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Expansion of 1/(1-x*(1-5*x)).
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%I #42 Feb 28 2024 15:59:47

%S 1,1,-4,-9,11,56,1,-279,-284,1111,2531,-3024,-15679,-559,77836,80631,

%T -308549,-711704,831041,4389561,234356,-21713449,-22885229,85682016,

%U 200108161,-228301919,-1228842724,-87333129,6056880491,6493546136,-23790856319,-56258586999,62695694596,343988629591

%N Expansion of 1/(1-x*(1-5*x)).

%C Row sums of Riordan array (1,x*(1-5*x)). In general, a(n) = Sum_{k=0..n}(-1)^(n-k)*binomial(k,n-k)*r^(n-k), yields the row sums of the Riordan array (1,x*(1-k*x)).

%H Seiichi Manyama, <a href="/A106854/b106854.txt">Table of n, a(n) for n = 0..2859</a>

%H Taras Goy and Mark Shattuck, <a href="https://doi.org/10.2478/amsil-2023-0027">Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries</a>, Ann. Math. Silesianae (2023). See p. 16.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,-5).

%F a(n) = ((1+sqrt(-19))^(n+1)-(1-sqrt(-19))^(n+1))/(2^(n+1)sqrt(-19)).

%F a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(k, n-k)*5^(n-k).

%F a(n) = 5^(n/2)(cos(-n*acot(sqrt(19)/19))-sqrt(19)sin(-n*acot(sqrt(19)/19))/19).

%F a(n) = a(n-1)-5*a(n-2), a(0)=1, a(1)=1. - _Philippe Deléham_, Oct 21 2008

%F a(n) = Sum_{k=0..n} A109466(n,k)*5^(n-k). - _Philippe Deléham_, Oct 25 2008

%F G.f.: Q(0)/2, where Q(k) = 1 + 1/( 1 - x*(2*k+1 -5*x)/( x*(2*k+2 -5*x) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Dec 07 2013

%t Join[{a=1,b=1},Table[c=b-5*a;a=b;b=c,{n,80}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 22 2011 *)

%t CoefficientList[Series[1/(1-x(1-5x)),{x,0,40}],x] (* or *) LinearRecurrence[ {1,-5},{1,1},40] (* _Harvey P. Dale_, Jan 21 2012 *)

%o (Sage) [lucas_number1(n,1,5) for n in range(1,35)] # _Zerinvary Lajos_, Jul 16 2008

%o (PARI) Vec(1/(1-x+5*x^2) + O(x^99)) \\ _Altug Alkan_, Sep 06 2016

%o (Magma) I:=[1,1]; [n le 2 select I[n] else Self(n-1) - 5*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jan 14 2018

%Y Cf. A106852, A106853, A145934.

%K easy,sign

%O 0,3

%A _Paul Barry_, May 08 2005