login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) = 5*2^(n-1) + 2^(2*n-1) + 6^n + 1.
2

%I #32 Dec 09 2017 03:19:41

%S 5,14,55,269,1465,8369,48865,288449,1713025,10210049,60993025,

%T 364899329,2185181185,13094268929,78498422785,470721937409,

%U 2823257554945,16935249707009,101594317062145,609497180274689,3656708198498305,21939149668876289,131630499945775105

%N a(n) = 5*2^(n-1) + 2^(2*n-1) + 6^n + 1.

%C Similar to A279511 Sierpinski square-based pyramid but with tetrahedral openings as found in the structure of the Sierpinski octahedron A279512.

%H Colin Barker, <a href="/A281698/b281698.txt">Table of n, a(n) for n = 0..1000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Sierpinski_triangle">Sierpinski triangle</a>, see section on higher dimensional analogs.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (13,-56,92,-48).

%F From _Colin Barker_, Jan 28 2017: (Start)

%F a(n) = 13*a(n-1) - 56*a(n-2) + 92*a(n-3) - 48*a(n-4) for n>3.

%F G.f.: (5 - 51*x + 153*x^2 - 122*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)).

%F (End)

%p A281698:=n->5*2^(n-1) + 2^(2*n-1) + 6^n + 1: seq(A281698(n), n=0..30); # _Wesley Ivan Hurt_, Apr 09 2017

%t Table[5*2^(n - 1) + 2^(2 n - 1) + 6^n + 1, {n, 0, 22}] (* or *)

%t LinearRecurrence[{13, -56, 92, -48}, {5, 14, 55, 269}, 23] (* or *)

%t CoefficientList[Series[(5 - 51 x + 153 x^2 - 122 x^3)/((1 - x) (1 - 2 x) (1 - 4 x) (1 - 6 x)), {x, 0, 22}], x] (* _Michael De Vlieger_, Jan 28 2017 *)

%o (PARI) Vec((5 - 51*x + 153*x^2 - 122*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)) + O(x^30)) \\ _Colin Barker_, Jan 28 2017

%o (PARI) a(n) = 5*2^(n-1) + 2^(2*n-1) + 6^n + 1 \\ _Charles R Greathouse IV_, Jan 29 2017

%Y Cf. A000330, A047999, A279511, A279512.

%K nonn,easy

%O 0,1

%A _Steven Beard_, Jan 27 2017