1,2

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

a(n+1) = a(n)^2 -a(n) + 1 for n >2.

1/3=1/4+1/13+1/157+1/24493+...1/a(n) n->Infinity or 1=3/4+3/13+3/157+3/24493+...3/a(n) n->Infinity [From Artur Jasinski (grafix(AT)csl.pl), Sep 21 2008]

1/3=Sum[1/a[n+2],{n,1,Infinity}]=1/4+1/13+1/157+1/24493+...1/a(n) n->Infinity or 1=Sum[3/a[n+2],{n,1,Infinity}]=3/4+3/13+3/157+3/24493+...3/a(n) n->Infinity. If we take segment of length 1 and we will be cut off in each step fragment maximal length such that numerator of fraction is 3, denominators of such fractions will be successive numbers of this sequence. [From Artur Jasinski (grafix(AT)csl.pl), Sep 22 2008]

a(n+2)=1.8806785436830780944921917650127503562630617563236301969047995953391\

4798717695395204087358090874194124503892563356447954254847544689332763...^(2^n) [From Artur Jasinski (grafix(AT)csl.pl), Sep 22 2008]

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 [From Artur Jasinski (grafix(AT)csl.pl), Sep 22 2008]

Cf. A000058.

A000058, A082732, A144743, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788 [From Artur Jasinski (grafix(AT)csl.pl), Sep 22 2008]

Sequence in context: A062165 A001056 A122151 this_sequence A009286 A076663 A079274

Adjacent sequences: A082729 A082730 A082731 this_sequence A082733 A082734 A082735

nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 14 2003

More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 15 2003