A191257 a(n) = A067368(n)/2.

1, 3, 5, 7, 8, 9, 11, 13, 15, 17, 19, 21, 23, 24, 25, 27, 29, 31, 33, 35, 37, 39, 40, 41, 43, 45, 47, 49, 51, 53, 55, 56, 57, 59, 61, 63, 64, 65, 67, 69, 71, 72, 73, 75, 77, 79, 81, 83, 85, 87, 88, 89, 91, 93, 95, 97, 99, 101, 103, 104, 105, 107, 109, 111, 113, 115, 117, 119, 120, 121, 123, 125, 127, 129, 131, 133, 135, 136, 137, 139, 141, 143

Offset: 1

Author

Clark Kimberling, May 28 2011

Keywords

• nonn

Comments

From Jianing Song, Sep 21 2018: (Start)

Numbers n such that A191255(n) = 0 or 3. Previous definition was numbers n such that A191255(2*n) = 1, that is, numbers of the form 2^(3t)*s where s is an odd number.

{+-a(n)} gives all nonzero cubes modulo all powers of 2, that is, nonzero cubes over the 2-adic integers. So this sequence is closed under multiplication. (End)

The old entry had the conjecture that a(n) = A067368(n)/2. Jianing Song, Sep 21 2018 showed that this is true, and gave us the simpler definition that we have now used. The conjecture is correct because {a(n)} lists the numbers of the form 2^(3t)*s, and {A067368(n)} lists the numbers of the form 2^(3t+1)*s, where s is an odd number. Note also that a(n) = A213258(n)/4.

The asymptotic density of this sequence is 4/7. - Amiram Eldar, May 31 2024

Links

• Recto Rex M. Calingasan and Alexander Vincent B. Policarpio, On the zeros of the OEIS A191257 zeta function, AIP Conference Proceedings 1905, 030011 (2017).

Crossrefs

Cf. A067368, A191255, A213258.

Perfect powers over the 2-adic integers:

Squares: positive: A234000; negative: A004215 (negated);

Cubes: this sequence;

Fourth powers: positive: A319281; negative: A319282 (negated).

Programs

Mathematica

t = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 2}, 2 -> {0, 3},
      3 -> {0, 1}}] &, {0}, 9] (* A191255 *)
Flatten[Position[t, 0]] (* A005408, the odds *)
a = Flatten[Position[t, 1]] (* A067368 *)
b = Flatten[Position[t, 2]] (* A213258 *)
a/2  (* A191257 *)
b/4  (* a/2 *)

PARI

isok(n) = valuation(2*n, 2)%3==1; \\ Altug Alkan, Sep 21 2018

Python

def A191257(n):
    def f(x): return n+x-sum(((x>>i)-1>>1)+1 for i in range(0,x.bit_length(),3))
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, Feb 17 2025

Extensions

Name corrected by Altug Alkan, Apr 03 2018

New name from Jianing Song, Sep 21 2018