A356556 Parity of A061418.

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

Offset

1

Comments

Can be described as follows: starting from a single pair of animals, and assuming any pair of animals can produce one offspring per day (as in the game Minecraft), a(n) = 0 on days with an even number of animals and 1 on days with an odd number.

While this sequence is easily generated from A061418, the reverse is also true. If we let r = sum of a(n)(2/3)^n = 0.755459... then the n-th term of A061418 is given by ceiling((4-r)/3*(3/2)^n).

The sequence is related to K(3) from the Josephus problem (A083286) via sum r = 4 - 2*K(3).

Links

Table of n, a(n) for n=1..108.

E. T. H. Wang and Phillip C. Washburn, Problem E2604, American Mathematical Monthly, 84 (1977), 821-822.

Formula

a(n) = A061418(n) mod 2.

Maple

b:= proc(n) option remember; iquo(3*b(n-1), 2) end: b(1):= 2:
a:= n-> irem(b(n), 2):
seq(a(n), n=1..200);  # Alois P. Heinz, Sep 20 2022

Mathematica

A061418[n_] := A061418[n] = If[n==1, 2, Quotient[3*A061418[n-1], 2]];
a[n_] := Mod[A061418[n], 2];
Table[a[n], {n, 1, 200}] (* Jean-François Alcover, Dec 26 2022 *)

Prog

(Python)
def a(n):
    val = 2
    for i in range(n):
        val += val//2
    return val%2

Crossrefs

Cf. A061418.

Sequence in context: A038219 A330731 A138710 * A330037 A255817 A071674

Adjacent sequences: A356553 A356554 A356555 * A356557 A356558 A356559

Keyword

easy,nonn

Author

Jacob Fauman, Aug 12 2022

Status

approved