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A112033 a(n) = 3 * 2^(floor(n/2) + 1 + (-1)^n). 8

%I #35 Jan 17 2023 15:41:16

%S 12,3,24,6,48,12,96,24,192,48,384,96,768,192,1536,384,3072,768,6144,

%T 1536,12288,3072,24576,6144,49152,12288,98304,24576,196608,49152,

%U 393216,98304,786432,196608,1572864,393216,3145728,786432,6291456,1572864

%N a(n) = 3 * 2^(floor(n/2) + 1 + (-1)^n).

%D G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chap. 4, Sect. 1, Problem 148.

%H Reinhard Zumkeller, <a href="/A112033/b112033.txt">Table of n, a(n) for n = 0..100</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,2).

%F a(n) = 1 / abs(A112031(n)/A112032(n) - 2/3). (previous name)

%F a(n) = 3*2^A084964(n) = 3*A112032(n).

%F From _Ralf Stephan_, Jul 16 2013: (Start)

%F Recurrence: a(n) = 2a(n-2), a(0)=12, a(1)=3.

%F G.f.: (6*x+24)/(1-2*x^2). (End)

%p A112033:=n->3*2^(floor(n/2) + 1 + (-1)^n); seq(A112033(k), k=0..50); # _Wesley Ivan Hurt_, Nov 01 2013

%t Table[3*2^(Floor[n/2] + 1 + (-1)^n), {n,0,50}] (* _Wesley Ivan Hurt_, Nov 01 2013 *)

%o (PARI) a(n) = 3 * 2^(n\2 + 1 + (-1)^n); \\ _Michel Marcus_, Nov 02 2013

%o (Python)

%o def A112033(n): return 3*(1<<(n>>1)+(int(not n&1)<<1)) # _Chai Wah Wu_, Jan 17 2023

%Y Cf. A112030.

%K nonn,easy

%O 0,1

%A _Reinhard Zumkeller_, Aug 27 2005

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