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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
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).
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] = 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
LinearRecurrence[{11, -28, -3, 1}, {0, 3, 31, 257}, 20] (* Harvey P. Dale, Jun 09 2022 *)
CROSSREFS
Cf. A000045.
Sequence in context: A057972 A221821 A198852 * A221899 A242134 A221894
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