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%I #36 May 07 2021 01:01:57
%S 5,6,9,20,100,2500,1562500,610351562500,93132257461547851562500,
%T 2168404344971008868014905601739883422851562500,
%U 1175494350822287507968736537222245677818665556772087521508751706278417259454727172851562500
%N a(n+1) = floor(a(n)/2) * ceiling(a(n)/2), a(0) = 5.
%C A194079(n) gives number of digits of a(n).
%F a(0) = 5; a(k+1) = floor(a(k)/2) * ceiling(a(k)/2).
%F a(n+1) = A002620(a(n)), a(0) = 5. - _Reinhard Zumkeller_, Oct 12 2011
%e 6 = 2*3; 9 = 3*3; 20 = 4*5; ...
%t a=5;a=Table[a=Ceiling[a/2]*Floor[a/2],{n,0,10}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 13 2010 *)
%t NestList[Floor[#/2]Ceiling[#/2]&,5,10] (* _Harvey P. Dale_, Jul 10 2012 *)
%o (Haskell)
%o a014980 n = a014980_list !! n
%o a014980_list = iterate a002620 5
%o -- _Reinhard Zumkeller_, Oct 12 2011
%o (Python)
%o from itertools import accumulate
%o def f(an, _): return (an//2)*((an+1)//2)
%o print(list(accumulate([5]*11, f))) # _Michael S. Branicky_, May 06 2021
%Y Cf. A002620, A194079.
%K nonn,nice,easy
%O 0,1
%A Colin Sandon (sandon(AT)together.net)
%E More terms from _James A. Sellers_, Feb 05 2000
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