... essentially -A002378
%C 3, 12, 24, 40, 60, 84, ... essentially A046092
%C -9, -24, -48, -80, -120, ... essentially -A033996
%C 15, 48, 96, 160, ...
%C -33, -96, -192, ...
%C 63, 192, ...
%C -129, ...
%C etc.
%C First column: (-1)^n*A062510(n).
%C The following columns are multiples of A122803(n)=(-2)^n. See A007283(n), A091629(n), A020714(n+1), A110286, A175805(n), 4*A005010(n).
%C An autosequence of the first kind is a sequence whose main diagonal is A000004 = 0's.
%C b(n) = 0, 0 followed by a(n) is an autosequence of the first kind.
%C The successive differences of b(n) are
%C 0, 0, 0, 3, 6, 10, 15, 21, ...
%C 0, 0, 3, 3, 4, 5, 6, 7, ... see A194880(n)
%C 0, 3, 0, 1, 1, 1, 1, 1, ...
%C 3, -3, 1, 0, 0, 0, 0, 0, ...
%C -6, 4, -1, 0, 0, 0, 0, 0, ...
%C 10, -5, 1, 0, 0, 0, 0, 0, ...
%C -15, 6, -1, 0, 0, 0, 0, 0, ...
%C 21, -7, 1, 0, 0, 0, 0, 0, ...
%C The inverse binomial transform (first column) is the signed sequence. This is general.
%C Also generalized hexagonal numbers without 1. - _Omar E. Pol_, Mar 23 2015
%H Muniru A Asiru, <a href="/A253145/b253145.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F Inverse Akiyama-Tanigawa transform of (-1)^n*A062510(n).
%F a(n) = (n+1)*(n+2)/2 for n > 0. - _Charles R Greathouse IV_, Mar 23 2015
%F a(n+1) = 3*A001840(n+1) + A022003(n).
%F a(n) = A161680(n+2) for n >= 1. - _Georg Fischer_, Oct 30 2018
%t Prepend[Table[(n + 1) (n + 2)/2, {n, 49}], 0] (* _Michael De Vlieger_, Mar 23 2015 *)
%o (PARI) a(n)=if(n,(n+1)*(n+2)/2,0) \\ _Charles R Greathouse IV_, Mar 23 2015
%o (GAP) Concatenation([0],List([1..50],n->(n+1)*(n+2)/2)); # _Muniru A Asiru_, Oct 31 2018
%Y Cf. A000217, A179865, A001045, A062510, A002378, A046092, A033996, A122803, A007283, A091629, A020714, A110286, A161680, A175805, A005010, A194880, A001840, A022003, A255935.
%K nonn,easy
%O 0,2
%A _Paul Curtz_, Mar 23 2015