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%I #27 Feb 17 2020 16:49:35
%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,
%U -1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1
%N The unique function f with f(1)=1 and f(jD!+k)=(-1)^j f(k) for all D, j=1..D, and k=1..D!.
%C sup_n |sum_{j=1}^n f(jd)| is finite (but not bounded) for all d, thus giving a counterexample to a strong form of the Erdos discrepancy conjecture [see Remark 1.13 in Tao link].
%H Terence Tao, <a href="http://arxiv.org/abs/1509.05363">The Erdős discrepancy problem</a>, arXiv:1509.05363 [math.CO], 2015.
%o (Sage)
%o A=[1,1]
%o for D in [1..4]:
%o j=1
%o while j<=D:
%o k=1
%o while k<=factorial(D):
%o A.append((-1)^j*A[k])
%o k+=1
%o j+=1
%o A[1:73] # _Tom Edgar_, Sep 29 2015
%Y Cf. A237695.
%K easy,sign
%O 1
%A _Terence Tao_, Sep 28 2015
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