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a(n) = (-1)^floor(n/5) * sign(mod(n, 5)).
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%I #15 Apr 19 2023 00:36:09

%S 0,1,1,1,1,0,-1,-1,-1,-1,0,1,1,1,1,0,-1,-1,-1,-1,0,1,1,1,1,0,-1,-1,-1,

%T -1,0,1,1,1,1,0,-1,-1,-1,-1,0,1,1,1,1,0,-1,-1,-1,-1,0,1,1,1,1,0,-1,-1,

%U -1,-1,0,1,1,1,1,0,-1,-1,-1,-1,0,1,1,1,1,0,-1

%N a(n) = (-1)^floor(n/5) * sign(mod(n, 5)).

%C This is a strong elliptic divisibility sequence t_n as given in [Kimberling, p. 16] where x = 1, y = 1, z = 1. - _Michael Somos_, Mar 17 2020

%H Clark Kimberling, <a href="http://www.fq.math.ca/Scanned/17-1/kimberling1.pdf">Strong divisibility sequences and some conjectures</a>, Fib. Quart., 17 (1979), 13-17.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1,-1).

%F Euler transform of length 10 sequence [1, 0, 0, -1, -1, 0, 0, 0, 0, 1].

%F G.f.: x * (1 + x) * (1 + x^2) / (1 + x^5).

%F a(n) = A099443(n-1). a(n) = A163812(n) except n=0.

%F a(n) = (-1)^floor(n/5) * A011558(n) for all n in Z.

%F 0 = a(n)*a(n+4) - a(n+1)*a(n+3) + a(n+2)^2) = a(n)*a(n+5) - a(n+1)*a(n+4) + a(n+2)*a(n+3) for all n in Z. - _Michael Somos_, Mar 17 2020

%e G.f. = x + x^2 + x^3 + x^4 - x^6 - x^7 - x^8 - x^9 + x^11 + x^12 + ...

%t a[ n_] := (-1)^Quotient[n, 5] Sign@Mod[n, 5];

%o (PARI) {a(n) = (-1)^(n\5) * sign(n%5)};

%Y Cf. A011558, A099443, A163812, A332828.

%K sign,easy

%O 0,1

%A _Michael Somos_, Nov 27 2019