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 Erdős discrepancy conjecture [see Remark 1.13 on p. 7 of Tao paper].

Links

Antti Karttunen, Table of n, a(n) for n = 1..40320

Terence Tao, The Erdős discrepancy problem, arXiv:1509.05363 [math.CO], 2015-2017.

Index entries for sequences related to factorial base representation

Formula

a(n) = (-1)^A034968(n-1). - Antti Karttunen and Peter Munn, Aug 09 2024

Mathematica

(* Generates terms recursively, using definition *)
Module[{dmax = 4, i = 2, a}, a = Table[1, (dmax+1)!]; For[d = 1, d <= dmax, d++, For[j = 1, j <= d, j++, For[k = 1, k <= d!, k++, a[[i++]] = (-1)^j*a[[k]]]]]; a] (* Paolo Xausa, Aug 10 2024 *)
(* Generates terms individually, via A034968 (slower) *)
A034968[n_] := Module[{a = n, i = 2}, While[i! <= n, a-=(i-1)*Floor[n/i++!]]; a];
Array[(-1)^A034968[#] &, 5!, 0] (* Paolo Xausa, Aug 10 2024 *)

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
(PARI)
A034968(n) = { my(s=0, b=2, d); while(n, d = (n%b); s += d; n = (n-d)/b; b++); (s); };
A262725(n) = ((-1)^A034968(n-1)); \\ Antti Karttunen, Aug 09 2024

Crossrefs

Cf. A034968, A237695, A374468.

Cf. also A343785 (another example from Tao paper).

Sequence in context: A332433 A374385 A057077 * A070748 A154990 A209615

Adjacent sequences: A262722 A262723 A262724 * A262726 A262727 A262728

Keyword

easy,sign,changed

Author

Terence Tao, Sep 28 2015

Status

approved