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%I #67 Sep 08 2022 08:45:08
%S 1,1,6,41,306,2426,20076,171481,1500666,13386206,121267476,1112674026,
%T 10318939956,96572168916,910896992856,8650566601401,82644968321226,
%U 793753763514806,7659535707782916,74225795172589006,722042370787826076
%N a(0)=1, for n>=1 a(n) = Sum_{k=0..n} 5^k*N(n,k) where N(n,k) = C(n,k)*C(n,k+1)/n are the Narayana numbers (A001263).
%C More generally coefficients of (1 + m*x - sqrt(m^2*x^2 - (2*m+2)*x + 1) )/( 2*m*x ) are given by a(n) = Sum_{k=0..n} (m+1)^k * N(n,k).
%C a(n) is the series reversion of x*(1-5*x)/(1-4*x). a(n+1) is the series reversion of x/(1 + 6*x + 5*x^2). a(n+1) counts (6,5)-Motzkin paths of length n, where there are 6 colors available for the H(1,0) steps and 5 for the U(1,1) steps. - _Paul Barry_, May 19 2005
%C The Hankel transform of this sequence is 5^C(n+1,2). - _Philippe Deléham_, Oct 29 2007
%C a(n) is the number of Schröder paths of semilength n in which there are no (2,0)-steps at level 0 and at a higher level they come in 4 colors. Example: a(2)=6 because we have UDUD, UUDD, UBD, UGD, URD, and UYD, where U=(1,1), D=(1,-1), while B, G, R, and Y are, respectively, blue, green, red, and yellow (2,0)-steps. - _Emeric Deutsch_, May 02 2011
%C Shifts left when INVERT transform applied five times. - _Benedict W. J. Irwin_, Feb 03 2016
%H Vincenzo Librandi, <a href="/A078009/b078009.txt">Table of n, a(n) for n = 0..200</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html">On Integer-Sequence-Based Constructions of Generalized Pascal Triangles</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
%H Xiaomei Chen, Yuan Xiang, <a href="https://arxiv.org/abs/2009.04900">Counting generalized Schröder paths</a>, arXiv:2009.04900 [math.CO], 2020.
%F G.f.: (1 + 4*x - sqrt(16*x^2 - 12*x + 1))/(10*x).
%F a(n) = Sum_{k=0..n} A088617(n, k)*5^k*(-4)^(n-k). - _Philippe Deléham_, Jan 21 2004
%F With offset 1 : a(1)=1, a(n) = -4*a(n-1) + 5*Sum_{i=1..n-1} a(i)*a(n-i). - _Benoit Cloitre_, Mar 16 2004
%F a(n+1) = Sum_{k=0..floor(n/2)} C(n, 2*k)*C(k)*6^(n-2k)*5^k; - _Paul Barry_, May 19 2005
%F a(n) = ( 6*(2*n-1)*a(n-1) - 16*(n-2)*a(n-2) ) / (n+1) for n >= 2, a(0) = a(1) = 1. - _Philippe Deléham_, Aug 19 2005
%F From _Gary W. Adamson_, Jul 08 2011: (Start)
%F a(n) = upper left term in M^n, M = the production matrix:
%F 1, 1
%F 5, 5, 5
%F 1, 1, 1, 1
%F 5, 5, 5, 5, 5
%F 1, 1, 1, 1, 1, 1
%F ... (End)
%F a(n) ~ sqrt(10+6*sqrt(5))*(6+2*sqrt(5))^n/(10*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 13 2012. Equivalently, a(n) ~ 2^(2*n) * phi^(2*n + 1) / (5^(3/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Dec 08 2021
%F a(n) = A127848(n) for n > 0. - _Philippe Deléham_, Apr 03 2013
%F G.f.: 1/(1 - x/(1 - 5*x/(1 - x/(1 - 5*x/(1 - x/(1 - ...)))))), a continued fraction. - _Ilya Gutkovskiy_, Apr 21 2017
%F a(n) = hypergeom([1 - n, -n], [2], 5). - _Peter Luschny_, Mar 19 2018
%p A078009_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
%p for w from 1 to n do a[w] := a[w-1]+5*add(a[j]*a[w-j-1],j=1..w-1) od;
%p convert(a, list) end: A078009_list(20); # _Peter Luschny_, May 19 2011
%t Table[SeriesCoefficient[(1+4*x-Sqrt[16*x^2-12*x+1])/(10*x),{x,0,n}],{n,0,30}] (* _Vaclav Kotesovec_, Oct 13 2012 *)
%t a[n_] := Hypergeometric2F1[1 - n, -n, 2, 5];
%t Table[a[n], {n, 0, 30}] (* _Peter Luschny_, Mar 19 2018 *)
%o (PARI) a(n)=sum(k=0,n,5^k/n*binomial(n,k)*binomial(n,k+1))
%o (Magma) R<x>:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1+4*x - Sqrt(16*x^2-12*x+1))/(10*x) )); // _G. C. Greubel_, Jun 28 2019
%o (Magma) [1] cat [&+[5^k*Binomial(n,k)*Binomial(n,k+1)/n:k in [0..n]]:n in [1..20]]; // _Marius A. Burtea_, Jan 21 2020
%o (Sage) a=((1+4*x -sqrt(16*x^2-12*x+1))/(10*x)).series(x, 30).coefficients(x, sparse=False); [1]+a[1:] # _G. C. Greubel_, Jun 28 2019
%Y Cf. A001003, A007564, A059231, A127848.
%K nonn
%O 0,3
%A _Benoit Cloitre_, May 10 2003
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