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a(n) = 1^n + 2^(n+1) + 3^(n+2).
2

%I #19 Dec 28 2024 19:34:21

%S 5,12,32,90,260,762,2252,6690,19940,59562,178172,533490,1598420,

%T 4791162,14365292,43079490,129205700,387551562,1162523612,3487308690,

%U 10461401780,31383156762,94147373132,282437925090,847305386660,2541899382762,7625664593852,22876926672690

%N a(n) = 1^n + 2^(n+1) + 3^(n+2).

%H Harry J. Smith, <a href="/A066280/b066280.txt">Table of n, a(n) for n = -1..200</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6, -11, 6).

%F a(n) = 1 + A000079(n+1) + A000244(n+2).

%F From _R. J. Mathar_, Feb 19 2010: (Start)

%F a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3).

%F G.f.: (-18*x+15*x^2+5)/((1-x) * (3*x-1) * (2*x-1) * x). (End)

%t Table[ 1^(n + 1) + 2^(n + 2) + 3^(n + 3), {n, -2, 24} ]

%o (PARI) a(n) = { 1 + 2^(n + 1) + 3^(n + 2) } \\ _Harry J. Smith_, Feb 08 2010

%Y Cf. A000079, A000244.

%K nonn

%O -1,1

%A _George E. Antoniou_, Dec 10 2001

%E More terms from _Robert G. Wilson v_, Dec 11 2001