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a(n) = 4th Euler polynomial evaluated at 2^n.
4

%I #19 Jun 13 2015 00:48:54

%S 0,2,132,3080,57360,983072,16252992,264241280,4261413120,68451041792,

%T 1097364145152,17575006177280,281337537761280,4502500115750912,

%U 72048797944922112,1152851135862702080,18446181123756195840,295143401579725586432,4722330454072626511872

%N a(n) = 4th Euler polynomial evaluated at 2^n.

%H Colin Barker, <a href="/A020524/b020524.txt">Table of n, a(n) for n = 0..830</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (26,-176,256).

%F From _Colin Barker_, May 04 2015: (Start)

%F a(n) = 2^n-2^(1+3*n)+16^n.

%F a(n) = 26*a(n-1)-176*a(n-2)+256*a(n-3) for n>2.

%F G.f.: -2*x*(40*x+1) / ((2*x-1)*(8*x-1)*(16*x-1)).

%F (End)

%p seq(euler(4, 2^n), n=0..24);

%t Table[EulerE[4,2^n],{n,0,40}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 03 2009 *)

%o (PARI) concat(0, Vec(-2*x*(40*x+1)/((2*x-1)*(8*x-1)*(16*x-1)) + O(x^100))) \\ _Colin Barker_, May 04 2015

%Y Cf. A020523 - A020526.

%K nonn,easy

%O 0,2

%A _Simon Plouffe_