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

1, 3, 4, 13, 157, 24493, 599882557, 359859081592975693, 129498558604939936868397356895854557, 16769876680757063368089314196389622249367851612542961252860614401811693

Offset: 1

Author

Amarnath Murthy, Apr 14 2003

Keywords

• nonn

Comments

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

Crossrefs

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

Programs

Mathematica

a[1] = 1; a[2] = 3; a[n_] := Apply[LCM, Table[a[i], {i, 1, n - 1}]] + 1; Table[ a[n], {n, 1, 10}]
c=1.8806785436830780944921917650127503562630617563236301969047995953391479871\
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 *)

Formula

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

For n>=3, a(n) = A004168(n-3) + 1. - Max Alekseyev, Aug 09 2019

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

a(n+2)=1.8806785436830780944921917650127503562630617563236301969047995953391\

4798717695395204087358090874194124503892563356447954254847544689332763...^(2^n). - Artur Jasinski, Sep 22 2008

Extensions

More terms from Robert G. Wilson v, Apr 15 2003