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Column 0 of triangle A118435.
3

%I #17 Oct 17 2024 11:22:06

%S 1,1,-3,-11,25,41,-43,29,-335,-1199,3117,6469,-10295,-8839,-16123,

%T -108691,354145,873121,-1721763,-2521451,1476985,-6699319,34182197,

%U 103232189,-242017775,-451910159,597551757,130656229,2465133865,10513816601,-29729597083,-66349305331

%N Column 0 of triangle A118435.

%C Binomial transform of A118434 = (1, 1, 3, 11, 25, 41, 43, -29, -335, -1199, ...). - _Gary W. Adamson_, Sep 19 2008

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0, -5, 0, -19, 0, 25).

%F G.f.: (1 + x + 2*x^2 - 6*x^3 + 29*x^4 + 5*x^5)/((1-x^2)*(1 + 6*x^2 + 25*x^4)).

%t LinearRecurrence[{0, -5, 0, -19, 0, 25}, {1, 1, -3, -11, 25, 41}, 32] (* _Jean-François Alcover_, Apr 08 2024 *)

%t CoefficientList[Series[(1+x+2x^2-6x^3+29x^4+5x^5)/((1-x^2)(1+6x^2+25x^4)),{x,0,40}],x] (* _Harvey P. Dale_, Oct 17 2024 *)

%o (PARI) {a(n)=polcoeff((1+x+2*x^2-6*x^3+29*x^4+5*x^5)/(1-x^2)/(1+6*x^2+25*x^4+x*O(x^n)),n)}

%Y Cf. A118434, A118435 (triangle), A118437 (row sums).

%K sign

%O 0,3

%A _Paul D. Hanna_, Apr 28 2006