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%I #44 Mar 09 2024 11:19:43
%S 0,0,2,2,2,10,18,26,66,138,242,506,1058,2026,4050,8282,16386,32586,
%T 65714,131258,261602,524458,1049490,2095898,4193730,8391690,16775282,
%U 33550202,67116962,134218090,268418898,536886746,1073759106,2147434698,4294981682,8590018106
%N a(0)=a(1)=0, a(2)=2; for n >= 3, a(n) = a(n-1) + 4*a(n-3).
%C See lemma 5.2 of Reznick's preprint.
%C Conjecture: count of even Markov numbers in generation n (with generations 0, 1 and 2 labeled as {5}, {13, 29} and {34, 194, 433, 169}. (Checked up to generation 20.) - _Wouter Meeussen_, Jan 16 2024
%C Wouter Meeussen's conjecture is true. Proof: label the Markov tree with Markov triples according to the scheme described at A368546. Mod 2, the triples are: row 0: (1,1,0); row 1: (1,1,1), (1,1,0); row 2: (1,0,1), (1,0,1), (1,1,1), (1,1,0); row 3: (1,1,0), (0,1,1), (1,1,0), (0,1,1), (1,0,1), (1,0,1), (1,1,1), (1,1,0); etc. Note that the Markov number labels of the tree (the center numbers of the triples) in rows 0 and 1 include no even numbers, while those in row 2 include two even numbers. Observing that the second triple in row 1 and the first four triples in row 3 are the same or the reverse of the root triple, and noting that every vertex in row 3 and beyond is in a subtree with one of these triples as root, the recurrence follows. - _William P. Orrick_, Mar 05 2024
%H A.H.M. Smeets, <a href="/A122946/b122946.txt">Table of n, a(n) for n = 0..300</a>
%H Bruce Reznick, <a href="https://arxiv.org/abs/math/0610601">Regularity properties of the Stern enumeration of the rationals</a>, arXiv:math/0610601 [math.NT], 2006.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,4).
%F a(n) = (1/7)*2^(-2 + n/2)*(7*2^(n/2) - 7*cos(n*(Pi - arctan(sqrt(7)))) + 5*sqrt(7)*sin(n*(-Pi + arctan(sqrt(7))))). - _Zak Seidov_, Oct 26 2006
%F G.f.: 2*x^2 / ((1-2*x)*(2*x^2+x+1)). - _Colin Barker_, Jun 20 2013
%F a(n) = 2 * A089977(n-2) for n >= 2. - _Alois P. Heinz_, Jan 16 2024
%F From _A.H.M. Smeets_, Jan 16 2024: (Start)
%F Limit_{n -> oo} a(n)/a(n-1) = 2.
%F a(n) = 2^(n-2) + A110512(n-2), for n >= 2. (End)
%t LinearRecurrence[{1,0,4},{0,0,2},36] (* _James C. McMahon_, Jan 16 2024 *)
%o (PARI) a0=a1=0;a2=2;for(n=3,50,a3=a2+4*a0;a0=a1;a1=a2;a2=a3;print1(a3,","))
%Y Cf. A002487, A002559, A089977, A110512, A368546.
%K nonn,easy
%O 0,3
%A _Benoit Cloitre_, Oct 24 2006
%E Entries checked by _Zak Seidov_, Oct 26 2006
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