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 A262725 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!. 0
 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, -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, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1 COMMENTS 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]. LINKS Terence Tao, The Erdos discrepancy problem, arXiv:1509.05363 [math.CO], 2015. PROG (Sage) A=[1, 1] for D in [1..4]:     j=1     while j<=D:         k=1         while k<=factorial(D):             A.append((-1)^j*A[k])             k+=1         j+=1 A[1:73] # Tom Edgar, Sep 29 2015 CROSSREFS Cf. A237695. Sequence in context: A160357 A186039 A057077 * A070748 A154990 A209615 Adjacent sequences:  A262722 A262723 A262724 * A262726 A262727 A262728 KEYWORD easy,sign AUTHOR Terence Tao, Sep 28 2015 STATUS approved

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Last modified January 29 08:12 EST 2020. Contains 331337 sequences. (Running on oeis4.)