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 A118434 Row sums of self-inverse triangle A118433. 8
 1, 0, 2, 4, -4, 0, -8, -16, 16, 0, 32, 64, -64, 0, -128, -256, 256, 0, 512, 1024, -1024, 0, -2048, -4096, 4096, 0, 8192, 16384, -16384, 0, -32768, -65536, 65536, 0, 131072, 262144, -262144, 0, -524288, -1048576, 1048576, 0, 2097152, 4194304, -4194304 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 LINKS Index entries for linear recurrences with constant coefficients, signature (0,0,0,-4). FORMULA O.g.f.: A(x) = (1+2*x^2+4*x^3)/(1+4*x^4). E.g.f.: A(x) = cos(x)*exp(-x) + sin(x)*exp(x). 2*a(n) = i*(1-i)^n-i*(1+i)^n + (-1-i)^n + (-1+i)^n with i=sqrt(-1). - R. J. Mathar, Jan 18 2011 MAPLE A118434 := proc(n) I*(1-I)^n-I*(1+I)^n+(-1-I)^n+(-1+I)^n ; expand(%/2) ; end proc: # R. J. Mathar, Jan 18 2011 MATHEMATICA a[n_] := 2^(Floor[(n+1)/2]-3)*(-2*Mod[n, 8] + Mod[n+2 , 8] - Mod[n+3, 8] + 2*Mod[n+4, 8] - Mod[n+6, 8] + Mod[n+7, 8]); Table[a[n], {n, 0, 44}] (* Jean-François Alcover, May 23 2013 *) PROG (PARI) {a(n)=polcoeff((1+2*x^2+4*x^3)/(1+4*x^4+x*O(x^n)), n)} (PARI) /* E.g.f.: */ {a(n)=local(x=X+X*O(X^n)); n!*polcoeff(cos(x)*exp(-x)+sin(x)*exp(x), n, X)} CROSSREFS Cf. A118433. Sequence in context: A112793 A009116 A146559 * A090132 A199051 A300190 Adjacent sequences:  A118431 A118432 A118433 * A118435 A118436 A118437 KEYWORD sign,easy AUTHOR Paul D. Hanna, Apr 28 2006 STATUS approved

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Last modified April 22 14:11 EDT 2021. Contains 343177 sequences. (Running on oeis4.)