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a(0) = 2, a(1) = 3, a(2) = 5; a(n) = a(n-1) + [a(n-1)-a(n-2)] * [a(n-2)-a(n-3)].
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%I #13 Sep 08 2022 08:45:08

%S 2,3,5,7,11,19,51,307,8499,2105651,17181974835,36028814200938803,

%T 618970019678718951650500915,

%U 22300745198530623760505737951367313156481331,13803492693581127574869511746854796103432841704846511061692361604079923

%N a(0) = 2, a(1) = 3, a(2) = 5; a(n) = a(n-1) + [a(n-1)-a(n-2)] * [a(n-2)-a(n-3)].

%H Robert Israel, <a href="/A079429/b079429.txt">Table of n, a(n) for n = 0..19</a>

%F Conjecture: a(n)=A011455(n-1)+5 where defined. - _R. J. Mathar_, Apr 26 2007

%F Proof of conjecture: if d(n) = log_2(a(n+1)-a(n)), we have d(0)=0, d(1)=1, d(n)=d(n-1)+d(n-2), so d(n) = Fibonacci(n). - _Robert Israel_, Oct 25 2017

%e a(3) = 7, since a(3) = a(2) + [(a(2)-a(1)) * (a(1)-a(0))] = 5 + ((5-3)*(3-2))

%p f:= proc(n) option remember; procname(n-1)+(procname(n-1)-procname(n-2))*(procname(n-2)-procname(n-3)) end proc:

%p f(0):= 2: f(1):= 3: f(2):= 5:

%p map(f, [$0..20]); # _Robert Israel_, Oct 25 2017

%t a[0] = 2; a[1] = 3; a[2] = 5; a[n_] := a[n] = a[n - 1] + (a[n - 1] - a[n - 2])*(a[n - 2] - a[n - 3]); Table[a[n], {n, 0, 14}]

%o (Magma) I:=[2,3,5]; [n le 3 select I[n] else Self(n-1)+(Self(n-1)-Self(n-2))*(Self(n-2)-Self(n-3)): n in [1..15]]; // _Vincenzo Librandi_, Oct 25 2017

%Y Cf. A011455.

%K nonn

%O 0,1

%A Ajay Chhabra (ajay(AT)cantab.net), Jan 08 2003

%E Edited, corrected and extended by _Robert G. Wilson v_, Jan 08 2002

%E Conjecture corrected by _Robert Israel_, Oct 25 2017