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Expansion of sqrt(1-4x)/(1-x).
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%I #18 Sep 26 2021 05:06:22

%S 1,-1,-3,-7,-17,-45,-129,-393,-1251,-4111,-13835,-47427,-164999,

%T -581023,-2066823,-7415703,-26805393,-97520733,-356810313,-1312087713,

%U -4846614093,-17974854933,-66907388973,-249872516253,-935991743553,-3515800038201,-13239692841105

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

%C Row sums of number triangle A106190. Partial sums of A002420.

%C For n >= 1, the absolute values also give the iterates of A122237, starting from 0. (A122237(0), A122237(A122237(0)), A122237(A122237(A122237(0))), ...), this stems from the fact that the sequence gives the positions of terms with binary expansion 1(10){n-1}0 in A014486 (see A080675).

%F a(n) = Sum_{k=0..n} binomial(2k, k)/(1-2k).

%F G.f.: (2/(1-x))/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 24 2013

%F D-finite with recurrence: a(0)=1, a(1)=-1; for n>1, a(n) = (1/n)*((5*n-6)*a(n-1) - (4*n-6)*a(n-2)). - _Tani Akinari_, Aug 25 2013

%Y |a(n)| = A080300(A080675(n)) = A075161(A001348(n)) (for n >= 1) = A075163(A000244(A008578(n-2))) = A014137(n-1)+A014138(n-2) = 2*A014137(n-1)-1, for n >= 2 (because binomial(2n+2, n+1)/(2n+1) = 2*A000108(n)).

%K easy,sign

%O 0,3

%A _Paul Barry_, Apr 24 2005

%E Barry's formula made more succinct, as well as comments regarding interpretation as absolute values added by _Antti Karttunen_, Sep 14 2006