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 2022-11-17, 11:27 #1 Charles Kusniec     Aug 2020 Guarujá - Brasil 10710 Posts The triangular sieve of primes. C002216 Introduction to the triangular sieve of primes from the oblong parabolic sieve. From the concepts seen in the oblong parabolic sieve of primes, we can sift the prime numbers on the number line at $$y=0$$ as follows: C002216.1 For clarity, to sift through the prime numbers on each horizontal number line, we are omitting all multiples of 2 and multiples of multiples of prime. So, on each horizontal number line, we will only cover with lines with the multiples of the odd prime divisors. Also, to differentiate, all "multiples of 1" are dashed. They define the odd primes in the number line. So, to sift the prime numbers on the number line at $$y=-1$$, we must move the Y-axis to the position $$x=-(-1)^2+(-1)=-2$$ and we get: C002216.2 Analogously, to sift the prime numbers on the number line at $$y=-2$$, we must shift the Y-axis to the position $$x=-(-2)^2+(-2)=-6$$ and we get: C002216.3 And we can do this successively for any number line at any value of $$y=integer \ positive \ or \ negative$$. For each number line in row $$y$$ we have a "triangular framework of divisors" with origin at the point $$0=(-y^2+y,y)$$. For the sake of clarity, let us draw the lines from each odd prime divisor $$P$$ up to the square of the next prime number minus 1 $$(P+1)^2-1$$. Since we are using only the odd primes $$3$$ and $$5$$ in this example, then $$48=7^2-1$$ is the largest composite we can detect. Remember that for the clarity of the figures, we consider by default all even integers as being composites. To detect the composite 49 and above, we need the prime $$7$$. We can no longer count on $$5$$ and $$3$$ only. In the examples above, we are omitting the trivial divisors. If we had placed the continuous lines of all the integer divisors along Y-axis, then:The prime numbers would be covered by only two lines representing the two trivial divisors. The square numbers would be covered by an odd>1 number of lines: one for the square root of the square and an even number of lines for the pair(s) of complementary divisor(s). All other composites would have an even>2 number of lines covering for all the pairs of complementary different divisors. Thus, the intersection of the lines in any integer number is the result of the product of two of their complementary divisors. For example, 15=3*5. Attached Thumbnails
 2022-11-17, 11:45 #2 Charles Kusniec     Aug 2020 Guarujá - Brasil 10710 Posts The pairs of triangular divisor frameworks. Now, see what happens when we group the horizontal number lines, two by two consecutively: C002216.4 The reference is the oblong parabola $$x=-y^2+y$$. In this case above, we show the non-negative integers of the number lines at $$y=0$$ in green, and $$y=-1$$ in pink. First, notice that all the lines in the triangular framework of the green divisors $$d$$ of $$y=0$$ act as $$(d+1)k+2$$ on the pink number line at $$y=-1$$ . For example, the green lines of divisor $$3$$ act as the divisors $$4k+2$$ in the pink number line. The green divisor $$5$$ acts as $$6k+2$$. And so on. In contrast, the triangular framework of the divisors of $$y=-1$$ in pink acts as $$(d-1)k-2$$ on the green number line at $$y=0$$. Thus, divisor $$5$$ pink acts on the green number line at $$y=0$$ as $$4k-2$$. That is, whatever odd divisor, we use on a row $$y$$, it will be an even divisor on adjacent rows $$y±1$$. The same geometric framework that generates the primes in the green number line generates the primes in the pink number line. Each framework by itself cannot cover all odd integers in its number line. Nor can one framework cover any odd numbers on the other number line. Disregarding the dashed lines that represent the trivial multiples of $$1$$, it is not possible to cover both number lines in all pairs of integers aligned in verticals between the two lines. Therefore, several pairs of prime numbers appear aligned vertically between $$y=-1$$ and $$y=0$$. An infinite number of vertically aligned pairs of odd numbers must appear between the two green and pink number lines. For example, prime $$7$$ in row $$y=-1$$ and prime $$5$$ in row $$y=0$$. Now, see the figure below: C002216.5 The non-negative integers of the number lines at $$y=-1$$ in pink, and $$y=-2$$ in brown. See that the same cover framework used before is now repeated. That is, the property which geometric lines cannot cover all integers is repeated here as well. The only difference is that there is now a different offset between the action of the two frameworks. The same neighborhood properties apply where one framework cannot cover the odd numbers of the other. Here, again, we will find an infinite number of vertically aligned pairs of odd numbers. The same property of coverage by the frameworks imposes to appear pairs of odd numbers aligned vertically without the coverage of the lines of odd prime divisors. For example, prime $$7$$ in row $$y=-2$$ and prime $$3$$ in row $$y=-1$$. And so on for all pairs of number lines in $$y$$ and $$y±1$$. It will always be the same frameworks that act as a sieve. Attached Thumbnails
 2022-11-17, 12:02 #3 Charles Kusniec     Aug 2020 Guarujá - Brasil 10710 Posts The semiprime numbers in the form of y(y+2k). The semiprime numbers in the form of y(y+2k) By way of example, let us understand the vertical alignment that exists between the elements of the number line at $$y=0$$ and $$y=-1$$ in figure C002216.4 above. See that, geometrically, we can align the two number lines $$y=0$$ and $$y=-1$$ on a single line. This alignment does not change the properties of each of the two triangular green and pink frameworks. Thus, we obtain: C002216.6 This figure possesses and brings us a lot of information. Let us use the green number line as a reference. Now, the parabola reference is not necessary. The pink number line moves left or right in positive or negative integers. In the figure above, the pink line is $$2$$ units to the left of the green line. As a result, every integer element on the new number line has two integer values. At all integer points, the pink integer is equal to the green integer plus 2. The multiplication of the two values that each point has, produces the quadratic composite generator of the (square minus 1) numbers https://oeis.org/A005563. Now the two green and pink frameworks act simultaneously on the same physical points on the same horizontal line. The covering lines of the frameworks only start from the odd prime divisors that lie along the green and pink verticals. For the sake of clarity: We avoid placing the covering lines with $$-1:1$$ or $$-45°$$ slope. They should be present on all prime numbers of each color. We avoid placing cover lines for divisors that are not prime. We start placing the covering lines from each square of prime $$P^2$$. For each prime, this multiplication occurs over the two dashed lines. We stop placing the covering lines of a prime divisor $$P$$ when we reach $$(P+1)^2-1$$. There just is not any framework line covering odd points, where the green and pink integer elements are both odd prime numbers. These pairs of odd prime numbers relate all prime pairs. They form the central pair of complementary positive divisors of the semiprimes in the form of $$y(y+2)$$. Attached Thumbnails   Last fiddled with by Charles Kusniec on 2022-11-17 at 12:16
 2022-11-24, 12:55 #4 Charles Kusniec     Aug 2020 Guarujá - Brasil 1538 Posts The central and inferior divisors in the triangular sieve of primes C002216.7 The central and inferior divisors in the triangular sieve of primes Let us consider all possible positive divisors in the triangular sieve of primes with two frameworks forming the quadratic CGs in the form of $$x=y(y±b_{dm})$$. See below example for $$x=y(y±2)$$: For each $$d_1=$$https://oeis.org/A341674 strictly inferior positive divisor of an integer $$x$$ there is its complementary divisor $$d_2=$$https://oeis.org/A341673 strictly superior positive divisor, such as $$x=d_1*d_2$$. For each smallest central positive divisor $$d_{c1}=$$https://oeis.org/A033676 (largest positive divisor $$≤√x$$) of an integer $$x$$ there is its complementary largest central positive divisor $$d_{c2}=$$https://oeis.org/A033677 (smallest positive divisor $$≥√x$$), such as $$x=d_{c1}*d_{c2}$$. The complementary pair $$(d_{c1};d_{c2})$$ form the central pair of complementary positive divisors. For each integer $$x$$ the central pair of complementary positive divisors has two divisors:$$d_{c1}$$ is the largest central positive divisor https://oeis.org/A033676 the largest divisor of $$x$$ that is $$≤√x$$, and $$d_{c2}$$ the smallest central positive divisor is https://oeis.org/A033677 the smallest divisor of $$x$$ that is $$≥√x$$. Only in the square numbers we have $$d_{c1}=d_{c2}$$. The difference between the divisors of the central pair of complementary divisors is: $$b_{dm}=d_{c2}-d_{c1}=$$https://oeis.org/A033677$$-$$https://oeis.org/A033676$$=$$https://oeis.org/A056737. The orange $$-1:1$$ diagonals connect the green $$y$$ divisors and the pink $$y+2$$ divisors. They cross the X-axis at the points that have the green $$y$$ and pink $$y+2$$ values. These points form the elements of the quadratic CG $$x=y(y+2)$$. These $$-1:1$$ diagonals are the most inclined and connect the two divisors of the central pair of complementary positive divisors. For example:element $$3=3*1$$ has only one diagonal covering the product $$3$$. A $$-1:1$$ diagonal covers the two complementary central divisors $$(d_{c1};d_{c2})=(1;3)$$. There are no other diagonals because we are not placing the trivial divisors. We are not placing the diagonal of the pink divisor $$1$$ of the factor $$3$$. The element $$8=4*2$$ has two diagonals covering the product $$8$$. A diagonal $$-1:1$$ covers the two complementary central divisors $$(d_{c1};d_{c2})=(2;4)$$, plus another diagonal in pink of divisor $$2$$ of the factor $$4$$. There are no other diagonals because we are not placing the trivial divisors, nor the strictly superior divisors. In this case the only strictly superior divisor is $$8$$, which is also trivial. This is because $$8$$ is a cube that has $$4$$ divisors. The element $$15=5*3$$ has a diagonal covering the product $$15$$. A $$-1:1$$ diagonal covers the two complementary central divisors $$(d_{c1};d_{c2})=(3;5)$$. We omit the diagonals of the trivial divisors. There are no other diagonals because $$3$$ and $$5$$ are prime numbers, we are not placing trivial divisors, nor strictly superior divisors. In this case the only strictly superior divisor is $$15$$, which is also trivial. This is because $$15$$ is a semiprime number that has $$4$$ divisors. The element $$24=6*4$$ has three diagonals covering the product $$24$$. A $$-1:1$$ diagonal covers the two complementary central divisors $$(d_{c1};d_{c2})=(6;4)$$. Another diagonal in pink of divisor $$3$$ of factor $$6$$ which is the same in green of divisor $$2$$ of factor $$4$$. Another diagonal in pink of divisor $$2$$ of factor $$6$$. There are no other diagonals because we are not placing the trivial divisors, nor the strictly superior divisors. The element $$35=7*5$$ has a diagonal covering the product $$35$$. A $$-1:1$$ diagonal covers the two complementary central divisors $$(d_{c1};d_{c2})=(7;5)$$. We omit the diagonals of the trivial divisors. There are no other diagonals because $$5$$ and $$7$$ are prime numbers, we are not placing trivial divisors, nor strictly superior divisors. In this case the only strictly superior divisor is $$35$$, which is also trivial. This is because $$35$$ is a semiprime that has $$4$$ divisors. The element $$48=8*6$$ has four diagonals covering the product $$48$$. A $$-1:1$$ diagonal covers the two complementary central divisors $$(d_{c1};d_{c2})=(8;6)$$. Another diagonal in pink of divisor $$4$$ of factor $$8$$ which is the same in green as divisor $$3$$ of factor $$6$$. Another diagonal in green of divisor $$2$$ of factor $$6$$. Another diagonal in pink of divisor $$2$$ of factor $$8$$. There are no other diagonals because we are not placing the trivial divisors, nor the strictly superior divisors. The element $$63=9*7$$ has two diagonals covering the product $$63$$. A $$-1:1$$ diagonal covers the two complementary central divisors $$(d_{c1};d_{c2})=(9;7)$$. Another diagonal in pink of the divisor $$3$$ of the factor $$9$$. There are no other diagonals because $$7$$ is a prime number, we are not placing trivial divisors, nor strictly superior divisors. And so on. Notice that there is always a complementarity between the inferior divisors and the superior ones, and vice versa. This complementarity is always hyperbolic. Because of the hyperbolic complementarity, each product formed by the $$-1:1$$ diagonals in orange has the smallest possible difference between the pairs of complementary divisors. The smallest difference of the complementary divisors of an integer is always the difference of the divisors of the central pair. In our triangular sieve C002216.6 the difference between the divisors of the central pair of complementary divisors is the distance between the verticals of the green and pink framework. All other diagonals with slope less than $$-1:1$$ show us the strictly inferior divisors of each product $$x=y(y \pm 2)$$. With $$b_{dm}$$ being the distance between the verticals of the green and pink frameworks, then all products on the X-axis will be in the form $$x=y(y \pm b_{dm})$$. We will do this analysis for $$b_{dm}=2k$$ even and $$b_{dm}=2k+1$$ odd. Conclusion: For any quadratic CG in the form of $$x=y(y \pm b_{dm})$$, then this means that for any $$y$$ the product $$x$$ will always be $$x=d_{c1}*d_{c2}=y(y \pm b_{dm})$$ such that $$b_{dm}=d_{c2}-d_{c1}=$$ https://oeis.org/A033677 $$-$$ https://oeis.org/A033676 $$=$$ https://oeis.org/A056737. This is the smallest possible difference between the complementary divisors. Semiprimes in the form of $$x=y(y \pm b_{dm})$$ only appear when $$y=prime1$$ and $$y \pm b_{dm}=prime2$$. Attached Thumbnails
 2022-11-24, 13:57 #5 Charles Kusniec     Aug 2020 Guarujá - Brasil 107 Posts The divisors clock. C002400 The divisors clock To help in understanding the integers and their hyperbolic complementarity divisors, let us put each integer and their divisors on a clock analogous to what we use in modular arithmetic. The fundamental theorem of arithmetic assures us that every integer has a set of different divisors. Every integer has unique pairs of complementary positive divisors, a unique central pair of complementary positive divisors, and a unique trivial pair of complementary positive divisors. Since we multiply complementary divisors always in pairs to result in the integer, then it is possible to make for each integer a distribution clock of the divisors. In these clocks, we will assign the top with $$0$$, which we will never multiply by another finite number to generate a positive or negative integer. Diametrically opposite to the $$0$$ we have at the bottom the square root of the integer $$x$$ in question. We multiply the square root by itself to result back in the original integer. We will distribute the inferior divisors $$d_1≤√x$$ sequentially from top to bottom on the right side of the clock. Thus, the closest divisor to the $$0$$ on the right side will always be the trivial divisor $$d_{t1}=1$$. The bottom-most divisor $$≤√x$$ is the smallest central complementary divisor $$d_{c1}≤√x$$ https://oeis.org/A033676. We will distribute the superior divisors $$d_2≥√x$$ sequentially from bottom to top on the left side of the clock. Thus, the closest divisor to the $$0$$ on the left side will always be the trivial divisor $$d_{t2}=x$$. The bottom-most divisor $$≥√x$$ is the largest central complementary divisor $$d_{c2}≥√x$$ https://oeis.org/A033677. The pair of complementary positive divisors at the bottom-most is the central pair of complementary positive divisors $$(d_{c1};d_{c2})$$ such as $$x=d_{c1}*d_{c2}$$, where $$x≥d_{c2}≥√x≥d_{c1}≥1$$. The pair of complementary positive divisors at upper-most is the trivial pair of complementary positive divisors $$(d_{t1};d_{t2})$$ such as $$x=d_{t1}*d_{t2}$$, where $$d_{t2}=x$$ and $$d_{t1}=1$$. See the first 12 integers as an example: Figure C002400. The divisors clock for $$1≤x≤12$$, where $$b_{dm}=d_{c2}-d_{c1}=$$https://oeis.org/A056737. We multiply the horizontals and get the value of the integer number. The top always has the number 0, so we never multiply it. The bottom has the square root of the positive integer. So, we multiply it by itself. The integer 1 has no horizontals. We only multiply it by itself (in the bottom). The prime numbers have only one horizontal multiplication with 2 distinct integer divisors. The primes are the only integers with the pair of distinct central divisors is the same pair of trivial divisors. The central pair of complementary divisors of any square number is the bottom point. They are the only integers with the central pair of complementary divisors in a point. The square of prime numbers has two multiplications: one multiplication between the trivial pair of distinct divisors and another multiplication of the prime by itself (the bottom point). All composite numbers that are not square of prime numbers have more than one horizontal (two distinct pairs of complementary distinct divisors). From this form of organization, we can create a table with all the possible multiplications in pairs of complementary divisors in arrays. These multiplications are the same that appear in the TMT - Triangular Multiplication Table in the study [xxx]. The table is based on the same equality: https://oeis.org/A161906 * https://oeis.org/A340791 = https://oeis.org/A340792 https://oeis.org/A319135 * https://oeis.org/A161908 = https://oeis.org/A340792 The number of horizontals H[n] is half the number of divisors d[n]: $$H[n]=floor(d[n]/2)=floor(($$https://oeis.org/A000005$$)/2)$$ Although these clocks provide us with valuable information about the divisors of each integer, their linear distribution around each circle is disproportionate. This is because in these clocks the divisor's positions always appear symmetrically. In real value, there is a hyperbolic symmetry between them. Hyperbolic symmetry occurs only using the transverse axis as reference. Attached Thumbnails
 2022-11-28, 16:52 #6 Charles Kusniec     Aug 2020 Guarujá - Brasil 107 Posts The tables of the positive semiprime numbers in the form of a x=(h^2-k^2) or x=y(y±2k) number C002218 to C002224 The tables of the positive semiprime numbers in the form of a x=(h^2-k^2)=(square sequence minus square number) or x=y(y±2k) number The understanding of the mechanism that occurs in Figure C002216.6 above provides us with a lot of information and hints as to how and why semiprime numbers appear. Specific to the study of the semiprimes, Figure C002216.6 shows us the smallest possible odd divisors at each point with odd values, except the trivial divisors $$(1;odd)$$. For example:At the point with value $$63=(7;9)$$ we have the line of the positive odd smallest divisor greater than $$1$$. This is the divisor $$3$$ in pink. The trivial green divisors 1 and 7 of the value 7 are omitted, as well as the trivial pink divisors 1 and 9 of the value 9. At the point with value $$120=(10;12)$$ also, the most inclined line that appears is the divisor 2. It reflects the positive odd smallest divisor greater than 1 of either 10 or 12. And so on. The only points that have no pink or green divisor lines, other than the two trivial divisors, are the points that have semiprime value $$SP=(Prime1;Prime2)$$. This leads us to investigate the operation of the smallest positive prime divisors $$d2$$ https://oeis.org/A020639 of the elements from the quadratic composite generator in the form of $$x=y(y+2k)$$. To do this study, we construct the tables C002218 up to C002224. These tables will be referenced in other threads such as the Goldbach conjecture. This is because to prove Goldbach it is enough to prove that semiprimes exist in all forms $$x=y(y \pm 2k)$$. As it seems to me that these tables get too heavy here, I am putting them in external links. If there is interest I will put them here. C002218 All the positive semiprime numbers. https://www.facebook.com/groups/snyp...8131561177121/ C002219 The positive semiprime numbers in the form of $$h^2-0$$ or $$y(y+0)$$ number. https://www.facebook.com/groups/snyp...8130287843915/ C002220 The positive semiprime numbers in the form of $$h^2-1$$ or $$y(y+2)$$ number. https://www.facebook.com/groups/snyp...8132211177056/ C002221 The positive semiprime numbers in the form of $$h^2-4$$ or $$y(y+4)$$ number. https://www.facebook.com/groups/snyp...8132844510326/ C002222 The positive semiprime numbers in the form of $$h^2-9$$ or $$y(y+6)$$ number. https://www.facebook.com/groups/snyp...38133817843562 C002223 The positive semiprime numbers in the form of $$h^2-16$$ or $$y(y+8)$$ number. https://www.facebook.com/groups/snyp...8134497843494/ C002224 The positive semiprime numbers in the form of $$h^2-25$$ or $$y(y+10)$$ number. https://www.facebook.com/groups/snyp...8135151176762/ See below for explanations of the semiprimes table columns for C002219 to C002224. Just to have one table as reference here, please see in the end the table C002220. Column 1: The index $$y$$. The Tally counting. Column 2: We have the quadratic CG elements in the form $$x=r^2-k^2=square-k^2$$. We calculate each element in its hyperbolic form $$x=y(y+2k)$$. This shows that the index $$y$$ is always a divisor of each element. $$column2=x=y(y+2k)=column1*(column1+2k)$$ Column 3: It is the square root of the quadratic CG element. $$column3=√x=√(y(y+2k))=√column2$$ Column 4: It is the round of the square root of the quadratic CG element. $$column4=round(√x)=round(√(y(y+2k) ))=round(√column2)$$ The reason we have columns 3 and 4 is because column 4 with the constant k can produce column 2. Particularly, also the semiprimes of columns 9 and 10 (see Where is the weakness of this reasoning?: $x=y(y+2k)$ $x=y^2+2ky$ $x=y^2+2ky+k^2-k^2$ $x=(y+k)^2-k^2$ $x=((y+k)-k)((y+k)+k)$ Let´s be $$r=y+k$$ $x=r^2-k^2$ $x=(r-k)(r+k)$ Because $$r=y+k$$, then: $r^2=(y+k)^2$ $r^2=y^2+2ky+k^2$ $r^2=y(y+2k)+k^2$ $r^2=x+k^2$ Because $$k=constant$$, for large $$y$$ we can say: $r=√(x+k^2 )≅√x=√(y(y+2k))$ And, $r=round(√x)=round(√(y(y+2k) ))$ Then, $x=(r-k)(r+k)$ $x=(round(√x)-k)(round(√x)+k)$ When $$(round(√x)-k)$$ is a prime record it will be $$d2$$ (column 6 or 9), then $$(round(√x)+k)$$ will also be a prime record and it will be $$d3$$ (column 7 or 10). Consequently, $$x$$ is a semiprime. Column 5: It is the 1st smallest positive divisor. It is the smallest trivial divisor. It is always 1: https://oeis.org/A000012. We will denote it as $$d1=d_{t1}=1$$. Column 6: It is the 2nd smallest positive divisor. It is always a prime number: https://oeis.org/A020639. We will denote it as $$d2=prime2$$. The records occur when $$y=prime$$. In such cases, $$d2=y$$. Column 7: It is the 3rd smallest positive divisor. It is either a prime or a square of primes (square of $$d2$$): https://oeis.org/A340768. We will denote it as $$d3=prime3$$, or $$d3=(d2)^2=(prime2)^2$$ If the 3rd smallest divisor of the n-th composite number https://oeis.org/A340768 is not a prime number, then necessarily it will be the square of a prime number that is necessarily the 2nd smallest divisor of that composite https://oeis.org/A020639. The records occur when $$y+2k=prime$$. In such cases, $$d3=y+2k$$. The records in columns 6 and 7 are the prime numbers of the semiprimes. Records of squares occur when $$y$$ is a square of prime?? Column 8: In column 8 we show the products: $$d2*d3=$$2nd smallest positive divisor$$*$$3rd smallest positive divisor $$d2*d3=column 6*column 7$$ When this product is a record, its value is a semiprime in the form of $$x=y(y+2k)=d2*d3=P2*P3$$. Because $$x=d2*d3$$ and $$d3$$ is a prime, the composite will have 4 divisors and will therefore be a semiprime number. Column 9: Column 9 shows us the smallest prime number of the semiprimes. They appear when $$d2=y$$. In this case: $$d2$$ is a record. $$d3$$ is a record. $$d2*d3=x=y(y+2k)$$ is a record. If $$d2≠y$$ then d$$2*d3≠x$$ and is not a record, nor a semiprime. Column 10: Column 10 shows us the largest prime number of the semiprimes. They appear when $$d3=y+2k$$. In this case: $$d2$$ is a record. $$d3$$ is a record. $$d2*d3=x=y(y+2k)$$ is a record. If $$d3≠y+2k$$ then $$d2*d3≠x$$ and is not a record, nor a semiprime. Column 11: In column 11 we have the semiprimes. $$d2=y$$, $$d2$$ is a record, $$d2$$ is a prime. $$d3=y+2k$$, $$d3$$ is a record, $$d3$$ is a prime. $$d2*d3=x=y(y+2k)$$, $$d2*d3$$ is a record, $$d2*d3$$ is a semiprime. Column 12: It shows us $$d´2$$ the complementary divisor of $$d2$$. We will denote it as: $$d´2=x/d2=y(y+2k)/d2$$ It is the largest proper divisor of $$x=y(y+2k)=d2*d´2$$. For each $$x=y(y+2k)$$, if $$d´2$$ is a composite, then it is larger than the previous $$d´2=prime$$. In such a case $$d´2>d3>d2$$. Record low peak only when $$d´2$$ is a prime number, where $$d´2=d3>d2$$. The CG $$x=y(y+2k)$$ is a semiprime if $$d´2=d3>d2$$. Column 13: It shows us $$d´3$$ the complementary divisor of $$d3$$. We will denote it as: $$d´3=x/d3=y(y+2k)/d3$$ It is the second largest proper divisor of $$x=y(y+2k)=d´3*d3$$. For each $$x=y(y+2k)$$, if $$d´3$$ is a composite, then it is larger than the previous $$d´3=prime$$. In such a case $$d´3>d3>d2$$. ($$d´3>d3$$) Record low peak only when $$d´3$$ is a prime number, where $$d´3=d2d3$$ only for the $$x$$ non-semiprime. If $$x=semiprime$$ then $$d'30$$. If $$b$$ is an odd number $$2k-1$$, then $$x=y(y±(2k-1))$$ is in the form of (oblong sequence minus oblong number). Only with these kinds of composite numbers can we find prime numbers from the “soviet CCCP” numbers https://www.mersenneforum.org/showthread.php?t=27328. If $$b$$ is an even number $$2k$$, then $$x=y(y±2k)$$ is in the form of (square sequence minus square number). Only with these kinds of composite numbers can we find semiprime numbers. If $$x=y(y±2k)$$ is a semiprime number, then $$d2=y$$ $$d´2=(y±2k)$$ $$b2=d´2-d2=±2k$$ Column 15: The column 15 shows us the difference between the complementary divisor of $$d3$$ and the divisor $$d3$$. We will denote it as: $$b3=d´3-d3=x/d3-d3=y(y±2k)/d3-d3$$. Considering only positive divisors, for all integers $$x$$, we have $$b3<0$$ for all number with 3 or 4 divisors. If $$x=y(y±2k)$$ is a semiprime number then $$b3<0$$. If $$x=y(y±2k)$$ is a semiprime number then $$b2=-b3$$ or $$d´2-d2=-(d´3-d3)$$. Attached Thumbnails

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