OFFSET
0,1
COMMENTS
Stella octangula with Sierpinski recursion.
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, Sierpinski triangle, see section on higher dimensional analogs.
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).
FORMULA
a(n) = 8*(2^(2*n+1)+2) - 6*(2^(n+1)+1).
From Colin Barker, Jan 28 2017: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3) for n>2.
G.f.: 2*(7 - 24*x + 32*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)).
(End)
MATHEMATICA
Table[8 (2^(2 n + 1) + 2) - 6 (2^(n + 1) + 1), {n, 0, 25}] (* or *)
LinearRecurrence[{7, -14, 8}, {14, 50, 218}, 26] (* or *)
CoefficientList[Series[2 (7 - 24 x + 32 x^2)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 25}], x] (* Michael De Vlieger, Jan 28 2017 *)
PROG
(PARI) Vec(2*(7 - 24*x + 32*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Jan 28 2017
(PARI) a(n) = 16*4^n - 12*2^n + 10 \\ Charles R Greathouse IV, Jan 29 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Steven Beard, Jan 27 2017
STATUS
approved