New interesting research about how to calculate products by simplified expression by combining the Catalan numbers. The author, Carlos López, has demonstrated several results for possible uses in science.

1. A NEW FORMULA FOR CALCULATING THE CATALAN NUMBERS AND ITS RELATION WITH THE
PRODUCT OF THE ODD NUMBERS
CARLOS HERNÁN LÓPEZ ZAPATA
ABSTRACT
A new formula written mainly by the operator baptized as the Catalan product or ∏ {𝑛}𝐶𝑎𝑡𝑎𝑙𝑎𝑛 lets
calculate every Catalan number by choosing an index 𝑛, for 𝑛 ≥ 4. The mathematical
recurrence 𝐶 𝑛+1 =
2(2𝑛+1)
𝑛+2
𝐶 𝑛, a well-known expression for these numbers, is used for
understanding that every next Catalan number, 𝐶 𝑛+1, is formed by the product of other previous
Catalan numbers. It results in the useful expression ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
𝑛−3
𝑖=1 that has been proved on this
article in order to get 𝐶 𝑛 = 5(2 𝑛−3
) ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
,𝑛−3
𝑖=1 𝑛 ≥ 4, which offers another new
representation of 𝐶 𝑛 that serves as a complement to the formula 𝐶 𝑛 = ∏ (
𝑛+𝑘
𝑘
)𝑛
𝑘=2 , 𝑛 ≥ 0. There is
also the official definition of the ∏ =𝐶𝑎𝑡𝑎𝑙𝑎𝑛 ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
=
5−1
2 𝑛−3
(2𝑛)!
(𝑛+1)!𝑛!
𝑛−3
𝑖=1 =
5−1
2 𝑛−3 𝐶 𝑛 =
8
(5)2 𝑛 𝐶 𝑛,
𝑛 ≥ 4, a so interesting way to represent the consequence of this natural recurrence for setting up
the formula with two special factors that rule the Catalan numbers product: 5 and 2 𝑛−3
. The most
valuable conclusion is that it is possible to write the product of all odd-numbers evaluated on that
formula from 𝑖 = 1 to 𝑖 = 𝑛 − 3, for 𝑛 ≥ 4, as a factorization of every 𝐶 𝑛: “the product among the
odd-numbers depends of the Catalan numbers on the expansion of a chosen 𝑛”. Furthermore, at the
end of this work, it is inferred an interesting additional expression, for evaluating a complicated
product that includes numerator and denominator (both expanded in 𝑘, from 𝑘 = 2 to 𝑘 = 𝑛 − 3
and according to some value of 𝑛, 𝑛 ≥ 5), by combining the results about the Catalan numbers that
were deduced here. It is another contribution in order to simplify long products that could be useful
in some applications in science and engineering.
Key words: Catalan numbers, Catalan product, odd-numbers, recurrence, Bernoulli numbers,
factorial
1. INTRODUCTION
At the beginning it came out as a simple curiosity by looking for some special connexion between
the Bernoulli numbers (see [1], [3]) and the Catalan numbers (see [2], [4]) by their well-known
historical formulas. The first idea emerged when, after expanding 𝐶 𝑛+1 =
2(2𝑛+1)
𝑛+2
𝐶 𝑛 [3], for some
values of 𝑛, for 𝑛 ≥ 1, appeared one valuable number that was defined as 𝐵1 = −
1
2
(one of the
first Bernoulli numbers on the well-known table!) (Table 1). The procedure that describes this
statement is given by:

2. 𝑛 = 0, 𝐶0+1 = 𝐶1 =
2(2(0)+1)
0+2
𝐶0 =
2
2
(1) = 𝐶1 = 1 (with 𝐶0 = 1, the initial Catalan number, and
the next one, 𝐶1 = 1. See the table 2)
𝑛 = 1, 𝐶1+1 = 𝐶2 =
2(2(1)+1)
1+2
𝐶1 = 2
3
3
𝐶1 = 2𝐶1 = −4𝐵1 = 2 (the next one using by 𝐵1 = −
1
2
).
By which it is possible to write: 𝐶1 =
−4𝐵1
2
= 1, since it is a valid expression easily to check.
Table 1. The first nine values of the Bernoulli numbers.
𝑛 0 1 2 3 4 5 6 7 8
𝐵𝑛 1 −1/2 1/6 0 -1/30 0 1/42 0 -1/30
For other consecutive numbers, the consideration remains, by 𝐵1 = −
1
2
, as follow below:
𝑛 = 2, 𝐶2+1 = 𝐶3 =
2(2(2)+1)
2+2
𝐶2 =
2(5)
4
𝐶2 =
5
2
𝐶2 =
5
2
(−4𝐵1) = 20𝐵1
2
= 5 (see the table 2)
𝑛 = 3, 𝐶3+1 = 𝐶4 =
2(2(3)+1)
3+2
𝐶3 =
2(7)
5
𝐶3 =
14
5
𝐶3 = (
14
5
) (20𝐵1
2
) = 14 (See the table 2)
At this moment, the previous step for 𝐶4 can be seen also as this equivalent formula in 𝑛 = 4:
𝐶 𝑛=4 = {
2[2(𝑛−1)+1]
(𝑛−1)+2
} . 20𝐵1
2
= {
2[2(4−1)+1]
(4−1)+2
} . 20𝐵1
2
= (
14
5
)(20𝐵1
2
), that is the same previous
value: 𝐶4 = 14.
Table 2. The first nine Catalan numbers.
𝑛 0 1 2 3 4 5 6 7 8
𝐶 𝑛 1 1 2 5 14 42 132 429 1.430
And for other consecutive numbers it is possible to deduce that it exists a trend, based on 20𝐵1
2
,
and with expansion on n about the products:
𝑛 = 5, 𝐶 𝑛=5 = {
2[2(𝑛−1)+1]
(𝑛−1)+2
} {
2[2(𝑛−2)+1]
(𝑛−2)+2
} 20𝐵1
2
= {
2[2(5−1)+1]
(5−1)+2
} {
2[2(5−2)+1]
(5−2)+2
} 20𝐵1
2
And, as a result, it is possible to calculate:

3. 𝐶5 = {
2(9)
(6)
} {
2(7)
(5)
} 20𝐵1
2
= 42 (See the table 2 again), when 𝐵1 is evaluated as 𝐵1 = −
1
2
.
If every 𝐶 𝑛 is calculated according to this statement, the inductive method lets conclude a general
expression whose number of factors, multiplied each other on it, is 𝑛 − 3.
For example, on the last Catalan number that was calculated, 𝐶 𝑛=5, the number of factors deduced
is 𝑛 − 3 = 5 − 3 = 2. These factors are:
{
2[2(𝑛−1)+1]
(𝑛−1)+2
} , {
2[2(𝑛−2)+1]
(𝑛−2)+2
} or evaluated as: {
2[2(5−1)+1]
(5−1)+2
} , {
2[2(5−2)+1]
(5−2)+2
} or {
2(9)
(6)
} , {
2(7)
(5)
}.
And for others, such as 𝐶 𝑛=6, this is 𝑛 − 3 = 6 − 3 = 3.
Pay attention especially when the Catalan number is 𝐶4, 𝑛 − 3 = 4 − 3 = 1, since the count of
factors is being analysed from one only factor, when 𝑛 = 4, to other ones greater than a unique
factor. This convention will let lead a clear understanding of the products on the final equations
expected on this work. The general version of this statement, for several factors, ∀ 𝑛 ≥ 4, is
represented by the next inductive expression:
𝐶 𝑛,𝑘=𝑛 = {
2[2(𝑛−1)+1]
(𝑛−1)+2
} {
2[2(𝑛−2)+1]
(𝑛−2)+2
} {
2[2(𝑛−3)+1]
(𝑛−3)+2
} … {
2[2(𝑛−(𝑘−3))+1]
(𝑛−(𝑘−3))+2
} 20𝐵1
2
(1)
Where k is set up according to the value of n, letting know how many factors has that product. This
formula lets generalize many inferred factors from now on.
So making a glance at the product {
[2(𝑛−1)+1]
(𝑛−1)+2
} {
[2(𝑛−2)+1]
(𝑛−2)+2
} {
[2(𝑛−3)+1]
(𝑛−3)+2
} … {
[2(𝑛−(𝑘−3))+1]
(𝑛−(𝑘−3))+2
} (notice
that without every 2 preceding the factors) it can be written by the operator ∏{𝑛} in a more
comprehensive way to identify this operation, what would avoid any confusion with the expansion
on n for several values of this index:
∏
2(𝑛 − 𝑖) + 1
𝑛 − 𝑖 + 2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
=
{
[2(𝑛−1)+1]
(𝑛−1)+2
} {
[2(𝑛−2)+1]
(𝑛−2)+2
} {
[2(𝑛−3)+1]
(𝑛−3)+2
} … {
[2(𝑛−𝑚)+1]
(𝑛−𝑚)+2
} {
[2(𝑛−(𝑚+1))+1]
(𝑛−(𝑚+1))+2
} … {𝑙𝑎𝑠𝑡 𝑓𝑎𝑐𝑡𝑜𝑟} (2)

4. It is clear that the expansion includes also an intermediate value on 𝑖 = 𝑚 and others after it until
reaching the final term, last factor in the equation (2), in 𝑛 − 3. On the next section, that expression
will be discussed with more relevance, for now it is known as the Catalan product or ∏ {𝑛}𝐶𝑎𝑡𝑎𝑙𝑎𝑛 ,
as the author of this article has wanted to baptized it.
2. THE CATALAN PRODUCT ∏ {𝑛}𝐶𝑎𝑡𝑎𝑙𝑎𝑛 : THE INTERESTING NUMBERS 20𝐵1
2
or 5 and 2 𝑛−3
.
Let us come back to the last result given by the equation 2 and combine it with the equation 1:
𝐶 𝑛 = {
2[2(𝑛 − 1) + 1]
(𝑛 − 1) + 2
} {
2[2(𝑛 − 2) + 1]
(𝑛 − 2) + 2
} {
2[2(𝑛 − 3) + 1]
(𝑛 − 3) + 2
} … {
2[2(𝑛 − (𝑘 − 3)) + 1]
(𝑛 − (𝑘 − 3)) + 2
} 20𝐵1
2
If the product of these factors is replaced by the equation 2) and the factorization of all these
numbers 2 that precede those factors is thought to be:
2 ∗ 2 ∗ 2 … 2 = 2 𝑛−3
, since the factors have been expanded from 𝑖 = 1 until 𝑖 = 𝑛 − 3 according
to the Catalan product operator ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
, this would let write the equation 1 as below:
𝐶 𝑛 = 20𝐵1
2
. 2 𝑛−3
. ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
(3)
Of course, more manipulations of the last expression, factorizing 20𝐵1
2
= 22
. 5 𝐵1
2
, lets lead to:
𝐶 𝑛 = 22
. 5 𝐵1
2
. 2 𝑛−3
. ∏
2(𝑛 − 𝑖) + 1
𝑛 − 𝑖 + 2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
And using the real value of 𝐵1 = −
1
2
it is possible to reduce the expression, step by step, to:

5. 𝐶 𝑛 = 22
. 5. (−
1
2
)2
. 2 𝑛−3
. ∏
2(𝑛 − 𝑖) + 1
𝑛 − 𝑖 + 2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
And simplifying: 22
. (−
1
2
)2
= 22
. (
1
2
)2
= 1.
𝐶 𝑛 = 5(2 𝑛−3
) ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
(4)
Notice that 5 and 2 𝑛−3
, or 5, 2 𝑛
and 2−3
=
1
8
rule this structure by preceding the Catalan product.
The expression 4 opens the possibility to summarize the Catalan numbers on the counting
techniques by a version that uses the product of these factors on the numerator and denominator.
This especial numerator is not whatever thing, it is the product of every odd-number from 7 until
the last odd given by the index in 𝑖 = 𝑛 − 3, since the term 2(𝑛 − 𝑖) + 1 = 2(4 − 1) + 1 = 2(3) +
1 = 7, when n=4. We will see, by a conclusion, that it is possible to summarize even the odd
numbers before 7, for getting the product: 1 ∗ 3 ∗ 5 and adding it to the current product from
7*9*11… until the wanted value included into the formula.
In the path of developing other relations for the Catalan numbers, the comprehension about how
to express the Catalan product in function of factorials was another useful result that comes to help
in the representation of this set of formulas for combinatorics. So using the referenced formula [4]
for 𝐶 𝑛:
𝐶 𝑛 =
(2𝑛)!
(𝑛+1)!𝑛!
, with 𝑛 ≥ 0 in general, and replacing it on the expression 4:
𝐶 𝑛 = 5(2 𝑛−3) ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
𝑛−3
𝑖=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
=
(2𝑛)!
(𝑛+1)!𝑛!
, 𝑛 ≥ 4 (5)
And factorizing the Catalan product after some manipulations as below:
∏ =𝐶𝑎𝑡𝑎𝑙𝑎𝑛 ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
=
5−1
2 𝑛−3
(2𝑛)!
(𝑛+1)!𝑛!
𝑛−3
𝑖=1 =
5−1
2 𝑛−3 𝐶 𝑛 =
8
(5)2 𝑛 𝐶 𝑛, 𝑛 ≥ 4 (6)
So not only the factorials help to calculate the Catalan numbers, but it is useful to find a way to
represent an operation such ∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
=
5−1
2 𝑛−3
(2𝑛)!
(𝑛+1)!𝑛!
𝑛−3
𝑖=1 by this exact formula.

6. 3. THE PRODUCT OF THE ODD-NUMBERS AND THE EXPANSION OF THE DENOMINATOR OF THE
CATALAN PRODUCT: 𝜌 𝑛
When the expression 6) is expanded in its denominator, it is possible to define an important product
𝜌 𝑛 given by the product of the factors on the denominator of (6), 𝑛 − 𝑖 + 2, as below:
∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
=
∏ 2(𝑛−𝑖)+1𝑛−3
𝑖=1
[ …(𝑛−2)(𝑛−1)𝑛(𝑛+1)]
=
8
(5)2 𝑛 𝐶 𝑛
𝑛−3
𝑖=1 (7)
∀ 𝑛 ≥ 4.
And naming the denominator as:
𝜌 𝑛 = ⋯ (𝑛 − 2)(𝑛 − 1)𝑛(𝑛 + 1) (8)
Notice, the expansion has been intentionally expressed by this way, in order to understand how the
product of the denominator runs for some election of 𝑛.
As a result, the equation 7,∀ 𝑛 ≥ 4, can be seen as:
∏
2(𝑛−𝑖)+1
𝑛−𝑖+2
=
∏ 2(𝑛−𝑖)+1𝑛−3
𝑖=1
𝜌 𝑛
=
8
(5)2 𝑛 𝐶 𝑛
𝑛−3
𝑖=1 (9)
From which it is gotten:
∏ [2(𝑛 − 𝑖) + 1]𝑛−3
𝑖=1 =
8
(5)2 𝑛 𝜌 𝑛 𝐶 𝑛 (10)
It is so valuable to have deduced the equation 10, since it represents the product among the odd-
numbers from the minimum odd 7 (in 𝑖 = 1, 𝑛 = 4) to some final odd (in 𝑛 − 3, for some election
of 𝑛). Notice that it has been written in a particular way in function of 𝜌 𝑛 𝐶 𝑛 and the interesting
numbers 8, 5 and 2 𝑛
.
The previous result would be more complete by adding the product of the first three odd-numbers,
which are 1, 3 and 5, being represented by this expression:
1. 3. 5 ∏ [2(𝑛 − 𝑖) + 1]𝑛−3
𝑖=1 = 1. 3. 5
8
(5)2 𝑛 𝜌 𝑛 𝐶 𝑛, ∀𝑛 ≥ 4 (11)

7. And it would be reduced as below by simple steps (cancelling the factor 5 in the numerator and
denominator, for example):
15 ∏ [2(𝑛 − 𝑖) + 1]𝑛−3
𝑖=1 = 3.
8
2 𝑛 𝜌 𝑛 𝐶 𝑛, ∀𝑛 ≥ 4 (12)
Or in this way:
∏ {𝑛}𝑂𝑑𝑑 = 15 ∏ [2(𝑛 − 𝑖) + 1]𝑛−3
𝑖=1 =
3.23
2 𝑛 𝜌 𝑛 𝐶 𝑛 = 3. 23−𝑛
𝜌 𝑛 𝐶 𝑛 =
3𝜌 𝑛 𝐶 𝑛
2 𝑛−3 , ∀𝑛 ≥ 4 (13)
The equation 13 summarizes the product of the all odd-numbers until a specified limit in this
product. It is very clear that exists an important relation between the odd numbers, and its product,
with the definition of the Catalan numbers, and vice versa. The Catalan numbers are connected to
the product of the odd-numbers in this way, a new connexion that supports combinatorics and
would be related to Arithmetic theory as well. Furthermore, a valuable conclusion of all this analysis
comes from the fact that the expression 8, 𝜌 𝑛 = ⋯ (𝑛 − 2)(𝑛 − 1)𝑛(𝑛 + 1), represents the
product of the natural numbers or the positive integers, except zero, starting from the natural 5,
since 𝜌 𝑛 = 𝑛 + 1 = 5, when 𝑛 = 4 (the minimum value of 𝑛 in these formulas), and has a
tremendous impact in all of this deduction, since in equation 13 happens a process of filtering by
the factor
3𝜌 𝑛 𝐶 𝑛
2 𝑛−3 which lets get the product of the set odd-numbers by the definition of a Catalan
number 𝐶 𝑛,∀𝑛 ≥ 4. So the existence of the expression
3𝐶 𝑛
2 𝑛−3, that multiplies 𝜌 𝑛, is the especial
transductor that converts the product of the positive integers from the minimum given 5, in the
definition of the formulas, until the final natural according to 𝜌 𝑛 = ⋯ (𝑛 − 2)(𝑛 − 1)𝑛(𝑛 + 1), for
the product of the odd numbers. It is extraordinary that the Catalan numbers, as numbers that
appear in several events of count, can be used for summarizing the ∏ {𝑛}𝑂𝑑𝑑 in equation 13, since
despite there are several alternatives for calculating the Catalan numbers and formulas, there is
almost nothing mentioned about how to transduce the product of the odd numbers in function of
the Catalan numbers. During the deduction of these formulas, it was so interesting to know also
about the role of the numbers 𝐶2 = 2 and 𝐶3 = 5 [4]. The only prime Catalan numbers are 2 and 5,
exactly those that define the factors 5 and 2 𝑛−3
in the formulas! This is crucial also to understand
that prime numbers appear for leading the connection between the Catalan numbers and natural
numbers. Moreover, the relations written on the previous pages let determine a new formula for
evaluating an especial long product that could be useful for simplifying some formulas of products
that appear in some fields of applied science.

8. 4. THE FORMULA 𝐶 𝑛 = ∏ (
𝑛+𝑘
𝑘
)𝑛
𝑘=2 AND ITS CONNECTION WITH THE EXPRESSION 𝐶 𝑛 =
5(2 𝑛−3
) ∏
2(𝑛−𝑘)+1
𝑛−𝑘+2
𝑛−3
𝑘=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
: HOW TO SIMPLIFY THE PRODUCT ∏
𝑘[2(𝑛−𝑘)+1]
(𝑛2−𝑘2)+2(𝑛+𝑘)
𝑛−3
𝑘=2 ?
On this section it will be introduced a deduction for evaluating the product ∏
𝑘[2(𝑛−𝑘)+1]
(𝑛2−𝑘2)+2(𝑛+𝑘)
𝑛−3
𝑘=2 .
That expression has been intentionally showed here in order to demonstrate that the Catalan
operator, and the other previous results deduced here, are very useful for reducing especial
operations whose formulas have not been defined yet. It would open more possibilities for
calculating products in several fields of science. The fundamental idea is to combine another well-
known formula:
𝐶 𝑛 = ∏ (
𝑛+𝑘
𝑘
)𝑛
𝑘=2 (14) [See 4], for 𝑛 ≥ 0, with the new result given by 5(2 𝑛−3
) ∏
2(𝑛−𝑘)+1
𝑛−𝑘+2
𝑛−3
𝑘=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
[15]. Notice that the index 𝑖 was substituted by the index 𝑘 in order to follow the deduction from
the reference in the original expression [see the equation on the reference 4]. It is only a change of
name.
The method used here consists in to consider the expansion of 14, when 𝑛 ≥ 5, in order to avoid
troubles about the limits of expansion of the new formula from 𝑘 = 2 to 𝑘 = 𝑛 − 3, in which the
minimum 𝑛 must be 𝑛 = 5, so the analysis of the combined formulas is set up by 𝑛 ≥ 5, and no for
𝑛 ≥ 0. The coincidence between (14) and (15) is given by:
𝐶 𝑛 = ∏ (
𝑛+𝑘
𝑘
)𝑛
𝑘=2 = 5(2 𝑛−3
) ∏
2(𝑛−𝑘)+1
𝑛−𝑘+2
𝑛−3
𝑘=1
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
(16)
With 𝑛 ≥ 5.
It can be written expanding the first factor given by 𝑘 = 1 on the right side of (16) as below:
𝐶 𝑛 = ∏ (
𝑛+𝑘
𝑘
)𝑛
𝑘=2 = 5(2 𝑛−3){
2(𝑛−1)+1
𝑛−1+2
} ∏
2(𝑛−𝑘)+1
𝑛−𝑘+2
𝑛−3
𝑘=2
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
, 𝑛 ≥ 5 (17)
After that, it is easy to manipulate the left side of the product by defining two products and other
factorizations as below:
∏ (
𝑛+𝑘
𝑘
) . ∏ (
𝑛+𝑘
𝑘
) =𝑛
𝑘=𝑛−2
𝑛−3
𝑘=2 5(2 𝑛−3)
2n−1
𝑛+1
∏
2(𝑛−𝑘)+1
𝑛−𝑘+2
𝑛−3
𝑘=2
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
, (18)
In 18) the main interest is to unify ∏ (
𝑛+𝑘
𝑘
)𝑛−3
𝑘=2 and ∏
2(𝑛−𝑘)+1
𝑛−𝑘+2
𝑛−3
𝑘=2
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
, which can be done by the
properties of the product by doing:

9. ∏ (
𝑛 + 𝑘
𝑘
) = 5(2 𝑛−3).
2n − 1
𝑛 + 1
𝑛
𝑘=𝑛−2
.
∏
2(𝑛 − 𝑘) + 1
𝑛 − 𝑘 + 2
𝑛−3
𝑘=2
𝑐𝑎𝑡𝑎𝑙𝑎𝑛
∏ (
𝑛 + 𝑘
𝑘
)𝑛−3
𝑘=2
And rearranging the products of the right side by the properties of product operator:
∏ (
𝑛 + 𝑘
𝑘
) = 5(2 𝑛−3).
2n − 1
𝑛 + 1
𝑛
𝑘=𝑛−2
. ∏ (
2(𝑛 − 𝑘) + 1
𝑛 − 𝑘 + 2
𝑛 + 𝑘
𝑘
)
𝑛−3
𝑘=2
After some simplifications on the right side of this expression the formula is written as below:
∏ (
𝑛 + 𝑘
𝑘
) = 5(2 𝑛−3).
2n − 1
𝑛 + 1
𝑛
𝑘=𝑛−2
. ∏ (
[2(𝑛 − 𝑘) + 1]
𝑛 − 𝑘 + 2
𝑘
(𝑛 + 𝑘)
)
𝑛−3
𝑘=2
Where the denominator (𝑛 − 𝑘 + 2)(𝑛 + 𝑘) can be written as:
(𝑛 − 𝑘 + 2)(𝑛 + 𝑘) = (𝑛2
− 𝑘2
) + 2(𝑛 + 𝑘)
∏ (
𝑛 + 𝑘
𝑘
) = 5(2 𝑛−3).
2n − 1
𝑛 + 1
𝑛
𝑘=𝑛−2
. ∏ (
k. [2(𝑛 − 𝑘) + 1]
(𝑛2 − 𝑘2) + 2(𝑛 + 𝑘)
)
𝑛−3
𝑘=2
This expression can be modified in other way by expanding the product from the left side:
∏ (
𝑛 + 𝑘
𝑘
) = (
n + n − 2
𝑛 − 2
) (
n + n − 1
𝑛 − 1
) (
n + n
𝑛
) = 2 (
2n − 2
𝑛 − 2
) (
2n − 1
𝑛 − 1
)
𝑛
𝑘=𝑛−2
And even more simple as below:
∏ (
𝑛 + 𝑘
𝑘
) = 22
(
n − 1
𝑛 − 2
) (
2n − 1
𝑛 − 1
)
𝑛
𝑘=𝑛−2
= 22
(
2n − 1
𝑛 − 2
)

10. That expression is substituted here:
∏ (
𝑛 + 𝑘
𝑘
) = 22
(
2n − 1
𝑛 − 2
) = 5(2 𝑛−3).
2n − 1
𝑛 + 1
𝑛
𝑘=𝑛−2
. ∏ (
k. [2(𝑛 − 𝑘) + 1]
(𝑛2 − 𝑘2) + 2(𝑛 + 𝑘)
)
𝑛−3
𝑘=2
From which it is possible to factorize and rearrange as below:
∏ (
k[2(𝑛 − 𝑘) + 1]
(𝑛2 − 𝑘2) + 2(𝑛 + 𝑘)
)
𝑛−3
𝑘=2
=
22
(
2n − 1
𝑛 − 2
)
5(2 𝑛−3). (
2n − 1
𝑛 + 1 )
Or:
∏ (
k[2(𝑛 − 𝑘) + 1]
(𝑛2 − 𝑘2) + 2(𝑛 + 𝑘)
) =
(n + 1)
5(2 𝑛−5). (n − 2)
𝑛−3
𝑘=2
With n ≥ 5
(19)
It is a very majestic result, which can be evaluated for several values of 𝑛, 𝑛 ≥ 5, again, the readers
can check that both sides of the expression are equivalent by the evaluation of some values of n and
it agrees to the algebra and the previous formulas from which emerges. That is an expression that
results from the consideration that exists a factorization in function of the few prime Catalan
numbers, 5 and 2, which still remain in this factor: 5(2 𝑛−5). The rules about products among integer
numbers, from the Catalan analysis, let conduct simplifications or factorizations that could be useful
in other deductions in Mathematics or other fields. This is an introductory work that sets new paths
about the importance of the products and integer numbers such as the Catalan set. In future, it
would open new doors for applications that demand more precise equations in some systems.

11. 5. CONCLUSIONS ABOUT THIS WORK
The mathematical recurrence offered the possibility to open a new door for connexions
between the Catalan numbers and the positive integers, except zero. In this work the
formulas showed interesting patterns associated with prime, odd and natural numbers,
except zero.
The wish to connect the Bernoulli numbers and the Catalan numbers played a valuable role
about the comprehension of Arithmetic and theory of numbers in general. This paper shows
that some formulas can lead new paths that could bring attractive methods or even
theorems that would be applied in all science, not only in Mathematics.
Some remarks are described below:
Remark 4.1. There are several ways to define the Catalan numbers by an operator named
product and expressions that use factorials, and, of course, other multiple operators in
Mathematics. For this work, an index 𝑖, from 𝑖 = 1 until a limit 𝑛 − 3 was introduced by the
author, by taking into consideration the expansion of the product and the recurrence given
by a well-known relation. The definition of recurrence leaded a Catalan product operator
∏ {}𝐶𝑎𝑡𝑎𝑙𝑎𝑛 that summarized the multiplication of all the factors involved here. The
difference between this way deduced here and other equations that specify expansion of
products is that it is possible expand a product from 𝑖 = 1 until a 𝑖 = 𝑛 − 3, instead of some
expansions like 𝑖 = 2 𝑜𝑟 𝑘 = 2 until 𝑖 𝑜𝑟 𝑘 = 𝑛 [4]. The fact of such definition came from
the basic supposition of using one of the Bernoulli numbers, 𝐵1, and the value 20𝐵1
2
, which
helped to follow the deduction the final formula given by equation 4 and 5. This formula
included numerical factors 5 and 2 that are prime numbers and that have been mentioned
before like 5 and 2 𝑛−3
. These elements show that only the unique prime Catalan numbers
plays a fundamental role in the perception of the deduced formulas. There is an intrinsic
nature that, for all the researches, should be topic of discussion in order to connect not only
prime numbers, odd numbers (or even number as well), Bernoulli numbers and natural
numbers, but also the Hypothesis of Riemann [3] or other great theorems that are
associated with theory of numbers. If in this work the author was able to deduce a simple
connexion between a Bernoulli number and the recurrence applied to the Catalan numbers,
so how many other possibilities could exist if the researchers tried to bring the Riemann
scenario or its deductions to the world of the Catalan numbers?
Remark 4.2. The possibility to factorize several types or sets of numbers each other into a
theory of Groups or others. It is fascinating to be able to open new doors about how the
factorization of elements of sets could be done. When something is done from a simple
element taken from a set, such like one element 𝐵1 from the Bernoulli’s set, and still it is
possible to look for more numbers and relations apparently unclear, the path of new kinds
of factorizations and formulas could be even related to very significant concepts like groups,

12. fields or rings. It is possible to examine polynomials or formulas from sums or products that
represent concepts into a formal scenario in the theory of groups.
Remark 4.3. Setting new processes of filtering by the factor
3𝜌 𝑛 𝐶 𝑛
2 𝑛−3 which lets get the product
of the set odd-numbers by the definition of a Catalan number 𝐶 𝑛, ∀𝑛 ≥ 4. The possibility
to make more complex mappings or transformations among sets of numbers that could be
used for concluding unique theories that summarize all the numbers. The filters could say
when it is possible to extract something valuable from some numbers enunciated like
parameters in equations for others. A clear scenario could contribute to organize equations
in other fields like differential equations or solving problems by filtering or transformations.
Remark 4.4. The possibility to correlate several kinds of infinites and set up some limits in
Mathematics that have not been so clear about the numbers. Which numbers are more
commons? Which ones have greater values of products until some limits of expansion in
comparison with other sets of numbers at the same limit? Many questions that emerge
from products and sums.
Remark 4.5. The possibility to conduct simplifications or getting concrete formulas from
products. The great dream in combinatorics and Arithmetic is to be able to characterize
precisely events of count by formulas that are not limited for large products or factorials.
The recognition of the Catalan numbers and their formulas could be connected with how to
reduce operations that take a lot of time in exponential time. Investigations could bring new
paths about complex processes of functions in NP vs. P problem [5, 6] and to take decisions
about how to find better options.
More considerations could be analysed in future works. The author of this paper wants to
offer paths that can be seen from non-visible methods or formulas.
6. ACKNOWLEDGEMENTS
The author is grateful to the channel on Youtube, “Derivando”, in which he has known the
importance of the Catalan numbers in several fields and how curious and important could be
these numbers, even in the same level of others: naturals, odd numbers, Bernoulli numbers,
even numbers and prime numbers.
The author would be very grateful if his formulas could be used in some research or taken into
consideration for studies in applications of several fields like in Physics, Chemistry or other
where the event of count is crucial using by the Catalan numbers. The new baptized formulas
could be an honour, in future, like ‘The Catalan-López product formulas’ in bibliography.

13. REFERENCES
[1] Los Números de Bernoulli, Wikipedia, Web. 1 June. 2019
https://es.wikipedia.org/wiki/N%C3%BAmero_de_Bernoulli
[2] Los Números de Catalan, Wikipedia, Web. 1 June. 2019
https://es.wikipedia.org/wiki/N%C3%BAmeros_de_Catalan
[3] Arakawa Tsuneo, Bernoulli Numbers and Zeta Functions, ISBN13 (EAN): 9784431549185,
2014
[4] Catalan Number, Wikipedia, Web. 1 June. 2019
https://en.wikipedia.org/wiki/Catalan_number
[5] Titchmarsh E.C., the Theory if Functions, Oxford Press 1939.
[6] Edwards H.M., Riemann’s zeta-function, Academic Press, 1974
This is a particular paper written by Carlos López on the 1 of July, 2019. He is an electronic
engineer graduated from ‘Pontifical Bolivarian University (UPB)’, 2017, Medellín, Colombia.
The author works in Szczecin, Poland, and his interest is to participate of fundamental solutions
for Mathematical problems or the connexion of those with applied science. These are his
personal ideas concerning combinatorics and Arithmetic. The author does not belong to any
institution or group for the purposes of this paper.
He thanks PhD. Johnson Garzón Reyes for his collaboration in the scientific dissemination of this
document and for those who take seriousness in this research for the advancement of
Mathematics and other related fields, despite the limited resources or not advance education in
very formal scenarios.
The use of this paper is not authorized without the confirmation by the author.
Author: Carlos Hernán López Zapata
Contact mobile: +48 579339337
E-mail: seniorcar@zoho.com
Linkedin: https://www.linkedin.com/in/carlos-l%C3%B3pez-b285aba8/
Szczecin, West Pomeranian province, Poland, 4 of July of 2019.