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%I #13 Mar 07 2017 11:32:14
%S 0,0,0,0,0,1,0,0,0,2,0,1,1,1,1,0,0,1,0,2,0,3,0,1,0,2,0,1,2,2,0,0,2,2,
%T 1,1,1,1,1,2,1,1,1,3,1,2,1,1,0,2,1,2,0,1,1,1,0,3,0,2,1,4,0,0,1,3,0,2,
%U 1,2,2,1,0,2,1,1,2,2,3,2,0,3,0,1,0,2,2,3,1,2,1,2,3,3,1,1,0,1,2,2,2,2,0,2,1,1,1,1,2,3
%N The difference between the levels of the highest and lowest leaves in the rooted tree with Matula-Goebel number n.
%C The Matula-Goebel number of a rooted tree is defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T.
%D F. Goebel, On a 1-1-correspondence between rooted trees and natural numbers, J. Combin. Theory, B 29 (1980), 141-143.
%D I. Gutman and A. Ivic, On Matula numbers, Discrete Math., 150, 1996, 131-142.
%D I. Gutman and Yeong-Nan Yeh, Deducing properties of trees from their Matula numbers, Publ. Inst. Math., 53 (67), 1993, 17-22.
%D D. W. Matula, A natural rooted tree enumeration by prime factorization, SIAM Review, 10, 1968, 273.
%H E. Deutsch, <a href="http://arxiv.org/abs/1111.4288">Tree statistics from Matula numbers</a>, arXiv preprint arXiv:1111.4288, 2011
%H <a href="/index/Mat#matula">Index entries for sequences related to Matula-Goebel numbers</a>
%F In A184154 one constructs for each n the generating polynomial P(n,x) of the leaves of the rooted tree with Matula-Goebel number n, according to their levels. a(n) = degree of the numerator of P(n,1/x) (see the Maple program).
%e a(7)=0 because the rooted tree with Matula-Goebel number 7 is the rooted tree Y with all leaves at level 2.
%e a(2^m)=0 because the rooted tree with Matula-Goebel number 2^m is the star with m edges; all leaves are at level 1.
%p with(numtheory): a := proc (n) local r, s, P: r := proc (n) options operator, arrow: op(1, factorset(n)) end proc: s := proc (n) options operator, arrow: n/r(n) end proc: P := proc (n) if n = 1 then 1 elif bigomega(n) = 1 then sort(expand(x*P(pi(n)))) else sort(P(r(n))+P(s(n))) end if end proc: degree(numer(subs(x = 1/x, P(n)))) end proc; seq(a(n), n = 1 .. 110);
%Y Cf. A184154
%K nonn
%O 1,10
%A _Emeric Deutsch_, Oct 17 2011
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