A118444 a(n) = A118443(n)/(n+1), where A118443 is the row sums of triangle A118441.

1, -1, 1, -9, -31, 79, 161, -249, -191, -481, -2879, 9111, 22049, -42641, -60319, 28071, -189311, 897599, 2643841, -6087369, -11130271, 14084239, 685601, 67678791, 274143169, -758178721, -1661999039, 2857102551, 3118415009, 1811852719, 22839485921, -82298680089, -214997290751

Offset

0,4

Comments

A118441 is the matrix log of triangle A118435.

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

Links

Table of n, a(n) for n=0..32.

Index entries for linear recurrences with constant coefficients, signature (0,-6,0,-25).

Formula

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.

For n > 3, a(n) = 4*a(n-1) - 5*a(n-2). - Gary W. Adamson, Aug 08 2006

E.g.f.: exp(x)*cos(2*x) - sin(2*x)*(cosh(x) - sinh(x)). - Stefano Spezia, Jul 01 2023

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

Mathematica

LinearRecurrence[{0, -6, 0, -25}, {1, -1, 1, -9}, 33] (* Jean-François Alcover, Apr 08 2024 *)

Prog

(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)}

Crossrefs

Cf. A118441, A118443, A118435.

Sequence in context: A177342 A224000 A373061 * A321598 A048374 A226274

Adjacent sequences: A118441 A118442 A118443 * A118445 A118446 A118447

Keyword

sign,easy

Author

Paul D. Hanna, Apr 28 2006

Status

approved