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A082732 a(1) = 1, a(2) = 3, a(n) = LCM of all the previous terms + 1. 21
1, 3, 4, 13, 157, 24493, 599882557, 359859081592975693, 129498558604939936868397356895854557, 16769876680757063368089314196389622249367851612542961252860614401811693 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
The LCM is in fact the product of all previous terms. From a(5) onwards the terms alternately end in 57 and 93.
LINKS
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
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 *)
CROSSREFS
Sequence in context: A122151 A294384 A216868 * A307893 A337297 A220846
KEYWORD
nonn
AUTHOR
Amarnath Murthy, Apr 14 2003
EXTENSIONS
More terms from Robert G. Wilson v, Apr 15 2003
STATUS
approved

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Last modified March 29 05:43 EDT 2024. Contains 371264 sequences. (Running on oeis4.)