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%I #67 Sep 08 2022 08:44:32
%S 1,5,20,80,320,1280,5120,20480,81920,327680,1310720,5242880,20971520,
%T 83886080,335544320,1342177280,5368709120,21474836480,85899345920,
%U 343597383680,1374389534720,5497558138880,21990232555520,87960930222080,351843720888320
%N Expansion of (1+x)/(1-4*x).
%C Coordination sequence for infinite tree with valency 5.
%C For n>=1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5} 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} we have f(x_i)<>y_i, (i=1,2,...,n). - _Milan Janjic_, May 10 2007
%C Number of length-n strings of 5 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - _Joerg Arndt_, Oct 11 2012
%C Create a rectangular prism with edges of lengths 2^(n-2), 2^(n-1), and 2^(n) starting at n=2; then the surface area = a(n). - _J. M. Bergot_, Aug 08 2013
%H T. D. Noe, <a href="/A003947/b003947.txt">Table of n, a(n) for n = 0..200</a>
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=306">Encyclopedia of Combinatorial Structures 306</a>
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%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 (4).
%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>
%F Binomial transform of A060925. Its binomial transform is A003463 (without leading zero). - _Paul Barry_, May 19 2003
%F From _Paul Barry_, May 19 2003: (Start)
%F a(n) = (5*4^n - 0^n)/4.
%F G.f.: (1+x)/(1-4*x).
%F E.g.f.: (5*exp(4*x) - exp(0))/4. (End)
%F a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 3. - _Philippe Deléham_, Jul 10 2005
%F a(n) = A146523(n)*A011782(n). - _R. J. Mathar_, Jul 08 2009
%F a(n) = 5*A000302(n-1), n>0.
%F a(n) = 4*a(n-1), n>1. - _Vincenzo Librandi_, Dec 31 2010
%F G.f.: 2+x- 2/G(0), where G(k)= 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 1/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 04 2013
%p k := 5; if n = 0 then 1 else k*(k-1)^(n-1); fi;
%t q = 5; Join[{a = 1}, Table[If[n != 0, a = q*a - a, a = q*a], {n, 0, 25}]] (* and *) Join[{1}, 5*4^Range[0, 25]] (* _Vladimir Joseph Stephan Orlovsky_, Jul 11 2011 *)
%t LinearRecurrence[{4},{1,5},30] (* _Harvey P. Dale_, Apr 19 2015 *)
%o (PARI) a(n)=5*4^n\4 \\ _Charles R Greathouse IV_, Sep 08 2011
%o (Magma) [1] cat [5*4^(n-1): n in [1..30]]; // _G. C. Greubel_, Aug 10 2019
%o (Sage) [1]+[5*4^(n-1) for n in (1..30)] # _G. C. Greubel_, Aug 10 2019
%o (GAP) Concatenation([1], List([1..30], n-> 5*4^(n-1) )); # _G. C. Greubel_, Aug 10 2019
%Y Cf. A003948, A003949. Column 5 in A265583.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
%E Edited by _N. J. A. Sloane_, Dec 04 2009
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