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a(n) = Product_{i=2..n} (tau(i)+1)/(tau(i)-1), where tau(.)=A000005(.).
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%I #15 Apr 17 2024 02:33:18

%S 1,3,9,18,54,90,270,450,900,1500,4500,6300,18900,31500,52500,78750,

%T 236250,330750,992250,1389150,2315250,3858750,11576250,14883750,

%U 29767500,49612500,82687500,115762500,347287500,446512500,1339537500,1875352500,3125587500

%N a(n) = Product_{i=2..n} (tau(i)+1)/(tau(i)-1), where tau(.)=A000005(.).

%C The value of the empty product at n=1 is 1.

%C Partial products of the sequence of fractions 3, 3, 2, 3, 5/3, 3, 5/3, 2, 5/3, 3, 7/5, ...

%C The sequence is declared "fini" because the products are no longer integers at n >= 72.

%H Nathaniel Johnston, <a href="/A181574/b181574.txt">Table of n, a(n) for n = 1..71</a> (full sequence)

%p A181574 := proc(n) mul( (numtheory[tau](i)+1)/(numtheory[tau](i)-1) ,i=2..n) ; end proc; # _R. J. Mathar_, Feb 01 2011

%t FoldList[Times, 1, (# + 1)/(# - 1) & /@ Array[DivisorSigma[0, #] &, 70, 2]] (* _Amiram Eldar_, Apr 17 2024 *)

%Y Cf. A000005.

%K nonn,easy,fini,full

%O 1,2

%A _Ctibor O. Zizka_, Jan 30 2011