

A122946


a(0)=a(1)=0, a(2)=2; for n >= 3, a(n) = a(n1) + 4*a(n3).


3



0, 0, 2, 2, 2, 10, 18, 26, 66, 138, 242, 506, 1058, 2026, 4050, 8282, 16386, 32586, 65714, 131258, 261602, 524458, 1049490, 2095898, 4193730, 8391690, 16775282, 33550202, 67116962, 134218090, 268418898, 536886746, 1073759106, 2147434698, 4294981682, 8590018106
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

See lemma 5.2 of Reznick's preprint.
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
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


LINKS



FORMULA

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
G.f.: 2*x^2 / ((12*x)*(2*x^2+x+1)).  Colin Barker, Jun 20 2013
Limit_{n > oo} a(n)/a(n1) = 2.
a(n) = 2^(n2) + A110512(n2), for n >= 2. (End)


MATHEMATICA



PROG

(PARI) a0=a1=0; a2=2; for(n=3, 50, a3=a2+4*a0; a0=a1; a1=a2; a2=a3; print1(a3, ", "))


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



