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%I A082732 #15 Aug 09 2019 14:28:39
%S A082732 1,3,4,13,157,24493,599882557,359859081592975693,
%T A082732 129498558604939936868397356895854557,
%U A082732 16769876680757063368089314196389622249367851612542961252860614401811693
%N A082732 a(1) = 1, a(2) = 3, a(n) = LCM of all the previous terms + 1.
%C A082732 The LCM is in fact the product of all previous terms. From a(5) onwards the terms alternately end in 57 and 93.
%F A082732 For n>=3, a(n+1) = a(n)^2 - a(n) + 1.
%F A082732 For n>=3, a(n) = A004168(n-3) + 1. - _Max Alekseyev_, Aug 09 2019
%F A082732 1/3 = Sum_{n=3..oo} 1/a(n) = 1/4 + 1/13 + 1/157 + 1/24493 + ... or 1 = Sum_{n=3..oo} 3/a(n) = 3/4 + 3/13 + 3/157 + 3/24493 + .... If we take segment of length 1 and cut off in each step fragment of maximal length such that numerator of fraction is 3, denominators of such fractions will be successive numbers of this sequence. - _Artur Jasinski_, Sep 22 2008
%F A082732 a(n+2)=1.8806785436830780944921917650127503562630617563236301969047995953391\
%F A082732 4798717695395204087358090874194124503892563356447954254847544689332763...^(2^n). -  _Artur Jasinski_, Sep 22 2008
%t A082732 a[1] = 1; a[2] = 3; a[n_] := Apply[LCM, Table[a[i], {i, 1, n - 1}]] + 1; Table[ a[n], {n, 1, 10}]
%t A082732 c=1.8806785436830780944921917650127503562630617563236301969047995953391479871\
%t A082732 7695395204087358090874194124503892563356447954254847544689332763; Table[c^(2^n),{n,1,6}] or a = {}; k = 4; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a (* _Artur Jasinski_, Sep 22 2008 *)
%Y A082732 Cf. A000058, A004168, A144743, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.
%K A082732 nonn
%O A082732 1,2
%A A082732 _Amarnath Murthy_, Apr 14 2003
%E A082732 More terms from _Robert G. Wilson v_, Apr 15 2003

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