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 A114849 F(4n+4)/F(4)-F(3n+3)/F(3) where F(n)=A000045(n). 0
 0, 3, 31, 257, 1950, 14164, 100464, 702919, 4878575, 33695365, 232040622, 1595043816, 10952137040, 75149854091, 515435467055, 3534332855753, 24230970910510, 166108203507452, 1138635489987488, 7804802111777935 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The limit as n -> infinity of a(n+1)/a(n) is (1+sqrt(5))*(2+sqrt(5))/2 = 6.8541019662... Old name was: Difference between two Fibonacci cycles A000045 (three's cycle minus two's cycle). LINKS Index entries for linear recurrences with constant coefficients, signature (11,-28,-3,1) FORMULA a(n) = F(4n+4)/F(4)-F(3n+3)/F(3) = (2*F(4n+4)-3*F(3n+3))/6, where F=A000045. G.f.: x*(2*x-3) / ((x^2-7*x+1)*(x^2+4*x-1)). - Colin Barker, Mar 15 2013 MATHEMATICA a = 0; a = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] b = Table[a[4*(n + 1)]/a, {n, 0, 25}]; c = Table[a[3*(n + 1)]/a, {n, 0, 25}]; aout = b - c CROSSREFS Cf. A000045. Sequence in context: A057972 A221821 A198852 * A221899 A242134 A221894 Adjacent sequences:  A114846 A114847 A114848 * A114850 A114851 A114852 KEYWORD nonn,easy AUTHOR Roger L. Bagula, Feb 20 2006 EXTENSIONS Edited by Bruno Berselli, Mar 15 2013 Put formula as clearer name, Joerg Arndt, Mar 15 2013 STATUS approved

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Last modified September 29 04:22 EDT 2020. Contains 337420 sequences. (Running on oeis4.)