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A082732 a(1) = 1, a(2) = 3, a(n) = LCM of all the previous terms + 1. 21

%I #15 Aug 09 2019 14:28:39

%S 1,3,4,13,157,24493,599882557,359859081592975693,

%T 129498558604939936868397356895854557,

%U 16769876680757063368089314196389622249367851612542961252860614401811693

%N a(1) = 1, a(2) = 3, a(n) = LCM of all the previous terms + 1.

%C The LCM is in fact the product of all previous terms. From a(5) onwards the terms alternately end in 57 and 93.

%F For n>=3, a(n+1) = a(n)^2 - a(n) + 1.

%F For n>=3, a(n) = A004168(n-3) + 1. - _Max Alekseyev_, Aug 09 2019

%F 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 a(n+2)=1.8806785436830780944921917650127503562630617563236301969047995953391\

%F 4798717695395204087358090874194124503892563356447954254847544689332763...^(2^n). - _Artur Jasinski_, Sep 22 2008

%t 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 c=1.8806785436830780944921917650127503562630617563236301969047995953391479871\

%t 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 Cf. A000058, A004168, A144743, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Apr 14 2003

%E More terms from _Robert G. Wilson v_, Apr 15 2003

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)