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Let u(1) = u(2) = v(1) = v(2) = 1, u(n+2) = u(n)+v(n+1), v(n+2) = abs(u(n)-v(n+1)), then a(n) = u(n).
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%I #15 Oct 08 2024 03:13:39

%S 1,1,2,1,3,2,3,5,4,7,7,8,13,9,20,15,23,32,25,53,36,67,75,70,139,81,

%T 198,161,199,358,235,521,430,561,911,612,1421,1013,1632,2223,1829,

%U 3658,2617,4699,5234,4699,9933,5234,14097,11003,14632,24565,17191,36638

%N Let u(1) = u(2) = v(1) = v(2) = 1, u(n+2) = u(n)+v(n+1), v(n+2) = abs(u(n)-v(n+1)), then a(n) = u(n).

%C For any initial values u(1), v(1), u(2), v(2), it seems that lim n ->infinity log(a(n))/n = C = 0.192...

%H Paolo Xausa, <a href="/A072515/b072515.txt">Table of n, a(n) for n = 1..1000</a>

%t a[1] = a[2] = v[1] = v[2] = 1; a[n_] := a[n] = a[n-2] + v[n-1]; v[n_] := v[n] = Abs[a[n-2] - v[n-1]];

%t Array[a, 100] (* _Paolo Xausa_, Oct 08 2024 *)

%K easy,nonn

%O 1,3

%A _Benoit Cloitre_, Aug 04 2002

%E Name corrected by _Sean A. Irvine_, Oct 07 2024