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F(4n+4)/F(4)-F(3n+3)/F(3) where F(n)=A000045(n).
0

%I #19 Jun 09 2022 13:23:04

%S 0,3,31,257,1950,14164,100464,702919,4878575,33695365,232040622,

%T 1595043816,10952137040,75149854091,515435467055,3534332855753,

%U 24230970910510,166108203507452,1138635489987488,7804802111777935

%N F(4n+4)/F(4)-F(3n+3)/F(3) where F(n)=A000045(n).

%C The limit as n -> infinity of a(n+1)/a(n) is (1+sqrt(5))*(2+sqrt(5))/2 = 6.8541019662...

%C Old name was: Difference between two Fibonacci cycles A000045 (three's cycle minus two's cycle).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (11,-28,-3,1)

%F a(n) = F(4n+4)/F(4)-F(3n+3)/F(3) = (2*F(4n+4)-3*F(3n+3))/6, where F=A000045.

%F G.f.: x*(2*x-3) / ((x^2-7*x+1)*(x^2+4*x-1)). - _Colin Barker_, Mar 15 2013

%t a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] b = Table[a[4*(n + 1)]/a[4], {n, 0, 25}]; c = Table[a[3*(n + 1)]/a[3], {n, 0, 25}]; aout = b - c

%t LinearRecurrence[{11,-28,-3,1},{0,3,31,257},20] (* _Harvey P. Dale_, Jun 09 2022 *)

%Y Cf. A000045.

%K nonn,easy

%O 0,2

%A _Roger L. Bagula_, Feb 20 2006

%E Edited by _Bruno Berselli_, Mar 15 2013

%E Put formula as clearer name, _Joerg Arndt_, Mar 15 2013