Zeroless Tetranacci Numbers

In a post titled, Sequences Formed By Removing Zeros, from January 2023,  I wrote that "It's interesting to consider what happens to a sequence if a certain rule is applied but with the stipulation that any zeros arising must be removed". In that post I looked at the zeroless Fibonacci sequence that falls into a repeating loop with a confirmed period of 912. 

The zeroless Tribonacci sequence falls into a much larger repeating loop with a confirmed period of 300,056,874. It reaches this cycle at index 208,666,297. However, it is not known whether the zeroless Tetranacci sequences cycles or not but, if it does, then \(s+p > 10^{10}\) where \(s\) and \(p\) are the starting index and period of the cycle, respectively.

 

 A371916: zeroless analog of tetranacci numbers.

The initial members are:

1, 1, 1, 1, 4, 7, 13, 25, 49, 94, 181, 349, 673, 1297, 25, 2344, 4339, 85, 6793, 13561, 24778, 45217, 9349, 9295, 88639, 1525, 1888, 11347, 13399, 28159, 54793, 17698, 11449, 11299, 95239, 135685, 253672, 495895, 98491, 983743, 183181, 176131, 1441546, 278461, 279319, 2175457

Figure 1 shows a plot of the first 100 terms:

Figure 1: permalink

Like the zeroless Fibonacci and Tribonacci sequences the ratio between successive terms of the zeroless Tetranacci sequence never approaches a limit. With no suppression of zeros, the following are the convergences:

• Fibonacci: $\phi = \frac{1+\sqrt{5}}{2} \text{ which is }\approx 1.61803$
• Tribonacci: the real root of $x^3 - x^2 - x - 1 = 0 \text{ which is } \approx 1.83929$
• Tetranacci: the real root of $x^4 - x^3 - x^2 - x - 1 = 0 \text{ which is }\approx 1.92756$

Pandigital Products

Yesterday I turned 28156 days old and this number has an interesting property:$$28156 = 4 \times 7039$$The factorisation shown is not the prime factorisation but, looking at both sides of the equation, it can be seen that each of the digits from 0 to 9 occurs exactly once. This makes the number a member of OEIS A370970:

A370970: numbers \(k\) which have a factorization \(k = f_1 \times f_2 \times \ldots \times f_n \) where the digits of \({k, f_1, f_2, \ldots, f_n}\) together give \(0,1, \ldots ,9\) exactly once.

Here is the complete list of terms:

8596 = 2 x 14 x 307

8790 = 2 x 3 x 1465

9360 = 2 x 4 x 15 x 78

9380 = 2 x 5 x 14 x 67

9870 = 2 x 3 x 1645

10752 = 3 x 4 x 896

12780 = 4 x 5 x 639

14760 = 5 x 9 x 328

14820 = 5 x 39 x 76

15628 = 4 x 3907

15678 = 39 x 402

16038 = 27 x 594 = 54 x 297

16704 = 9 x 32 x 58

17082 = 3 x 5694

17820 = 36 x 495 = 45 x 396

17920 = 8 x 35 x 64

18720 = 4 x 5 x 936

19084 = 52 x 367

19240 = 8 x 37 x 65

20457 = 3 x 6819

20574 = 6 x 9 x 381

20754 = 3 x 6918

21658 = 7 x 3094

24056 = 8 x 31 x 97

24507 = 3 x 8169

25803 = 9 x 47 x 61

26180 = 4 x 7 x 935

26910 = 78 x 345

27504 = 3 x 9168

28156 = 4 x 7039

28651 = 7 x 4093

30296 = 7 x 8 x 541

30576 = 8 x 42 x 91

30752 = 4 x 8 x 961

31920 = 5 x 76 x 84

32760 = 8 x 45 x 91

32890 = 46 x 715

34902 = 6 x 5817

36508 = 4 x 9127

47320 = 8 x 65 x 91

58401 = 63 x 927

65128 = 7 x 9304 

65821 = 7 x 9403

These numbers are few and far between as can be seen and 28156 in particular recurs with permuted digits as 15628, 21658, 28651, 65128 and 65821.

Cartesian Product versus Self-Convolution

I asked Gemini the following:

How does the self convolution of a set A differ from the Cartesian product of the set with itself (A x A)?

This was the response: 

To understand the difference between the Cartesian product of a set and its self-convolution, we have to look at what mathematical object each operation produces. In short: the Cartesian product generates a new set of raw pairs, while the convolution generates a sequence that counts the frequencies of their sums.

Here is a full breakdown of how these two concepts differ and how they ultimately connect.

1. The Cartesian Product: Structural Pairing

The Cartesian product of a set $A$ with itself, denoted as $A \times A$, is an operation strictly within set theory.

It takes the elements of $A$ and creates a new set containing every possible ordered pair $(x, y)$, where both $x$ and $y$ are elements of $A$.

$$A \times A = \{(x, y) \mid x \in A \text{ and } y \in A\}$$

• Nature: It is purely structural. It does not add, multiply, or combine the values of the elements; it merely groups them side-by-side in a 2D grid.
• Size: If set $A$ contains $k$ elements, the Cartesian product $A \times A$ will always contain exactly $k^2$ elements.

2. The Self-Convolution: Frequency of Sums

Strictly speaking, you cannot convolve a "set" directly; convolution is an operation applied to functions or sequences. When mathematicians refer to the convolution of a set $A$, they are actually referring to the discrete convolution of the set's indicator sequence (also called the characteristic function).

Let $1_A(x)$ be a sequence where the value is $1$ if $x$ is in set $A$, and $0$ if it is not. The self-convolution of this sequence produces a new sequence, $c_n$:

$$c_n = \sum_{i+j=n} 1_A(i) \cdot 1_A(j)$$

• Nature: It is an arithmetic and combinatorial operation. The resulting sequence $c_n$ tells you exactly how many times the integer $n$ can be expressed as the sum of two elements from set $A$.
• Size: The result is not a finite set of pairs, but an infinite sequence of integers (though it will be mostly zeros if $A$ is a finite set).

3. The Core Difference

The Cartesian product asks: "What are all the possible ways to pick two elements from this set?"

The Convolution asks: "If I add those two elements together, how many pairs produce a sum of $n$?"

If you imagine the Cartesian product as a square grid of pairs, the convolution is the result of drawing diagonal lines through that grid (where $x + y = n$) and counting how many pairs fall on each diagonal.

4. A Concrete Example

Let’s define a set of integers: $A = \{1, 2, 4\}$

The Cartesian Product ($A \times A$) creates a set of $3^2 = 9$ ordered pairs:

$A \times A = \{(1,1), (1,2), (1,4), (2,1), (2,2), (2,4), (4,1), (4,2), (4,4)\}$

The Self-Convolution maps the sums of these pairs to their frequencies. Let's add the elements of each pair together:

• Sums to 2: $(1,1) \rightarrow$ 1 pair
• Sums to 3: $(1,2), (2,1) \rightarrow$ 2 pairs
• Sums to 4: $(2,2) \rightarrow$ 1 pair
• Sums to 5: $(1,4), (4,1) \rightarrow$ 2 pairs
• Sums to 6: $(2,4), (4,2) \rightarrow$ 2 pairs
• Sums to 7: None $\rightarrow$ 0 pairs
• Sums to 8: $(4,4) \rightarrow$ 1 pair

The resulting convolution sequence $c_n$ (starting from index $n=0$) would look like this:

$0, 0, 1, 2, 1, 2, 2, 0, 1, 0, 0, 0...$

5. The Generating Function Connection

Because you are working with tools like SageMath, the most elegant way to see the relationship between these two concepts is through generating functions.

If you turn set $A$ into a polynomial where the elements of the set are the exponents, you get:

$$f(x) = x^1 + x^2 + x^4$$

When you square this polynomial (which is the algebraic equivalent of the Cartesian product), you get:

$$f(x)^2 = (x^1 + x^2 + x^4) \cdot (x^1 + x^2 + x^4)$$

$$f(x)^2 = x^2 + 2x^3 + x^4 + 2x^5 + 2x^6 + x^8$$

Look at the coefficients of the expanded polynomial: $1, 2, 1, 2, 2, 1$. These are the exact non-zero values of our convolution sequence. The Cartesian product dictates the terms being multiplied, but the mathematical simplification of gathering like-terms is the act of convolution.

Convolutions Revisited

On the 12th October 2025, I created a post titled A Practical Example of a Convolution which began as follows:

Over the past decade, for some reason, I've chosen to ignore convolutions. Whenever they were mentioned in an OEIS sequence, I simply skipped over the reference. However, I'm now attempting to redress that neglect and to that end I was lucky to find two excellent YouTube videos about convolutions made by 3Blue1Brown (this guy has 7.76 million subscribers and for good reason).

I finished the post by saying:

This post is simply the first in what I hope will be a series of posts relating to convolutions. As I've already discovered, convolutions linked to Fourier transforms and Laplace transformations so it's a big topic to investigate but at least I've finally made a start.

Unfortunately there was no future series of posts relating to convolutions but I was reminded of them once again thanks to the number associated with my diurnal age today: \( \textbf{28152}\). It is a member of OEIS A014342:

 A014342: Convolution of primes with themselves.

 The initial members are:

4, 12, 29, 58, 111, 188, 305, 462, 679, 968, 1337, 1806, 2391, 3104, 3953, 4978, 6175, 7568, 9185, 11030, 13143, 15516, 18177, 21150, 24471, \( \textbf{28152}\), 32197, 36678, 41543, 46828, 52621, 58874, 65659, 73000, 80949, 89462, 98631, 108396, 118869, 130102, 142071

This time around I got Gemini to explain how the convolution was calculated:

In the context of integer sequences like those found in the OEIS, a convolution (specifically, a discrete Cauchy convolution) is a mathematical way of combining two sequences to create a third. When a sequence is convolved with itself, you are multiplying its terms together in a symmetrical, sliding window and summing the results.

Here is the full breakdown of how this works for OEIS A014342.

The Formula

First, let's establish our base sequence, which is the sequence of prime numbers, denoted as p_n:

• p₁ = 2
• p₂ = 3
• p₃ = 5
• p₄ = 7
• p₅ = 11
• ...and so on.

The n-th term of the convolved sequence, let's call it c_n, is calculated by taking the sum of the products of two primes whose indices add up to n + 1.

The formal equation looks like this:

c_n = Σ (p_i × p_n-i+1)

(summed from i = 1 to n)

In simpler terms: to find the n-th number in your sequence, you line up the first n primes going forward, line up the exact same n primes going backward, multiply the overlapping pairs, and add those products together.

Step-by-Step Calculation

Here is how the first few members of the sequence are calculated using this method.

1st Term (n = 1):

We only use the first prime (p₁ = 2).

c₁ = (p₁ × p₁)

c₁ = (2 × 2)

c₁ = 4

2nd Term (n = 2):

We use the first two primes (2 and 3). We pair the first with the second, and the second with the first.

c₂ = (p₁ × p₂) + (p₂ × p₁)

c₂ = (2 × 3) + (3 × 2)

c₂ = 6 + 6

c₂ = 12

3rd Term (n = 3):

We use the first three primes (2, 3, and 5). Notice how the indices always add up to 4 (n + 1).

c₃ = (p₁ × p₃) + (p₂ × p₂) + (p₃ × p₁)

c₃ = (2 × 5) + (3 × 3) + (5 × 2)

c₃ = 10 + 9 + 10

c₃ = 29

4th Term (n = 4):

We use the first four primes (2, 3, 5, and 7).

c₄ = (p₁ × p₄) + (p₂ × p₃) + (p₃ × p₂) + (p₄ × p₁)

c₄ = (2 × 7) + (3 × 5) + (5 × 3) + (7 × 2)

c₄ = 14 + 15 + 15 + 14

c₄ = 58

5th Term (n = 5):

We use the first five primes (2, 3, 5, 7, and 11).

c₅ = (p₁ × p₅) + (p₂ × p₄) + (p₃ × p₃) + (p₄ × p₂) + (p₅ × p₁)

c₅ = (2 × 11) + (3 × 7) + (5 × 5) + (7 × 3) + (11 × 2)

c₅ = 22 + 21 + 25 + 21 + 22

c₅ = 111

The Pattern

As you calculate further into the sequence, this symmetrical "folding" action continues. You are always starting at the outer edges of your chosen set of primes and working your way inward to the middle, multiplying the pairs and summing them up. This specific operation is incredibly common in polynomials, probability theory, and signal processing, but when applied purely to integer sequences like the primes, it yields the exact sequence you provided.

Remember that SageMath makes it very easy to determine self-convolutions due to its formula convolutions(A, A) where A is any given set.