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a(n) = [x^n] (2*x^4 + 2*x^3 + 2*x^2 + x + 1)/(x^2 + 1).
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%I #12 Jan 25 2023 08:07:05

%S 1,1,1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,

%T 1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,

%U 1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1

%N a(n) = [x^n] (2*x^4 + 2*x^3 + 2*x^2 + x + 1)/(x^2 + 1).

%H Madeline Beals-Reid, <a href="https://journals.calstate.edu/pump/article/view/3549">A Quadratic Relation in the Bernoulli Numbers</a>, The Pump Journal of Undergraduate Research, 6 (2023), 29-39.

%F Let B(x) = x/(1 - exp(-x)), the e.g.f. of the Bernoulli numbers with B(1) = 1/2.

%F a(n) = signum([x^n] B(x)^2)) = signum([x^n] z^2 / (exp(-z) - 1)^2).

%F a(n) = signum([x^n] (x + 1)*B(x) - x*B'(x)).

%p ogf := (2*z^4 + 2*z^3 + 2*z^2 + z + 1)/(z^2 + 1):

%p ser := series(ogf, z, 100): seq(coeff(ser, z, n), n = 0..74);

%Y Cf. A266591, A100615.

%K sign

%O 0

%A _Peter Luschny_, Jan 23 2023