

A079583


a(n) = 3*2^n  n  2.


17



1, 3, 8, 19, 42, 89, 184, 375, 758, 1525, 3060, 6131, 12274, 24561, 49136, 98287, 196590, 393197, 786412, 1572843, 3145706, 6291433, 12582888, 25165799, 50331622, 100663269, 201326564, 402653155, 805306338, 1610612705, 3221225440
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OFFSET

0,2


COMMENTS

Consider the infinite sequence of strings x(1) = a, x(2) = aba, x(3) = ababbaba, ..., where x(n+1) = x(n).b^{n+1}.x(n), for n >= 1. Each x(n), for n >= 2, has borders x(1), x(2), ..., x(n1), none of which cover x(n). The length of x(n+1) is 3*2^nn2.  William F. Smyth, Feb 29 2012
Number of edges in the rooted tree g[n] (n>=0) defined recursively in the following manner: denoting by P[n] the path on n vertices, we define g[0] =P[2] while g[n] (n>=1) is the tree obtained by identifying the roots of 2 copies of g[n1] and one of the endvertices of P[n+1]; the root of g[n] is defined to be the other endvertex of P[n+1]. Roughly speaking, g[4], for example, is obtained from the planted full binary tree of height 5 by replacing the edges at the levels 1,2,3,4 with paths of lengths 4, 3, 2, and 1, respectively.  Emeric Deutsch, Aug 08 2013


REFERENCES

T. Flouri, C. S. Iliopoulos, T. Kociumaka, S. P. Pissis, S. J. Puglisi, W. F. Smyth, W. Tyczynski, New and efficient approaches to the quasiperiodic characterization of a string, Proc. Prague Stringology Conf., 2012, 7588.


LINKS



FORMULA

a(0)=1, a(n) = 2*a(n1) + n;
G.f.: (x^2x+1)/((12*x)*(1x)^2) = 3*U(0)x, where U(k) = 1  (k+2)/(3*2^k  18*x*4^k/(6*x*2^k  (k+2)/U(k+1))); (continued fraction, 3step).  Sergei N. Gladkovskii, Jul 04 2012


MATHEMATICA



PROG

(Magma) I:=[1, 3, 8]; [n le 3 select I[n] else 4*Self(n1)5*Self(n2)+2*Self(n3): n in [1..40]]; // Vincenzo Librandi, Jun 23 2012


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



