First, we define a concise strategy by selecting the roots of primes greater than 10 and composites without factors of 2, 3, 5, and 7, and then form the Cyclic Table of Composites (CTC) by identifying the locations of such composites, followed by advancing the Formula of Primes and building the PTP. The procedures are summarized in 4 steps:

1) Selecting the roots and cycles

Adopting the first four primes 2, 3, 5, and 7, we take 2 × 3 × 5 × 7 = 210 as the length of a period. Within the interval [11, 211], we sort out 48 integers of primes or composites that do not contain factors of 2, 3, 5, and 7. These 48 integers, to generate primes and composites, are considered as the roots denoted by r 1 = 11 , r 2 = 13 , r 3 = 17 , ⋯ , r 23 = 103 , r 24 = 107 , ⋯ , r 48 = 211 . These r_is are placed on the left column of a table.

2) Developing the Cyclic Table of Composites (CTC)

The CTC consists of multiple 48 × 48 tables [ l ( i , j ) ] derived from the positions of composites without factors of 2, 3, 5, and 7.

Within an interval of [ 48 ( θ − 1 ) + 1 , 48 θ ] , where θ ∈ N + is considered a cycle, there are 48 integers which do not contain factors of 2, 3, 5, and 7. These 48 × θ integers are either primes or composites, denoted as q 1 = r 1 = 11 , q 2 = r 2 = 13 , q 3 = r 3 = 17 , ⋯ , q 48 = r 48 = 211 , for cycle 1; q 49 = q 1 + 210 = 221 , q 50 = q 2 + 210 = 223 , ⋯ , q 96 = q 48 + 210 = 421 , for cycle 2; ⋯ ; q 48 ( θ − 1 ) + 1 = q 48 ( θ − 2 ) + 1 + 210 = q 1 + 210 ( θ − 1 ) , q 48 ( θ − 1 ) + 2 = q 48 ( θ − 2 ) + 2 + 210 = q 2 + 210 ( θ − 1 ) , ⋯ , and q 48 θ = q 48 ( θ − 1 ) + 210 = q 48 + 210 ( θ − 1 ) , for cycle θ. We call q 1 to q 48 segments of cycle 1, q 49 to q 96 segments of cycle 2, ⋯ , and q 48 × ( θ − 1 ) + 1 to q 48 θ segments of cycle θ. The q_js are placed on the top row of a CTC table of θ cycles.

Formulating the CTC of the first cycle, namely, CTC(1), we observe the dual effect existing in [ l ( i , j ) ] , where index i refers to r i and j refers to q j , such that in each row of this table, pairs of entries share the same value of elements l ( i , j ) s. We also find for each row in the CTC(1), there is another row complementary to it, calling it the mirror effect. Specifically, there exist the mirror effects between the 23 top rows and the next 23 rows as well as between the 47^th row and the 48^th row of CTC(1). We define the basic CTC (CTC(basic)) to be a 24 × 48 table of [ l ( i , j ) ] , i = 1 , 2 , ⋯ , 23 , 47 and j = 1 , 2 , ⋯ , 48 .

Utilizing the mirror effect, we find the complementary CTC (CTC (complementary)) a 24 × 48 table made of [ l ( i , j ) ] for i = 24 , 25 , ⋯ , 46 , 48 and j = 1 , 2 , ⋯ , 48 . Combining the CTC (basic) and the CTC(complementary), we form CTC(1).

We observe further an intermedia effect in CTC(1) between different cycles of composite tables where each is 48 × 48 in size.

3) Advancing the Formula of Primes

By deleting all composites located by CTC in an interval, we develop the Formula of Primes.

Given a positive integer b with b – 211 being a multiplier of 210, an integer α ∈ [ 0 , b ] without factors of 2, 3, 5, and 7 is a prime if and only if there exists an i ∈ { 1 , 2 , ⋯ , 48 } such that α = r i + 210 k , for some integer k with k + 1 ∉ L b ( i ) , where L b ( i ) is a set of l ( i , j ) s which satisfy some conditions of b and θ described in the Results section.

4) Building the Periodic Table of Primes (PTP)

Denote PTP( k 0 , k ′ ) as a periodic table of primes from period k 0 to k ′ , k 0 < k ′ ∈ N + . PTP( k 0 , k ′ ) is a 48 × ( k ′ − k 0 + 1 ) table composed of all primes within an interval [ a , b ] , for a = 11 + 210 × k 0 and b = 211 + 210 × k ′ . Denote PTP⁺( k 0 , k ′ ) as a table composed of primes in PTP( k 0 , k ′ ) plus composites without factors 2, 3, 5, and 7 within the interval [ a , b ] .

The steps of building a PTP are depicted in Figure 1. Starting from the CTC (initial), we form a CTC (initial dual). By combining the CTC (initial) and the CTC (initial dual), we obtain the CTC (basic) which is outlined in the Supplement under Establishing the CTC (basic) by the CTC (initial). Once CTC (1) is deduced from CTC (basic), all subsequent CTC (2), CTC (3), …, CTC (θ) are

readily available. Utilizing CTCs, we generate PTP⁺( k 0 , k ′ ), where we delete all composites to obtain PTP( k 0 , k ′ ). The red and green points in Figure 1 represent the diagonal and intermediate effects, respectively, and also show the cyclical pattern on CTC and the periodical pattern on PTP⁺.

Lei et al. [11] held that the evolution of Chinese historical dynasties can be mapped by the properties of elements’ electrons in the Periodic Table of Chemical Elements. Wang [12] claimed that irregular primes are the survivors of natural numbers after regular decimations by primes.

There are various sieve methods to locate primes. In coming up with a solution for predicting primes, many have searched for a particular pattern of primes distributions. For example, Holt [13] mentioned structure to cycles due to recursion and created exact relative population models for all gaps up to a certain level. Dastych found a “mirroring” effect of numbers such as 2 × 3, 2 × 3 × 5 and 2 × 3 × 5 × 7 being represented by a wheel which were already in existence [14] . His algorithm originated from playing with Goldbach’s conjecture, but became useless when the numbers grew large. Others may have expressed their wishes to organize primes listing over the years. Although we tried to be thorough and are blessed by many professional colleagues in reaching this unique PTP, we could miss non-disclosed endeavors made by individuals beyond the publicly accessible domain.

Note that r_is and q_js are integers in [ 11 ,   211 ] without factors of 2, 3, 5, and 7. Let ( i , j ) be at a position in the CTC and l ( i , j ) the corresponding entry of point ( i , j ) , where 1 ≤ i , j ≤ 48 , determined by

l ( i , j ) = 1 + ( q j × q j ^ − r i ) / 210 . (1)

Given the i^th row and j^th column, we observe a unique q j ^ ∈ S that couples with q j such that q j × q j ^ − r i is a multiplier of 210. Both q j and q j ^ depend on r i or the i^th row.

An outline of CTC is shown in Figure 2. Various effects in CTC are described by the following statements. Presented in Figure S1(a) along with an illustrative example are given in the Supplement under Example for Figure S1(a). Observing Figure S1(a), note that l ( i , i ) = q i + 1 , i = 1 , 2 , ⋯ , 23 , 47 . We identify the diagonal effect in CTC in Statement 1 below.

Analyzing the distribution of l ( i , j ) in the i^th row of Figure S1(a), we find the dual effect between l ( i , j ) and l ( i , j ^ ) , described in Statement 2 below.

Statement 2 implies that given l ( i , j ) , we know l ( i , j ^ ) , under which situation j and j ^ are dual to each other. Take i = 1 and r 1 = 11 for instance, l ( 1 , 5 ) = l ( 1 , 8 ) = 5 , where q 5 × q 8 = 23 × 37 = 851 = 11 + 210 ( 5 − 1 ) . Notice that when j ^ = j , q j 2 is a descendant of r i and l ( i , j ) appears in the i^th row only once. For instance, l ( 48 , 9 ) = 8 appears only once and 41 2 = 1681 = 211 + 210 × ( 8 − 1 ) .

Further elaboration of the dual effect and associated examples are given in the Supplement under Elaboration of the Dual Effect of the CTC (basic). Analyzing Figure S1(a) and Figure S1(b), we also identify the mirror effect between each row of the CTC(basic) and its complementary row in CTC deduced below.

Statement 3 implies that if we know l ( i , j ) for 1 ≤ i ≤ 23 , then we can find l ( 47 − i , j ) . Similarly, if we know l ( 47 , j ) , then we find l ( 48 , j ) . Following Statement 3, we form the CTC (complementary), the complementary of the CTC (basic).

CTC (complementary) is presented as Figure S1(b), in the Supplement. Merging the CTC (basic) and the CTC (complementary), we obtain the CTC (1) as the CTC for cycle 1, referring to Figure 2.

Consider CTC in Figure 2. There are two red diagonals shown in Figure 2, which further elaborates Figure S1(a) and Figure S1(b), The first diagonal is l ( 1 , 1 ) = 12 , l ( 2 , 2 ) = 14 , ⋯ , and l ( 46 , 46 ) = 200 , and the second one is l ( 46 , 1 ) = 11 , l ( 45 , 2 ) = 13 , ⋯ , and l ( 1 , 46 ) = 199 . The two cross lines in Figure 2 are obvious due to the mirror effect. l ( 47 , j ) and l ( 48 , j ) , j = 1 , 2 , 3 , ⋯ , 48 , are mirrors with each other. An example is given in the Supplement under An Associated Example for Statement 3.

Figure S1(a) and Figure S1(b) show another distribution of l ( i , j ) on CTC, named the intermedia effect, which happens at some symmetrical columns, described in the Supplement under Statement and Proof on the Intermedia effect in the CTC. Both the diagonal effect and the intermedia effect on the CTC are helpful in showing how to transfer CTC to PTP. In Figure 2, the red and green numbers represent the diagonal and the intermediate effects, respectively.

Proofs for the above Statements are given in the Supplement under Proofs of the Statements.

We summarize the Formula of Primes below:

A list of L b ( 1 ) for various b up to 44521 and θ ^ = 1 is given in Figure S2, which is presented in the Supplement.

According to CTC(θ) and the Formula of Primes, we specify the Periodic Table of Primes below.

Denote PTP⁺( k 0 , k ′ ) as the table composed of primes in PTP( k 0 , k ′ ) plus composites without factors of 2, 3, 5 and 7, within an interval [ 11 + 211 × k 0 ,   211 ( k ′ + 1 ) ] .

Figure 3 is an outline of PTP⁺(0, 210), which includes PTP(0, 210) and composites without factors of 2, 3, 5, and 7 within the interval [ 11 ,   1 + 210 2 ] converted from Figure 2. Figure 3 has 48 rows and 211 columns, for i = 1 , 2 , ⋯ , 48 , and k = 0 , 1 , 2 , 3 , ⋯ , 210 . λ ( i , k ) is an integer of the entry ( i , k ) , computed from l ( i , j ) of Figure 2. The two dotted red curves in Figure 3 are configured from two diagonal red lines of Figure 2, i.e., 12–14–18∙∙∙198–200 and 11–13–17∙∙∙ 197–199. The two dotted green diamond-shape lines in Figure 3 come from two green lines of Figure 2 of 24–75–74∙∙∙29–80 and 84–31–32∙∙∙77–24.

Figure 4 is a realization of PTP⁺(0, 10) including primes and composites without factors of 2, 3, 5, and 7. Further elaboration is seen in the Supplement under Numerical Illustrations for The Periodic Table of Primes. According to Gauss [8] [9] , the number of primes no more than b is approximately b / ln b . We find further the exact Counting Function of primes π ( b ) when b – 211 is a multiplier of 210.

We use π ( b ) and π * ( b ) to represent the Counting Function of Primes and the Counting Function of Twin Primes in [ 1 , b ] for a natural number b, respectively.

Currently, few predict the number of twin-prime pairs in an interval [2, b]. For b – 211 as a multiplier of 210, denote T ( i , i + 1 ) as the set of twin-prime pairs on PTP( 0 , ⌈ b / 211 ⌉ ), i.e., a set of k with k + 1 ∉ L b ( i ) and k + 1 ∉ L b ( i + 1 ) , where i and i+1 are for the i^th and the (i+1)^st row on the PTP, respectively. Our study finds the exact π * ( b ) below:

Elaborations of Expressions (3) and (4) are given in the Supplement under Elaborations of Expressions (3) and (4), and examples of computing π ( b ) and π * ( b ) are illustrated in the Supplement under Examples for the Predictions of Primes by the PTP.

Specifically, we adopt the following steps to establish the CTC (basic):

a) From the CTC (basic) of Figure 1, we find 23 × 23 CTC (initial) table for the first cycle. In the CTC (initial), its entries of [ l ( i , j ) ] , for i = 1 , 2 , ⋯ , 23 and j = 1 , 2 , ⋯ , 23 , are both starting from 11; namely, the left column contains r 1 , r 2 , ⋯ , r 23 and the top row contains q 1 , q 2 , ⋯ , q 23 .

b) Applying the dual effect, we form another 23 × 23 CTC (initial dual) table from CTC (initial). In this new table, its entries of [ l ( i , j ) ] , for i = 1 , 2 , ⋯ , 23 and j = 24 , 25 , ⋯ , 46 , are the first 23 roots starting from 11 for “i”s and the 24^th to 46^th roots for “j”s, respectively. Namely, the left column contains r 1 , r 2 , ⋯ , r 23 , and the top row contains q 24 , q 25 , ⋯ , q 46 .

c) Add r₄₇ as the end of the 24^th row and add q₄₇ and q₄₈ as the end of the 47^th and the 48^th columns. Namely, the left column contains r₄₇ as the last index and the top row contains q₄₇, q₄₈ as the last two indices.

d) Combine tables developed by a, b, and c steps to complete CTC(basic).

As shown in Figure 1, there are the CTC(initial), the CTC(initial-dual), the CTC(basic), the CTC(complementary), and CTC(1).

CTC ( initial ) = { l ( i , j ) | i = 1 , 2 , ⋯ , 23 ; j = 1 , 2 , ⋯ , 23 } ,

CTC ( initial-dual ) = { l ( i , j ) | i = 1 , 2 , ⋯ , 23 ; j = 24 , ⋯ , 46 } ,

CTC ( basic ) = { l ( i , j ) | i = 1 , 2 , ⋯ , 23 , 47 ; j = 1 , 2 , ⋯ , 48 } ,

CTC ( complementary ) = { l ( i , j ) | i = 24 , 25 , ⋯ , 46 , 48 ; j = 1 , 2 , ⋯ , 48 } ,

CTC ( 1 ) = { l ( i , j ) | i , j = 1 , 2 , ⋯ , 48 } .

Given q 1 = 11 , compute l ( 1 , 1 ) as follows:

Since l ( 1 , 1 ) = 1 + q 1 × q 1 ^ − r 1 210 , and the unique q 1 ^ to let 11 × q 1 ^ − 11 210 be an integer is

q 1 ^ = 211 ,   l ( 1 , 1 ) = 1 + 11 × 211 − 11 210 = 1 + 11 = 12.

Similarly, we find

l ( 2 , 1 ) , l ( 3 , 1 ) , l ( 4 , 1 ) , l ( 5 , 1 ) , l ( 6 , 1 ) , ⋯ , l ( 23 , 1 ) , l ( 47 , 1 ) = ( 10 , 6 , 4 , 11 , 5 , ⋯ , 8 , 1 ) .

By utilizing Statement 2, we can generate a new table CTC(initial dual) in the Introduction from CTC(initial). Taking the first row, i.e. i = 1 and r 1 = 11 as an example (Figure 2):

(i) Since l ( 1 , 2 ) = l ( 1 , 3 ) = 2 , we have

l ( 1 , 47 − 2 ) = l ( 1 , 47 − 3 ) = 1 + q 45 × q 44 q 2 × q 3 ( 11 210 + l ( 1 , 2 ) − 1 ) − 11 210 = 182

Similarly, since l ( 1 , 5 ) = l ( 1 , 8 ) = 5 , we have

l ( 1 , 47 − 5 ) = l ( 1 , 47 − 8 ) = 1 + q 42 × q 39 q 5 × q 8 ( 11 210 + l ( 1 , 5 ) − 1 ) − 11 210 = 155

(ii) For i = 1 and j = 1 , 6 , 10 , 12 , ⋯ , 19 , 21 , 22 , we do not have 1 ≤ j ^ ≤ 23 to fit l ( 1 , j ) = l ( 1 , j ^ ) , then we compute l ( 1 , j ^ ) for 24 ≤ j ^ ≤ 46 .

When j = 1 , q 1 = 11 and l ( 1 , 1 ) = 12 ,

since r 1 + 210 ( l ( 1 , 1 ) − 1 ) q 1 = 11 + 210 ( 12 − 1 ) 11 = 211 = q 48 , we have

1 ^ = 48 and l ( 1 , 48 ) = l ( 1 , 1 ) = 12

When j = 6 , q 6 = 29 and l ( 1 , 6 ) = 16 ,

since r 1 + 210 ( l ( 1 , 6 ) − 1 ) q 6 = 109 = q 25 , we have

6 ^ = 25 and l ( 1 , 25 ) = l ( 1 , 6 ) = 16 .

Similarly, we can predict l ( 1 , 26 ) = l ( 1 , 15 ) = 37 , l ( 1 , 28 ) = l ( 1 , 12 ) = 33 , l ( 1 , 29 ) = l ( 1 , 14 ) = 39 , l ( 1 , 30 ) = l ( 1 , 10 ) = 29 , ⋯ , l ( 1 , 41 ) = l ( 1 , 22 ) = 88 .

Comparing the CTC (basic) with the CTC (complementary) and referring to Figure S1(a) and Figure S1(b), it is clear that:

Given j = 1 and q 1 = 11 , we have l ( 1 , 1 ) + l ( 46 , 1 ) = 12 + 11 = 2 q 1 + 1 .

Given j = 2 and q 2 = 13 , we have l ( 2 , 2 ) + l ( 45 , 2 ) = 14 + 13 = 2 q 2 + 1 .

Given j = 1 and q 1 = 11 , for i = 2 , we have l ( 45 , 1 ) + l ( 2 , 1 ) = 2 + 10 = 12 = q 1 + 1 .

Given j = 12 and q 12 = 53 , for i = 1 , l ( 46 , 12 ) + l ( 1 , 12 ) = 21 + 33 = 54 = q 12 + 1 ; for i = 23 , l ( 24 , 12 ) + l ( 23 , 12 ) = 28 + 26 = 54 = q 12 + 1 ; for i = 47 , l ( 47 , 12 ) + l ( 48 , 12 ) = 26 + 27 = 53 = q 12 .

Figure S1(a) and Figure S1(b) show another distribution of l ( i , j ) on the CTC, named the intermedia effect, which happens at some symmetrical columns, described below.

For instances, in the case of green columns 23 and 24 of Figure 2, l ( 19 , 23 ) = l ( 28 , 24 ) = 6 , ⋯ , l ( 2 , 23 ) + 2 = l ( 45 , 24 ) = 77 . Other column pairs also exhibit the intermedia effect. For example, in the case of columns 22 and 25, if l ( i , 22 ) + l ( i , 25 ) = 110 , then l ( i , 22 ) = l ( 47 − i , 25 ) .

Proof of the intermedia effect in the CTC

For the case of columns 23 and 24,

(i) From the mirror effect, we have l ( i , 24 ) + l ( 47 − i , 24 ) = q 24 + 1 = 108 .

If l ( i , 23 ) + l ( i , 24 ) = 108 , then l ( i , 23 ) = 108 − l ( i , 24 ) = l ( 47 − i , 24 ) .

(ii) If l ( i , 23 ) + l ( i , 24 ) = 106 , then l ( i , 23 ) + 2 = 108 − l ( i , 24 ) = l ( 47 − i , 24 ) .

For the case of columns 22 and 25, from the mirror effect, we know l ( i , 25 ) + l ( 47 − i , 25 ) = q 25 + 1 = 109 + 1 = 110

If l ( i , 22 ) + l ( i , 25 ) = 110 , then l ( i , 22 ) = l ( 47 − i , 25 ) .

Proof of Statement 1

For each i , j = 1 , 2 , ⋯ , 48 , notice that in CTC(1),

l ( j , j ) = 1 + q j × q j ^ − r j 210 = 1 + q j ( 211 − 1 ) 210 = 1 + q j ,   as   r j = q j   and   we   choose   q j ^ = q 48 = 211.

For a given j, r i = q j × q j ^ − ( l ( i , j ) − 1 ) × 210 ,for some j ^ .This means that a fixed r i is the remainder of q j × q j ^ subtracting the maximum multiple of 210 smaller than q j × q j ^ . Since q j ^ = q 48 = 211 is the largest q j ^ hence l ( j , j ) ≥ l ( i , j ) .

Proof of Statement 2

By definition, l ( i , j ) = 1 + q j q j ^ − r i 210 , l ( i , j ^ ) = 1 + q j ^ q j − r i 210 .

Therefore, we have l ( i , j ) = l ( i , j ^ ) = 1 + q j × q j ^ − r i 210

Proof of Statement 3

(i):

In this proof, we use q j ^ | i to differentiate q j ^ given different r i .

Notice that r 47 − i + r i = 210 , ∀ i = 1 , ⋯ , 23 ,

l ( 47 − i , j ) = 1 + q j | i q j ^ | ( 47 − i ) − r 47 − i 210 for some q j ^ | 47 − i and

l ( i , j ) = 1 + q j | i q j ^ | i − r i 210 forsome q j ^ | i .

Hence,

l ( 47 − i , j ) + l ( i , j ) = 2 + q j | i ( q j ^ | ( 47 − i ) + q j ^ | i ) − 210 210 = 1 + q j | i ( q j ^ | ( 47 − i ) + q j ^ | i ) 210

Since q j | i ( q j ^ | ( 47 − i ) + q j ^ | i ) 210 must be an integer, and q j | i has no factors of 2, 3, 5, 7, q j ^ | ( 47 − i ) + q j ^ | i must be a multiple of 210. This happens only when

q j ^ | ( 47 − i ) + q j ^ | i = { 210 ,     if     j ^ | ( 47 − i ) + j ^ | i = 47 420 ,     if     j ^ | ( 47 − i )   or   j ^ | i = 48

Therefore,

l ( 47 − i , j ) + l ( i , j ) = { q j | i + 1 ,     if     i ≠ j   and   47 − i ≠ j 2 q j | i + 1 ,     if     i = j   or   47 − i = j

(ii):

Notice that r 47 + r 48 = 209 + 211 = 420 . Hence

l ( 47 , j ) + l ( 48 , j ) = ( 1 + q j | 47 q j ^ | 47 − r 47 210 ) + ( 1 + q j 48 q j ^ | 48 − r 48 210 ) = q j | i ( q j ^ | 47 + q j ^ | 48 ) 210 = { q j | i ,     if   j ^ | 47 + j ^ | 48 = 47 2 q j | i ,     if     j ^ | 47   or   j ^ | 48 = 48

The 2^nd case happens only when j = 47 , 48 .

Starting with θ = 1 , compute q j , q j ^ and l ( 1 , j ) using Expression (1)

l ( 1 , j ) = 1 + q j × q j ^ − 11 210 , for j = 1 , 2 , ⋯ , 48 .

Then for 2 ≤ θ ≤ 19 and θ ^ = 1 , we have:

l ( 1 , j + 48 ) = l ( 1 , j ) + q j ^

l ( 1 , j + 96 ) = l ( 1 , j ) + 2 q j ^

⋮

l ( 1 , j + 864 ) = l ( 1 , j ) + 18 q j ^

We then list l ( i , j + 48 ( θ − 1 ) ) for θ = 1 , 2 , 3 , ⋯ , 19 in Figure S2, corresponding to Expression (2’). The bottom row of Figure S3 demonstrates that Δ 1 , 1 = 2 , Δ 1 , 2 = 15 , Δ 1 , 3 = 28 , ⋯ , Δ 1 , 19 = 210 . Notably,

min j { l ( i , j + 48 ( θ − 1 ) ) } < min j { l ( i , j + 48 θ ) } for all θ = 1 , 2 , 3 , ⋯ , 19 , and Δ i , 1 < Δ i , 2 < Δ i , 3 < ⋯ , for all i.

For instance, given b = 2311 = 211 + 210 × 10 , compute L 2311 ( 1 ) by applying Expression (2)to obtain L 2311 ( 1 ) = { 2 , 5 , 7 , 9 } .

According to the Formula of Primes, in order for α = 11 + 210 k ≤ 2311 to be primes, it requires k + 1 ∉ L 2311 ( 1 ) for k ∈ { 1 , 2 , ⋯ , 10 } . Therefore, we let

k ∈ { 0 , 2 , 3 , 5 , 7 , 9 , 10 } and get the following primes: 11, 11 + 210 × 2 = 431 , 11 + 210 × 3 = 641 , 11 + 210 × 5 = 1061 , 11 + 210 × 7 = 1481 , 11 + 210 × 9 = 1901 and 11 + 210 × 10 = 2111 . They are all shown in the first row of Figure 4.

Given b = 211 + 210 × 28 = 6091 , we get

L 6091 ( 1 ) = { 2 , 5 , 7 , 9 , 12 , 15 , 16 , 17 , 19 , 23 , 24 , 28 } .

In order for α = 11 + 210 k ≤ 6091 to be primes, we let k ∈ { 0 , 2 , 3 , 5 , 7 , 9 , 10 , 12 , 13 , 17 , 19 , 20 , 21 , 24 , 25 , 26 } such that k + 1 ∈ L 6091 ( 1 ) to obtain the following primes: 11, 431, 641, 1061, 1481, 1901, 2111, 2531, 2741, 3581, 4001, 4211, 4421, 5051, 5261, 5471.

Proof of Expression (3)

The total number of elements in PTP⁺(0, k) is 48 × ( k + 1 ) , for k ∈ N 0 . Within the interval [ 11 ,   211 2 θ ] , the number of different l ( i , j ) is | L b ( i ) | for b = 211 + 210 k . Hence, the total number of composites without 2, 3, 5, and 7 within the above interval is | L b ( i ) | . Note that 4 primes are less than 10. Therefore, the number of primes within [ 2 , b ] is

π ( b ) = 4 + 48 × ( k + 1 ) − ∑ i = 1 48 | L b ( i ) | .

Proof of Expression (4)

For an integer i ∈ W = { 1 , 3 , 6 , 9 , 13 , ⋯ , 4 5 , 47 } , ( r i , r i + 1 ) are twin primes for r i ≥ 11 . If there exists a k, k ∈ N 0 , such that k + 1 ∉ L b ( i ) and k + 1 ∉ L b ( i + 1 ) , then r i + 210 k and r i + 1 + 210 k are twin primes. Since (2, 3), (3, 5), (5, 7) are the only three twin-prime pairs less than 11. Therefore, the number of twin-prime pair in interval [ 2 , b ] is π * ( b ) = 3 + ∑ i ∈ W | T ( i , i + 1 ) | , where T ( i , i + 1 ) is a set of k such that k + 1 ∉ L b ( i ) and k + 1 ∉ L b ( i + 1 ) .

Let λ ( i , k ) be the integer value of the entry ( i , k ) in PTP⁺(0, 210) if Figure 3, which is computed based on CTC (1) and the Formula of Primes. For instance, λ ( i , k ) on the diagonal dotted points (1, 11), (2, 13), (3, 17), (4, 19) and (5, 23) are computed below:

Now return to columns 22 and 25 in Figure 2. We note that (i, 23) and (i, 24), for i = 1 , 2 , ⋯ , 48 being the intermediate dotted points. There are two other green lines 12–10–9–7–6 for column 22, and 6–7–9–10–12 for column 25. Both are caused by the intermedia effect, with respect to 47. These two lines are also converted to Figure 4, shown as 11 × 101 – 13 × 101 – 17 × 101 – 19 × 101 – 23 × 101, and 11 × 109 – 13 × 109 – 17 × 109 – 19 × 109 – 23 × 109, respectively.

The above conversions between l ( i , j ) in the CTC and λ ( i , k ) in the PTP⁺ demonstrate that (i) many composites such as the digonal and intermediate dotted points which allocate cyclically at the CTC are also allocated periodically at the PTP⁺. (ii) the PTP is obtained by removing all composites periodically allocated at PTP⁺. Therefore, we claim that prime numbers are also generated periodically since they come from removing all periodically distributed composites.

Figure 4 contains the first 11 columns for the PTP⁺(0, 210), which is composed of 48 × 11 = 528 elements, represented by λ ( i , k ) and elaborated below.

(i) Each entry λ ( i , k ) is either a prime or a composite.

(ii) If λ ( i , k ) is a prime, then λ ( i , k ) = r i + 210 k , and vice versa. If λ ( i , k ) is a composite, then there are q j and q j ^ such that

λ ( i , k ) = q j × q j ^ = r i + 210 × ( l ( i , j ) − 1 ) .

Take i = 1 , r i = 11 for instance,

λ ( 1 , 1 ) = 11 + 210 × ( l ( 1 , 2 ) − 1 ) = q 2 × q 3 = 13 × 17 ,

λ ( 1 , 4 ) = 11 + 210 × ( l ( 1 , 5 ) − 1 ) = q 5 × q 8 = 23 × 37 .

Then, we have λ ( 1 , 0 ) = 11 , λ ( 1 , 2 ) = 11 × 210 ( 3 − 1 ) = 431 ,

λ ( 1 , 3 ) = 11 + 210 × ( 4 − 1 ) = 641 , ⋯ ,

λ ( 1 , 9 ) = 11 + 210 × ( 10 − 1 ) = 1901 .

(iii) Primes λ ( i , k ) and λ ′ ( i , k ) = λ ( i , k ) + 1 are twin-prime pairs, if and only if r ′ i = 2 + r i , where λ ( i , k ) = r i + 210 × k , λ ′ ( i , k ) = r ′ i + 210 × k .

For instance, (431, 433), (641, 643), (1061, 1063), (1481, 1483) are twin-prime pairs.

(iv) The numbers of primes in column k of the PTP⁺(0, 210) are shown as (43, 35, 32, 31, 31, 28, 28, 30, 26, 27, ⋯). The sum of primes within interval [11, 2311] is 43 + 35 + 32 + ⋯+ 18 = 329. The number of twin-prime pairs of a column k are listed as (12, 6, 4, 4, 4, 8, 6, 4, 4, 4, ⋯, 3).The maximum prime gap for the k^th column can also be found from Figure 4. For instance, the maximum prime gap for column 2 is 10, occurring at 409 and 419 of r₄₆ and r₄₇, respectively.

(v) The PTP(k⁰,k') is useful in predicting primes. Examples are given by Examples for the Predictions of Primes by the PTP.

In summary, the Formula of Primes, for the case θ ^ = 1 , is checked below:

From Figure S1(a), we know that

L b ( 1 ) = { 2 , 5 , 7 , 9 , 12 , ⋯ }

L b ( 2 ) = { 7 , 10 , 12 , 13 , 14 , ⋯ }

L b ( 3 ) = { 3 , 6 , 11 , 12 , 13 , 14 , ⋯ }

⋮

From Figure 4, we know that

α = r 1 + 210 k is prime for k = 0 , 2 , 3 , 5 , 7 , 9 , 10

α = r 2 + 210 k is prime for k = 0 , 1 , 2 , 3 , 4 , 5 , 7 , 8 , 10

α = r 3 + 210 k is prime for k = 0 , 1 , 3 , 4 , 6 , 7 , 8 , 9

⋮

In general, α = r i + 210 k is prime, if k + 1 ∉ L b ( i ) and θ ^ = 1 .