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%I #43 Sep 08 2022 08:45:08
%S 1,1,7,55,469,4237,39907,387739,3858505,39130777,402972031,4202705311,
%T 44299426717,471189693925,5051001609115,54513542257795,
%U 591858123926545,6459813793353265,70837427884259575,780073647992404615
%N a(n) = Sum_{k=0..n} 6^k*N(n,k), with a(0)=1, 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+4)*x + 1) )/( (2*m+2)*x ) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k).
%C The Hankel transform of this sequence is 6^C(n+1,2). - _Philippe Deléham_, Oct 29 2007
%C Shifts left when INVERT transform applied six times. - _Benedict W. J. Irwin_, Feb 07 2016
%H Vincenzo Librandi, <a href="/A078018/b078018.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.
%F G.f.: (1 + 5*x - sqrt(25*x^2-14*x+1))/(12*x).
%F a(n) = Sum_{k=0..n} A088617(n, k)*6^k*(-5)^(n-k). - _Philippe Deléham_, Jan 21 2004
%F a(n) = ( 7*(2*n-1)*a(n-1) - 25*(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 6, 6, 6;
%F 1, 1, 1, 1;
%F 6, 6, 6, 6, 6;
%F 1, 1, 1, 1, 1, 1;
%F ... (End)
%F a(n) ~ sqrt(12+7*sqrt(6))*(7+2*sqrt(6))^n/(12*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 13 2012
%F G.f.: 1/(1 - x/(1 - 6*x/(1 - x/(1 - 6*x/(1 - x/(1 - ...)))))), a continued fraction. - _Ilya Gutkovskiy_, Apr 21 2017
%F a(n) = hypergeom([1 - n, -n], [2], 6). - _Peter Luschny_, Mar 19 2018
%p A078018_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]+6*add(a[j]*a[w-j-1],j=1..w-1) od;
%p convert(a, list) end: A078018_list(19);
%p # _Peter Luschny_, May 19 2011
%t Table[SeriesCoefficient[(1+5*x-Sqrt[25*x^2-14*x+1])/(12*x),{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 13 2012 *)
%t a[n_]:= Hypergeometric2F1[1 - n, -n, 2, 6]; Table[a[n], {n, 0, 20}] (* _Peter Luschny_, Mar 19 2018 *)
%o (PARI) a(n)=if(n<1,1,sum(k=0,n,6^k/n*binomial(n,k)*binomial(n,k+1)))
%o (Magma) R<x>:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1 + 5*x - Sqrt(25*x^2-14*x+1))/(12*x) )); // _G. C. Greubel_, Jun 29 2019
%o (Sage) a=((1 + 5*x - sqrt(25*x^2-14*x+1))/(12*x)).series(x, 30).coefficients(x, sparse=False); [1]+a[1:] # _G. C. Greubel_, Jun 29 2019
%Y Cf. A001003, A007564, A059231.
%K nonn
%O 0,3
%A _Benoit Cloitre_, May 10 2003
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