OFFSET
0,2
COMMENTS
The full triangle of the inverse Akiyama-Tanigawa transform applied to (-1)^n*A062510(n)=3*(-1)^n*A001045(n) yielding a(n) is
0, 3, 6, 10, 15, 21, 28, 36, ...
-3, -6, -12, -20, -30, -42, -56, ... essentially -A002378
3, 12, 24, 40, 60, 84, ... essentially A046092
-9, -24, -48, -80, -120, ... essentially -A033996
15, 48, 96, 160, ...
-33, -96, -192, ...
63, 192, ...
-129, ...
etc.
First column: (-1)^n*A062510(n).
The following columns are multiples of A122803(n)=(-2)^n. See A007283(n), A091629(n), A020714(n+1), A110286, A175805(n), 4*A005010(n).
An autosequence of the first kind is a sequence whose main diagonal is A000004 = 0's.
b(n) = 0, 0 followed by a(n) is an autosequence of the first kind.
The successive differences of b(n) are
0, 0, 0, 3, 6, 10, 15, 21, ...
0, 0, 3, 3, 4, 5, 6, 7, ... see A194880(n)
0, 3, 0, 1, 1, 1, 1, 1, ...
3, -3, 1, 0, 0, 0, 0, 0, ...
-6, 4, -1, 0, 0, 0, 0, 0, ...
10, -5, 1, 0, 0, 0, 0, 0, ...
-15, 6, -1, 0, 0, 0, 0, 0, ...
21, -7, 1, 0, 0, 0, 0, 0, ...
The inverse binomial transform (first column) is the signed sequence. This is general.
Also generalized hexagonal numbers without 1. - Omar E. Pol, Mar 23 2015
LINKS
Muniru A Asiru, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
Inverse Akiyama-Tanigawa transform of (-1)^n*A062510(n).
a(n) = (n+1)*(n+2)/2 for n > 0. - Charles R Greathouse IV, Mar 23 2015
a(n) = A161680(n+2) for n >= 1. - Georg Fischer, Oct 30 2018
MATHEMATICA
Prepend[Table[(n + 1) (n + 2)/2, {n, 49}], 0] (* Michael De Vlieger, Mar 23 2015 *)
PROG
(PARI) a(n)=if(n, (n+1)*(n+2)/2, 0) \\ Charles R Greathouse IV, Mar 23 2015
(GAP) Concatenation([0], List([1..50], n->(n+1)*(n+2)/2)); # Muniru A Asiru, Oct 31 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Mar 23 2015
STATUS
approved