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a(n) = 6*a(n-2) for n > 2; a(1) = 1, a(2) = 4.
4

%I #14 Sep 08 2022 08:45:47

%S 1,4,6,24,36,144,216,864,1296,5184,7776,31104,46656,186624,279936,

%T 1119744,1679616,6718464,10077696,40310784,60466176,241864704,

%U 362797056,1451188224,2176782336,8707129344,13060694016,52242776064,78364164096

%N a(n) = 6*a(n-2) for n > 2; a(1) = 1, a(2) = 4.

%C Interleaving of A000400 and A067411 without initial term 1.

%C Binomial transform is apparently A123011. Fourth binomial transform is A154235.

%H G. C. Greubel, <a href="/A164532/b164532.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,6).

%F a(n) = (5 - (-1)^n)*6^(1/4*(2*n - 5 + (-1)^n)).

%F G.f.: x*(1+4*x)/(1-6*x^2).

%F a(n+3) = a(n+2)*a(n+1)/a(n). - _Reinhard Zumkeller_, Mar 04 2011

%F a(n) = ((1-(-1)^n)*sqrt(6)/2 + 2*(1+(-1)^n))*6^(n/2 -1). - _G. C. Greubel_, Jul 16 2021

%t LinearRecurrence[{0,6}, {1,4}, 40] (* _G. C. Greubel_, Jul 16 2021 *)

%o (Magma) [ n le 2 select 3*n-2 else 6*Self(n-2): n in [1..29] ];

%o (Sage) [((1 - (-1)^n)*sqrt(6)/2 + 2*(1 + (-1)^n))*6^(n/2 -1) for n in (1..40)] # _G. C. Greubel_, Jul 16 2021

%Y Cf. A000400 (powers of 6), A067411, A123011, A154235.

%K nonn

%O 1,2

%A _Klaus Brockhaus_, Aug 15 2009