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a(n) = 2^(4*n + 3) + 2*4^n + 3.
1

%I #32 Mar 28 2024 13:19:04

%S 13,139,2083,32899,524803,8390659,134225923,2147516419,34359869443,

%T 549756338179,8796095119363,140737496743939,2251799847239683,

%U 36028797153181699,576460752840294403,9223372039002259459,147573952598266347523,2361183241469182345219

%N a(n) = 2^(4*n + 3) + 2*4^n + 3.

%C Binary numbers of form 1j1i11 where j and i are the number of 0's, n is the index, i = 2*n+1, j = 2*n+3.

%H Vincenzo Librandi, <a href="/A195464/b195464.txt">Table of n, a(n) for n = 0..800</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (21,-84,64).

%F a(n) = 2^(4*n+3) + A188161(n).

%F From _Alexander R. Povolotsky_, Sep 19 2011: (Start)

%F a(n+2) = 20*a(n+1) - 64*a(n) + 135.

%F G.f.: (-13 + 134*x - 256*x^2)/(-1 + 21*x - 84*x^2 + 64*x^3). (End)

%F a(n) = 3 + A026244(n+1). - _Bruno Berselli_, Sep 19 2011

%e Terms starting from n=1 written in binary are 10001011, 100000100011, 1000000010000011, 10000000001000000011.

%o (Magma) [2^(4*n + 3) + 2*4^n + 3: n in [0..20]]; // _Vincenzo Librandi_, Sep 30 2011

%Y Cf. A188161.

%K nonn,easy

%O 0,1

%A _Brad Clardy_, Sep 19 2011