A079429 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)].
2, 3, 5, 7, 11, 19, 51, 307, 8499, 2105651, 17181974835, 36028814200938803, 618970019678718951650500915, 22300745198530623760505737951367313156481331, 13803492693581127574869511746854796103432841704846511061692361604079923
Offset: 0
Keywords
Examples
a(3) = 7, since a(3) = a(2) + [(a(2)-a(1)) * (a(1)-a(0))] = 5 + ((5-3)*(3-2))
Links
- Robert Israel, Table of n, a(n) for n = 0..19
Crossrefs
Cf. A011455.
Programs
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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
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Maple
f:= proc(n) option remember; procname(n-1)+(procname(n-1)-procname(n-2))*(procname(n-2)-procname(n-3)) end proc: f(0):= 2: f(1):= 3: f(2):= 5: map(f, [$0..20]); # Robert Israel, Oct 25 2017
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Mathematica
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}]
Formula
Conjecture: a(n)=A011455(n-1)+5 where defined. - R. J. Mathar, Apr 26 2007
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
Extensions
Edited, corrected and extended by Robert G. Wilson v, Jan 08 2002
Conjecture corrected by Robert Israel, Oct 25 2017