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

%I #41 Jan 13 2021 10:41:51

%S 0,16,64,192,512,1280,3072,7168,16384,36864,81920,180224,393216,

%T 851968,1835008,3932160,8388608,17825792,37748736,79691776,167772160,

%U 352321536,738197504,1543503872,3221225472,6710886400,13958643712

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

%C A hierarchical sequence (S(W'3{2,2}*cc) - see A059126).

%C Generating floretion: AB + BA with A = .5'i + .5'ii' + .5'ij' + .5'ik' and B = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'. - _Creighton Dement_, Dec 19 2004

%H Harry J. Smith, <a href="/A059165/b059165.txt">Table of n, a(n) for n = -1..200</a>

%H Jonas Wallgren, <a href="/A059126/a059126.txt">Hierarchical sequences</a>, 2001.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).

%F G.f.: 16/(2*x-1)^2.

%F a(n) = 4*A058922(n+2) = 16*A001787(n+1). - _Philippe Deléham_, Apr 21 2009

%F From _Amiram Eldar_, Jan 13 2021: (Start)

%F Sum_{n>=0} 1/a(n) = log(2)/8.

%F Sum_{n>=0} (-1)^n/a(n) = log(3/2)/8. (End)

%t Table[2^(n+4)*(n+1),{n,-1,100}] (* _Vladimir Joseph Stephan Orlovsky_, Jan 15 2011 *)

%t LinearRecurrence[{4,-4},{0,16},30] (* _Harvey P. Dale_, Oct 29 2019 *)

%o (PARI) for(n=1,40,print1(shift(n,n+3),","))

%o (PARI) { for (n = -1, 200, write("b059165.txt", n, " ", (n + 1)*2^(n + 4)); ) } \\ _Harry J. Smith_, Jun 25 2009

%Y Cf. A001787, A058922.

%K easy,nonn

%O -1,2

%A _Jonas Wallgren_, Feb 02 2001

%E More terms from _Benoit Cloitre_, Apr 07 2002

%E Edited by _N. J. A. Sloane_, Apr 16 2008 at the suggestion of _Vim Wenders_