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A109845 a(1) = 2; a(2n) = least common multiple of all previous terms + 1. a(2n+1) = least common multiple of all previous terms - 1. 3

%I #17 Feb 23 2018 23:02:19

%S 2,3,5,31,929,863971,746445024869,557180175152428473492031,

%T 310449747582890872093779269721785644810947012929,

%U 96379045774280656880008037888192772255684941220159788508646084243678677683026025975278640171971

%N a(1) = 2; a(2n) = least common multiple of all previous terms + 1. a(2n+1) = least common multiple of all previous terms - 1.

%C For n > 2 a(2n) == 1 mod 10 and a(2n+1) == 9 mod 10. Conjecture : There are infinitely many primes in this sequence.

%C Number of decimal digits of a(n): 1, 1, 1, 2, 3, 6, 12, 24, 48, 95, 190, 380, 760, 1520, 3040, 6079, 12158, 24316, 48632, 97264, …, . - _Robert G. Wilson v_, Jun 19 2016

%H Robert G. Wilson v, <a href="/A109845/b109845.txt">Table of n, a(n) for n = 1..13</a>

%F a(2n) = 1+ Product_{n=1..2n-1} a(k). a(2n+1) = -1 + Product_{n=1..2n} a(k).

%F a(1)=2, a(2)=3 and a(n)=a(n-1)^2+(-1)^n*a(n-1)+(-1)^n for n>=3. - _Walter Kehowski_, Aug 15 2005

%p a[1]:=2: for n from 1 to 5 do a[2*n]:=1+lcm(seq(a[i],i=1..2*n-1)); a[2*n+1]:=-1+lcm(seq(a[j],j=1..2*n)): od: seq(a[n],n=1..10); # _Emeric Deutsch_, Aug 06 2005

%p a := proc(n) option remember; if n=1 then 2 elif n=2 then 3 else a(n-1)^2 + (-1)^n*a(n-1) + (-1)^n fi end: # _Walter Kehowski_, Aug 15 2005

%t f[n_] := If[OddQ@ n, Fold[LCM, 1, Array[f, n - 1]] - 1, Fold[LCM, 1, Array[f, n - 1]] + 1]; f[1] = 2; Array[f, 10] (* _Robert G. Wilson v_, Jun 19 2016 *)

%K easy,nonn

%O 1,1

%A _Amarnath Murthy_, Jul 06 2005

%E More terms from _Emeric Deutsch_, Aug 06 2005

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Last modified April 24 18:17 EDT 2024. Contains 371962 sequences. (Running on oeis4.)