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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

Table of n, a(n) for n=1..72.

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.)