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Inverse binomial transform of A001651.
3

%I #37 Feb 15 2021 07:00:31

%S 1,1,1,-2,4,-8,16,-32,64,-128,256,-512,1024,-2048,4096,-8192,16384,

%T -32768,65536,-131072,262144,-524288,1048576,-2097152,4194304,

%U -8388608,16777216,-33554432,67108864,-134217728,268435456,-536870912,1073741824,-2147483648

%N Inverse binomial transform of A001651.

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (-2).

%F a(n) = A123344(n+1), n > 0.

%F a(n) = (-2)^n/4 = (-1)^n*A000079(n-2), n > 1.

%F O.g.f.: (1 + 3*x + 3*x^2)/(1 + 2*x). - _R. J. Mathar_, Aug 27 2008

%F a(n) = -2*a(n-1) for n >= 3; a(0)=1, a(1)=1, a(2)=1. - _Harvey P. Dale_, May 04 2012

%F G.f.: x+1/Q(0) where Q(k) = 1 + x*(k+1)/(1 - 1/(1 - (k+1)/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Sep 23 2012

%F G.f.: 1+x/U(0) where U(k) = 1 - x*(k+4) + x*(k+3)/U(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Oct 11 2012

%F a(n) = A122803(n-2) for n >= 2. - _Georg Fischer_, Nov 03 2018

%F E.g.f.: (3/4) + (3/2)*x + (1/4)*exp(-2*x). - _Alejandro J. Becerra Jr._, Feb 15 2021

%t CoefficientList[Series[(1+3x+3x^2)/(1+2x),{x,0,40}],x] (* or *) Join[ {1,1},NestList[-2#&,1,38]] (* _Harvey P. Dale_, May 04 2012 *)

%t Join[{1, 1},LinearRecurrence[{-2},{1},32]] (* _Ray Chandler_, Aug 12 2015 *)

%o (PARI) Vec((1 + 3*x + 3*x^2)/(1 + 2*x) + O(x^40)) \\ _Andrew Howroyd_, Nov 03 2018

%Y Cf. A000079, A122803, A123344.

%K sign

%O 0,4

%A _Paul Curtz_, Aug 12 2008

%E Edited and extended by _R. J. Mathar_, Aug 28 2008