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%I #103 Sep 08 2022 08:44:32
%S 1,6,30,150,750,3750,18750,93750,468750,2343750,11718750,58593750,
%T 292968750,1464843750,7324218750,36621093750,183105468750,
%U 915527343750,4577636718750,22888183593750,114440917968750,572204589843750,2861022949218750,14305114746093750
%N Expansion of (1+x)/(1-5*x).
%C Coordination sequence for infinite tree with valency 6.
%C The n-th term of the coordination sequence of the infinite tree with valency 2m is the same as the number of reduced words of size n in the free group on m generators. In the five sequences A003946, A003948, A003950, A003952, A003954, m is 2, 3, 4, 5, 6. - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001 and Ola Veshta (olaveshta(AT)my-deja.com), Mar 30 2001
%C Hamiltonian path in S_4 X P_2n.
%C For n>=1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5,6} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+1} and fixed y_1, y_2,...,y_n in {1,2,3,4,5,6} we have f(x_i)<>y_i, (i=1..n). - _Milan Janjic_, May 10 2007
%C For n>=1, a(n) equals the numbers of words of length n over the alphabet {0..5} with no two adjacent letters identical. - _Milan Janjic_, Jan 31 2015 [Corrected by _David Nacin_, May 30 2017]
%C a(n) equals the numbers of sequences of length n on {0,...,5} where no two adjacent terms differ by three. - _David Nacin_, May 30 2017
%H T. D. Noe, <a href="/A003948/b003948.txt">Table of n, a(n) for n = 0..200</a>
%H F. Faase, <a href="http://www.iwriteiam.nl/counting.html">Counting Hamiltonian cycles in product graphs</a>
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=307">Encyclopedia of Combinatorial Structures 307</a>
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%H A. M. Nemirovsky et al., <a href="http://dx.doi.org/10.1007/BF01049010">Marriage of exact enumeration and 1/d expansion methods: lattice model of dilute polymers</a>, J. Statist. Phys., 67 (1992), 1083-1108.
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>
%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (5).
%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>
%F G.f.: (1+x)/(1-5*x).
%F a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 4. - _Philippe Deléham_, Jul 10 2005
%F The Hankel transform of this sequence is [1,-6,0,0,0,0,0,0,0,0,...]. - _Philippe Deléham_, Nov 21 2007
%F a(n) = 6*5^(n-1) for n>0, a(0)=1. - _Vincenzo Librandi_, Nov 18 2010
%F G.f.: 2/x - 5 - 8/(x*U(0)) where U(k)= 1 + 2/(3^k - 3^k/(2 + 1 - 12*x*3^k/(6*x*3^k + 1/U(k+1)))) ; (continued fraction, 4-step). - _Sergei N. Gladkovskii_, Oct 30 2012
%F E.g.f.: (6*exp(5*x) - 1)/5. - _Ilya Gutkovskiy_, Dec 10 2016
%F Sum_{n>=0} 1/a(n) = 29/24. - _Bernard Schott_, Oct 25 2021
%p k := 6; if n = 0 then 1 else k*(k-1)^(n-1); fi;
%t q = 6; Join[{a = 1}, Table[If[n != 0, a = q*a - a, a = q*a], {n, 0, 25}]] (* and *) Join[{1}, 6*5^Range[0, 25]] (* _Vladimir Joseph Stephan Orlovsky_, Jul 11 2011 *)
%t Join[{1},NestList[5#&,6,30]] (* _Harvey P. Dale_, Dec 31 2013 *)
%t CoefficientList[Series[(1+x)/(1-5x), {x,0,30}], x] (* _Michael De Vlieger_, Dec 10 2016 *)
%o (PARI) Vec((1+x)/(1-5*x)+O(x^30)) \\ _Charles R Greathouse IV_, Nov 20 2012
%o (Magma) [1] cat [6*5^(n-1): n in [1..30]]; // _G. C. Greubel_, Sep 24 2019
%o (Sage) [1]+[6*5^(n-1) for n in (1..30)] # _G. C. Greubel_, Sep 24 2019
%o (GAP) Concatenation([1], List([1..30], n-> 6*5^(n-1) )); # _G. C. Greubel_, Sep 24 2019
%Y Cf. A003946, A003947, A003949, A003950, A003952, A003954, A029653.
%K nonn,easy,nice,walk
%O 0,2
%A _N. J. A. Sloane_
%E Definition corrected by _Frans J. Faase_, Feb 07 2009
%E Edited by _N. J. A. Sloane_, Dec 04 2009
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