login
Inverse binomial transform of Fibonacci oblongs.
2

%I #9 Nov 24 2020 19:02:21

%S 0,1,0,3,-1,10,-7,35,-36,127,-165,474,-715,1807,-3004,6995,-12393,

%T 27370,-50559,107883,-204820,427351,-826045,1698458,-3321891,6765175,

%U -13333932,26985675,-53457121,107746282,-214146295,430470899,-857417220,1720537327,-3431847189

%N Inverse binomial transform of Fibonacci oblongs.

%C Inverse binomial transform of A001654.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1,3,2).

%F a(n)=((1/2+sqrt(5)/2)^(n+1)+(1/2-sqrt(5)/2)^(n+1)-(-2)^n)/5;

%F G.f.: x(1+x)/(1+x-3x^2-2x^3)=x(1-x)/((1+2x)(1-x-x^2)).

%F a(n) = A084179(n)+A084179(n-1). - _R. J. Mathar_, Dec 10 2014

%t LinearRecurrence[{-1,3,2},{0,1,0},40] (* _Harvey P. Dale_, Nov 24 2020 *)

%Y Cf. A000045, A052964, A084179.

%K easy,sign

%O 0,4

%A _Paul Barry_, May 18 2003