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a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3; a(0) = 1, a(1) = 4, a(2) = 12, a(3) = 32.
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%I #17 Jan 03 2023 04:20:44

%S 1,4,12,32,72,144,272,512,992,1984,4032,8192,16512,33024,65792,131072,

%T 261632,523264,1047552,2097152,4196352,8392704,16781312,33554432,

%U 67100672,134201344,268419072,536870912,1073774592,2147549184

%N a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3; a(0) = 1, a(1) = 4, a(2) = 12, a(3) = 32.

%C Conjecture: a(n) = 2*A038503(n+3) if n > 0. - _R. J. Mathar_, Oct 23 2007

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4).

%F Sequence is identical to its fourth differences.

%F From _R. J. Mathar_, Nov 18 2007: (Start)

%F G.f.: -(1 + 2*x^2 + 4*x^3)/((2*x - 1)*(2*x^2 - 2*x + 1)). - [Corrected by _Georg Fischer_, May 12 2019]

%F a(n) = -2*(-1)^n*A009116(n)+3*2^n. (End)

%F E.g.f.: exp(x)*(3*cosh(x) - 2*(cos(x) + sin(x)) + 5*sinh(x)). - _Stefano Spezia_, Jan 03 2023

%p A133212 := proc(n) option remember ; if n <= 3 then op(n+1,[1,4,12,32]) ; else 4*A133212(n-1)-6*A133212(n-2)+4*A133212(n-3) ; fi ; end: seq(A133212(n),n=0..50) ; # _R. J. Mathar_, Oct 23 2007

%t Join[{1},LinearRecurrence[{4, -6, 4},{4, 12, 32},29]] (* _Ray Chandler_, Sep 23 2015 *)

%Y Cf. A009116, A038503, A099087.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Oct 11 2007

%E More terms from _R. J. Mathar_, Oct 23 2007