A070563 a(n) = 0 if 3 divides the Ramanujan number tau(n) (A000594(n)), otherwise 1.

1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 1

Author

N. J. A. Sloane, May 07 2002

Keywords

• nonn
• mult

Comments

Multiplicative because A000594 is. Conjecture: a(3^k) = 0, if p == 1 mod 3, a(p^2k) = 0 and a(p^(2k+1)) = 1, if p == -1 mod 3, a(p^2k) = 1 and a(p^(2k+1)) = 0. - Christian G. Bower, Jun 10 2005

From Antti Karttunen, Jul 03 2024: (Start)

The above conjecture is not correct. The first counterexample occurs at n = 2401 = 7^4. My improved conjecture is that this is actually a characteristic function of nonmultiples of 3 whose sum of divisors is also a nonmultiple of 3, that is, having a following multiplicative formula: a(3^k) = 0, if p == 1 mod 3, a(p^e) = 1 if e != 2 (mod 3), otherwise 0, and if p == -1 mod 3, a(p^2k) = 1 and a(p^(2k+1)) = 0. This conjecture has now been proved correct by Seiichi Manyama.

Bower's formula is now submitted as A374053.

(End)

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..131072 (first 100000 terms from Antti Karttunen)
• P. Moree and H. J. J. te Riele, The hexagonal versus the square lattice, arXiv:math/0204332 [math.NT], 2002.
• P. Moree and H. J. J. te Riele, The hexagonal versus the square lattice, Math. Comp. 73 (2004), no. 245, 451-473.
• Index entries for characteristic functions

Crossrefs

Characteristic function of A374135, nonmultiples of 3 whose sum of divisors is also a nonmultiple of 3.

Cf. A000203, A000594, A011655, A070564, A126825, A353815, A374053.

Programs

Mathematica

a[n_] := Boole[!Divisible[RamanujanTau[n], 3]]; Array[a, 92] (* Jean-François Alcover, Jul 05 2017 *)

PARI

A070563(n) = !!(ramanujantau(n)%3); \\ Antti Karttunen, Jul 02 2024

PARI

A070563(n) = ((n%3) && (sigma(n)%3)); \\ Antti Karttunen, Jul 03 2024

PARI

A070563(n) = { my(f=factor(n)); prod(i=1, #f~, if(3==f[i, 1], 0, 1==(f[i, 1]%3), 2!=(f[i, 2]%3), (1+f[i, 2])%2)); }; \\ Antti Karttunen, Jul 03 2024

Formula

a(n) = A011655(n) * A353815(n), conjectured by Antti Karttunen, proved by Seiichi Manyama, Jul 03 2024