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 A123761 Let k(n) = mod(3,n)-1. Then a(n) = 4*a(n-1) if n is odd, otherwise ((5+k(n))/4)*a(n-1), with a(0) = 1, a(1) = 2. 0
 1, 2, 3, 12, 15, 60, 60, 240, 360, 1440, 1800, 7200, 7200, 28800, 43200, 172800, 216000, 864000, 864000, 3456000, 5184000, 20736000, 25920000, 103680000, 103680000, 414720000, 622080000, 2488320000, 3110400000, 12441600000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS A double modulo switch recursion with four basic ratio states: {4,1,5/4,3/2}. Surprisingly, the function behaves very much like the factorial function. LINKS Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,120). FORMULA a(n) = 120*a(n-6) for n>=7. G.f.: (60*x^6-60*x^5-15*x^4-12*x^3-3*x^2-2*x-1) / (120*x^6-1). - Colin Barker, May 08 2014 MATHEMATICA k[n_] := Mod[n, 3] - 1; f = 1; f = 2; f[n_] := f[n] = If[Mod[n, 2] == 1, 4*f[n - 1], ((5 + k[n])/4)*f[n - 1]]; a = Table[f[n], {n, 0, 30}] CROSSREFS Sequence in context: A290168 A124486 A260908 * A181121 A047163 A046486 Adjacent sequences:  A123758 A123759 A123760 * A123762 A123763 A123764 KEYWORD nonn AUTHOR Roger L. Bagula, Nov 16 2006 EXTENSIONS Edited by N. J. A. Sloane, Nov 19 2006 STATUS approved

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Last modified July 22 15:12 EDT 2019. Contains 325224 sequences. (Running on oeis4.)