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Expansion of sqrt(8*x + sqrt(1 + 64*x^2)).
1

%I #30 Sep 08 2022 08:46:13

%S 1,4,8,-32,-160,896,5376,-33792,-219648,1464320,9957376,-68796416,

%T -481574912,3408068608,24343347200,-175272099840,-1270722723840,

%U 9268801044480,67971207659520,-500840477491200,-3706219533434880,27531916534087680,205237923254108160

%N Expansion of sqrt(8*x + sqrt(1 + 64*x^2)).

%C Signs are important to distinguish this from (for example) A098579.

%C Sqrt(A(x)) = 1 + 2*x + 2*x^2 - 20*x^3 - 42*x^4 + 572*x^5 ... defines another (new) integer sequence.

%H G. C. Greubel, <a href="/A261196/b261196.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f. A(x) satisfies: A(x)^4 = 1 + 16*x*A(x)^2.

%F A(x) = sqrt(G(4*x)), where G(x) is the g.f. of A182122.

%F A(x) * A(-x) = 1.

%F A(x) = sqrt(1 + 8*x + 32*x^2*C(-16*x^2)), where C(x) is the g.f. of A000108.

%F a(n) = A002420(n)*2^n*(-1)^(n*(n+1)/2). - _M. F. Hasler_, Aug 14 2015

%F Conjecture D-finite with recurrence: n*(n-1)*a(n) +(n-1)*(n-2)*a(n-1) +16*(2*n-3)*(2*n-5)*a(n-2) +16*(2*n-5)*(2*n-7)*a(n-3)=0. - _R. J. Mathar_, Jun 07 2016

%e A(x) = 1 + 4*x + 8*x^2 - 32*x^3 - 160*x^4 + 896*x^5 + 5376*x^6 ...

%e A(x)^2 = 1 + 8*x + 32*x^2 + 0*x^3 - 512*x^4 + 0*x^5 +16384*x^6 ...

%e A(x)^4 = 1 + 16*x + 128*x^2 + 512*x^3 + 0*x^4 -8192*x^5 + 0*x^6 ...

%t CoefficientList[Series[Sqrt[8 x + Sqrt[1 + 64 x^2]], {x, 0, 45}], x] (* _Vincenzo Librandi_, Aug 12 2015 *)

%o (PARI) Vec(sqrt(8*x + sqrt(1 + 64*x^2))) \\ _M. F. Hasler_, Aug 14 2015

%o (Magma) m:=50; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!(Sqrt(8*x+Sqrt(1+64*x^2)))); // _G. C. Greubel_, Aug 12 2018

%Y Cf. A000108, A002420, A098579, A182122.

%K sign,easy

%O 0,2

%A _Werner Schulte_, Aug 11 2015