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%I #30 Apr 08 2024 12:16:47
%S 1,-1,1,-9,-31,79,161,-249,-191,-481,-2879,9111,22049,-42641,-60319,
%T 28071,-189311,897599,2643841,-6087369,-11130271,14084239,685601,
%U 67678791,274143169,-758178721,-1661999039,2857102551,3118415009,1811852719,22839485921,-82298680089,-214997290751
%N a(n) = A118443(n)/(n+1), where A118443 is the row sums of triangle A118441.
%C A118441 is the matrix log of triangle A118435.
%C Given the series S = (1, -i)^n, n>0: (1, -1), (0, -2), (-2, -2), ...; the real part of the binomial transform of S = (1, 1, -1, -9, -31, -79, -161, -249, -191, 481, ...). - _Gary W. Adamson_, Sep 19 2008
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,-6,0,-25).
%F G.f.: (1 - x + 13*x^2 - 21*x^3 + 67*x^4 - 115*x^5 + 175*x^6 - 375*x^7) / (1 + 6*x^2 + 25*x^4)^2.
%F For n > 3, a(n) = 4*a(n-1) - 5*a(n-2). - _Gary W. Adamson_, Aug 08 2006
%F E.g.f.: exp(x)*cos(2*x) - sin(2*x)*(cosh(x) - sinh(x)). - _Stefano Spezia_, Jul 01 2023
%F a(n) = (-1)^floor((n+1)/2)*(1+i)*((2+i)^n-i*(2-i)^n)/2, where i is the imaginary unit. - _Gerry Martens_, Mar 31 2024
%t LinearRecurrence[{0, -6, 0, -25}, {1, -1, 1, -9}, 33] (* _Jean-François Alcover_, Apr 08 2024 *)
%o (PARI) {a(n)=polcoeff((1-x+13*x^2-21*x^3+67*x^4-115*x^5+175*x^6-375*x^7) /(1+6*x^2+25*x^4 +x*O(x^n))^2,n)}
%Y Cf. A118441, A118443, A118435.
%K sign,easy,changed
%O 0,4
%A _Paul D. Hanna_, Apr 28 2006
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