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%I #16 Feb 09 2024 11:02:03
%S 1,-1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N a(0) = -a(1) = a(2) = 1, a(n) = 0 for n>2.
%C The binomial transform is 1, 0, 0, 1, 3, 6, 10, 15,..., i.e. A161680 with a 1 in front. The inverse binomial transform is 1, -2, 4, -7, 11, -16, 22, -29, 37, -46, 56,.. a variant of A000124. - _R. J. Mathar_, Feb 16 2023
%F Euler transform of length 6 sequence [ -1, 1, 1, 0, 0, -1].
%F Given g.f. A(x), then B(q) = A(q) / q satisfies 0 = f(B(q), B(q^2)) where f(u, v) = 2 + v - u * (u + 2).
%F G.f.: (1 + x^3) / (1 + x).
%F a(n) = (-1)^n * A130716(n).
%F G.f. is the sixth cyclotomic polynomial.
%F Convolution inverse is A010892.
%e G.f. = 1 - x + x^2.
%e G.f. = 1/q - 1 + q.
%t PadRight[{1, -1, 1}, 100] (* _Paolo Xausa_, Feb 09 2024 *)
%o (PARI) {a(n) = (-1)^n * (n>=0 && n<=2)};
%Y Cf. A000124, A010892, A130716, A161680.
%K sign,easy
%O 0
%A _Michael Somos_, Oct 07 2015
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