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A055101 Expansion of square of continued fraction 1/ ( 1+q/ ( 1+q^2/ ( 1+q^3/ ( 1+q^4/... )))).

%I A055101 #28 Jul 29 2024 06:16:34
%S A055101 1,-2,3,-2,-1,4,-6,6,-3,-2,9,-16,17,-10,-5,24,-36,36,-21,-10,46,-74,
%T A055101 77,-42,-22,94,-144,142,-78,-38,172,-266,266,-146,-73,312,-471,464,
%U A055101 -251,-122,534,-814,801,-432,-213,910,-1364,1328,-713,-344,1485,-2234,2178
%N A055101 Expansion of square of continued fraction 1/ ( 1+q/ ( 1+q^2/ ( 1+q^3/ ( 1+q^4/... )))).
%H A055101 Seiichi Manyama, <a href="/A055101/b055101.txt">Table of n, a(n) for n = 0..10000</a>
%H A055101 For the third power see G. E. Andrews, <a href="http://www.jstor.org/stable/2974472">Simplicity and surprise in Ramanujan's "Lost" Notebook</a>, Amer. Math. Monthly, 104 (No. 10, Dec. 1997), 918-925.
%F A055101 a(0) = 1, a(n) = -(2/n)*Sum_{k=1..n} A109091(k)*a(n-k) for n > 0. - _Seiichi Manyama_, Apr 16 2017
%F A055101 Euler transform of period 5 sequence [-2, 2, 2, -2, 0, ...]. - _Georg Fischer_, Aug 18 2020
%F A055101 From _Seiichi Manyama_, Jul 29 2024: (Start)
%F A055101 G.f.: ( Sum_{k in Z} x^(3*k) / (1 - x^(5*k+1)) ) / ( Sum_{k in Z} x^k / (1 - x^(5*k+1)) ).
%F A055101 G.f.: ( Sum_{k in Z} x^(2*k) / (1 - x^(5*k+2)) ) / ( Sum_{k in Z} x^k / (1 - x^(5*k+2)) ). (End)
%Y A055101 Product_{k>0} ((1-x^{5k-1}) * (1-x^{5k-4})/((1-x^{5k-2}) * (1-x^{5k-3})))^m: A285444 (m=-4), A285443 (m=-3), A285442 (m=-2), A003823 (m=-1), A007325 (m=1), this sequence  (m=2), A055102 (m=3), A055103 (m=4).
%Y A055101 Cf. A109091, A340453, A340454, A340455, A340456.
%K A055101 sign,easy
%O A055101 0,2
%A A055101 _N. J. A. Sloane_, Jun 14 2000
%E A055101 More terms from Kok Seng Chua (chuaks(AT)ihpc.nus.edu.sg), Jun 20 2000