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a(n) = 4^n + 8^n.
40

%I #32 Aug 18 2024 14:04:29

%S 2,12,80,576,4352,33792,266240,2113536,16842752,134479872,1074790400,

%T 8594128896,68736253952,549822922752,4398314946560,35185445830656,

%U 281479271677952,2251816993554432,18014467228958720,144115462953762816

%N a(n) = 4^n + 8^n.

%C Shift 2^n+1 left 2n bits.

%H Harry J. Smith, <a href="/A063481/b063481.txt">Table of n, a(n) for n=0,...,200</a>

%H D. Suprijanto and Rusliansyah, <a href="http://dx.doi.org/10.12988/ams.2014.4140">Observation on Sums of Powers of Integers Divisible by Four</a>, Applied Mathematical Sciences, Vol. 8, 2014, no. 45, 2219 - 2226.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,-32).

%F G.f.: 1/(1-4*x)+1/(1-8*x). E.g.f.: e^(4*x)+e^(8*x). - _Mohammad K. Azarian_, Jan 11 2009

%F a(n)=12*a(n-1)-32*a(n-2) with a(0)=2, a(1)=12. - _Vincenzo Librandi_, Jul 21 2010

%F G.f.: G(0), where G(k)= 1 + 2^k/(1 - 4*x/(4*x + 2^k/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jul 22 2013

%e n=3: 23+1 shifted 2*3 bits to the left is 576 because 23+1 in binary is [1, 0, 0, 1] and 576 is [1, 0, 0, 1, 0, 0, 0, 0, 0, 0].

%t Table[4^n + 8^n, {n, 0, 25}]

%o (PARI) for(n=0,22,print(shift(2^n+1,2*n)))

%o (PARI) { for (n=0, 200, write("b063481.txt", n, " ", shift(1, 2*n) + shift(1, 3*n)) ) } \\ _Harry J. Smith_, Aug 23 2009

%Y Cf. A000051, A034472, A052539, A034474, A062394, A034491, A062395, A062396, A007689, A063376, A074600 - A074624.

%K easy,nonn

%O 0,1

%A _Jason Earls_, Jul 28 2001

%E Edited by _Robert G. Wilson v_, Aug 25 2002