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Expansion of (1+x)^4/(1-x).
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%I #48 Oct 02 2023 11:45:05

%S 1,5,11,15,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,

%T 16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,

%U 16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16

%N Expansion of (1+x)^4/(1-x).

%C For n>=4 a(n)=2^4=16. This sequence is the transform of A115291 by the following transform T: T(u_0,u_1,u_2,u_3,u_4,...)=(u_0, u_0+u_1, u_1+u_2,u_2+u_3, ...); we observe that T(A040000)=A113311 and also T(A113311)=A115291.

%C Also continued fraction expansion of (55305+sqrt(65))/46231. - _Bruno Berselli_, Sep 23 2011

%H Vincenzo Librandi, <a href="/A171418/b171418.txt">Table of n, a(n) for n = 0..200</a>

%H Richard Choulet, <a href="https://mp.sbpm.be/MP157.PDF">Une nouvelle formule combinatoire ?</a>, Mathématique et Pédagogie, 157 (2006), p. 53-60.

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (1).

%F a(n) = Sum_{k=0..floor(n/2)} binomial(5,n-2*k).

%e a(3) = C(5,3-0)+C(5,3-2) = 10+5 = 15.

%p m:=5:for n from 0 to m+1 do a(n):=sum('binomial(m,n-2*k)',k=0..floor(n/2)): od : seq(a(n),n=0..m+1);

%Y Cf. A040000, A113311, A115291, A171440, A171441, A171442, A171443.

%K nonn,easy

%O 0,2

%A _Richard Choulet_, Dec 08 2009

%E Definition rewritten by _Bruno Berselli_, Sep 23 2011