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!.

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

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 Erdős 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: A186039 A332433 A057077 * A070748 A154990 A209615

Adjacent sequences: A262722 A262723 A262724 * A262726 A262727 A262728

Keyword

easy,sign

Author

Terence Tao, Sep 28 2015

Status

approved