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a(1) = 1, a(2) = 3; for n>2, a(n) = {a(n-1)*a(n+1)}^(1/n) or a(n+1) = a(n)^n/a(n-1).
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%I #6 Sep 08 2022 08:45:06

%S 1,3,9,243,387420489,35917545547686059365808220080151141317043

%N a(1) = 1, a(2) = 3; for n>2, a(n) = {a(n-1)*a(n+1)}^(1/n) or a(n+1) = a(n)^n/a(n-1).

%p A073889:= proc(n) option remember;

%p if n<3 then 2*n-1;

%p else a(n-1)^(n-1)/a(n-2);

%p fi; end;

%p seq(A073889(n), n = 1..7); # _G. C. Greubel_, May 18 2020

%t a[n_]:= a[n]= If[n<3, 2*n-1, a[n-1]^(n-1)/a[n-2]]; Table[a[n], {n, 7}] (* _G. C. Greubel_, May 18 2020 *)

%o (Magma) I:=[1,3]; [n le 2 select I[n] else Self(n-1)^(n-1)/Self(n-2): n in [1..7]]; # _G. C. Greubel_, May 18 2020

%o (Sage)

%o def a(n):

%o if (n<3): return 2*n-1

%o else: return a(n-1)^(n-1)/a(n-2)

%o [a(n) for n in (1..7)] # _G. C. Greubel_, May 18 2020

%Y Cf. A058798, A073888.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Aug 17 2002

%E The next term is too large to include.