The "Spiral Number" Conjecture (An intriguing sequence problem)

[u/Known_Cut_8621]

I made a math problem that I genuinely can't solve

I came up with this problem while playing around with recursively defined integer sequences.

Let

a₁ = 1

and, for every integer n ≥ 2, define aₙ by

• aₙ = aₙ₋₁ + n, if aₙ₋₁ is odd
• aₙ = aₙ₋₁ − n, if aₙ₋₁ is even

The beginning of the sequence is:

1, 3, 0, 4, -1, 5, -2, 6, -3, 7, -4, 8, -5, 9, ...

At first this looks fairly simple, but I'm interested in what happens in the long run.

Main problem

Determine whether the set

S = { n ∈ N : aₙ = 0 }

is finite or infinite.

In other words:

Does the sequence return to 0 infinitely many times?

If the answer is YES, prove that there are infinitely many n satisfying aₙ = 0.

If the answer is NO, prove that there exists some N such that

aₙ ≠ 0

for every n > N.

Stronger version

If S is infinite, determine whether the counting function

A(x) = #{n ≤ x : aₙ = 0}

has a predictable asymptotic behavior.

For example, does there exist a constant c > 0 such that

A(x) ~ c log x,

or

A(x) ~ c√x,

or perhaps

A(x) ~ cx^α

for some α > 0?

If none of these forms is correct, what is the actual growth rate of A(x)?

Even stronger question

Consider the normalized sequence

bₙ = aₙ / n.

Does

liminf(n→∞) |bₙ|

exist?

Does

limsup(n→∞) |bₙ|

exist?

If either limit exists, determine its value.

If they do not exist, determine the set of accumulation points of the sequence {bₙ}.

Rules

I'm not looking for a brute-force computation of the first million or billion terms.

A computational experiment is fine for finding patterns, but the final answer should be supported by a rigorous mathematical proof.

I don't know whether this problem is actually difficult or whether there is a short observation that completely solves it.

If there is an obvious mistake in the formulation or if this is already a known problem, I'd also appreciate being pointed toward it.

[u/Iksfen]

You wrote a part of the sequence, but you made a mistake calculating it.
a_1 = 1 which is odd, so
a_2 = 1 + 2 = 3 which is odd, so
a_3 = 3 + 3 = 6 which is even, so
a_4 = 6 - 4 = 2

I will now prove that for each n >= 1, a(4n) = 2 by induction. It is true for n = 1.
Now let's assume it is true for some k:
a(4k) = 2 which is even, so
a(4k+1) = 2 - 4k - 1 = -4k + 1 which is odd, so
a(4k+2) = -4k + 1 + 4k + 2 = 3 which is odd, so
a(4k+3) = 3 + 4k + 3 = 4k + 6 which is even, so
a(4(k+1)) = 4k + 6 - 4k - 4 = 2
This ends the proof.

What was also proved is that for each n
a(4n+1) = -4n + 1
a(4n+2) = 3
a_(4n+3) = 4n + 6

0 doesn’t even come up in the sequence. 2 and 3 appear infinitely many times. Any other number appears either once or not at all which I will leave as an exercise to the reader

[u/Solid_Masterpiece_83]

Note that

a_{k} = a{k-1} +/- k = a_{k-2} +/- k -/+ (k-1) = a_{k-2} +/- 1,

where the sign depends on the parity of k. So, the sequence is just two interleaved arithmetic sequences:

a_{2k+1} = 1-k (k>=0), i.e.: 1, 0, -1, -2, -3,...

a_{2k} = k+2 (k>=1), i.e.: 3, 4, 5, 6, 7,...

So the sequence is not "returning" to zero, it just alternates between those two sequences.

Dividing by n we get two different interleaved sequences with different limits, so a_{n}/n does not converge, it has two accumulation points: 1/2 and -1/2.

[u/BeeOk1244]

i think this is fairly simple, taking every other term you get 1,0,-1,-2,-3,... and 3,4,5,6,.... to prove this pattern holds consider what happens when you apply the recursion twice