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Multiplicative with a(p^k) = 1 or -1 so as to minimize abs(Sum_{m = 1..p^k} a(m)); in case of a tie, a(p^k) = a(p^k-1).
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%I #10 Feb 19 2023 12:31:31

%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,

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

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

%N Multiplicative with a(p^k) = 1 or -1 so as to minimize abs(Sum_{m = 1..p^k} a(m)); in case of a tie, a(p^k) = a(p^k-1).

%C The partial sums (A360711) show oscillations of increasing magnitude.

%e For n = 1, a(1) = 1 (as this sequence is multiplicative).

%e For n = 2, abs(1 + 1) > abs(1 - 1), so a(2) = -1.

%e For n = 3, abs(1 - 1 + 1) = abs(1 - 1 - 1), so a(3) = a(3-1) = -1.

%o (PARI) { my (s=0, f); for (n=1, #a=vector(75), f=factor(n); print1 (a[n]=if (#f~==1, if (s, -sign(s), a[n-1]), prod(k=1, #f~, a[f[k, 1]^f[k, 2]]))", "); s+=a[n]) }

%Y Cf. A360711 (partial sums).

%K sign,mult

%O 1

%A _Rémy Sigrist_, Feb 17 2023