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%I #20 Sep 08 2022 08:45:28
%S 1,2,3,12,15,60,60,240,360,1440,1800,7200,7200,28800,43200,172800,
%T 216000,864000,864000,3456000,5184000,20736000,25920000,103680000,
%U 103680000,414720000,622080000,2488320000,3110400000,12441600000
%N 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.
%C A double modulo switch recursion with four basic ratio states: {4,1,5/4,3/2}.
%C Surprisingly, the function behaves very much like the factorial function.
%C 10^floor(n/6) | a(n). - _G. C. Greubel_, Aug 10 2019
%H G. C. Greubel, <a href="/A123761/b123761.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,120).
%F a(n) = 120*a(n-6) for n>=7.
%F G.f.: (1+2*x+3*x^2+12*x^3+15*x^4+60*x^5-60*x^6)/(1-120*x^6). - _Colin Barker_, May 08 2014
%p seq(coeff(series((1+2*x+3*x^2+12*x^3+15*x^4+60*x^5-60*x^6)/(1-120*x^6), x, n+1), x, n), n = 0 .. 35); # _G. C. Greubel_, Aug 10 2019
%t k[n_]:= Mod[n, 3] -1; f[0]=1; f[1]=2; f[n_]:= f[n] = If[Mod[n, 2] == 1, 4*f[n-1], ((5 +k[n])/4)*f[n-1]]; Table[f[n], {n, 0, 35}]
%t LinearRecurrence[{0,0,0,0,0,120}, {1,2,3,12,15,60,60}, 35] (* _G. C. Greubel_, Aug 10 2019 *)
%o (PARI) my(x='x+O('x^35)); Vec((1+2*x+3*x^2+12*x^3+15*x^4+60*x^5-60*x^6 )/(1-120*x^6)) \\ _G. C. Greubel_, Aug 10 2019
%o (Magma) I:=[2,3,12,15,60,60]; [1] cat [n le 6 select I[n] else 120*Self(n-6): n in [1..35]]; // _G. C. Greubel_, Aug 10 2019
%o (Sage)
%o def A123761_list(prec):
%o P.<x> = PowerSeriesRing(ZZ, prec)
%o return P((1+2*x+3*x^2+12*x^3+15*x^4+60*x^5-60*x^6)/(1-120*x^6)).list()
%o A123761_list(35) # _G. C. Greubel_, Aug 10 2019
%o (GAP) a:=[2,3,12,15,60,60];; for n in [7..35] do a[n]:=120*a[n-6]; od; Concatenation([1], a); # _G. C. Greubel_, Aug 10 2019
%K nonn
%O 0,2
%A _Roger L. Bagula_, Nov 16 2006
%E Edited by _N. J. A. Sloane_, Nov 19 2006
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