Pattern Recursion as an Approach to Goldbach's Conjecture

In other pages we looked at pattern recursion. In this page we will review how pattern recursion is so prevalent, with every Goldbach sequence sharing patterns with many other Goldbach sequences, that recursion suggests that Goldbach's Conjecture is probably true. We have not yet succeeding at proving Goldbach. We are just examining a possible approach.

Last Update April 2011

  Introduction: Reviewing Pattern Recursion
recursion example Here we see 4 terms of three Goldbach sequences starting from the symmetry point. The prime factors smaller than 11 are listed at the bottom. The pattern is the same for each sequence. This is the concept of pattern recursion. Multiple sets of sequences will have same pattern of critical primes. The challenge, then, is to determine whether numbers which share the same prime factor pattern, have useful relationships that can be used to prove, or disprove, Goldbach's Conjecture.
Definitions:
  • Goldbach Sequence: a series of number pairs that all add to a given even number, G
  • Goldbach Number (G): an even generating a Goldbach Sequence
  • Symmetry point (S): the point around which the critical prime pattern is symmetrical: S = 1/2*G
  • Critical Primes: All primes being considered in the Goldbach Sequence. Typically this would be all primes smaller than sqrt(G).
  • Number of Primes (n): how many critical primes are being considered.
  • Hole: a number pair in a Goldbach sequence not containing any critical prime as a factor. When both numbers are smaller than the largest critical prime squared, the hole with be a prime pair.
  • P(p): the product of all the critical primes (eg: for primes 7 and smaller, P(p) = 2*3*5*7 = 210. P(p) is the pattern length for all Goldbach sequences with those critical primes.

Part 1: Complete Pattern Recurrence
For any set of critical primes, every Goldbach sequence has pattern recurrence with other Goldbach sequences.  Each Goldbach sequence will have 2^n recurrents smaller than P(p) (including itself) where n = the count of critical primes which are not factors of G.
Every Goldbach sequence will contain holes. Thus complete pattern recurrence may provide a means to develop a solution.  The distribution of recurrents may provide a method to determine a relatively smooth distribution of holes, consequently a relatively smooth distribution of primes, as suggested by Reimann&';s Hypothesis.

Method for Identifying Goldbach sequences showing pattern recurrence with a given Goldbach Sequence

  1. Identify all the critical primes
  2. Show all possible ways to break the critical prime list into two groups.  Primes which are factors of G must be shown in both groups.
  3. Multiply together the primes in each group.
  4. Find two integers which are multiples of those two products which add to G
  5. Switch the sign of one of the numbers in each pair and add. 
  6. This will give you more numbers that have the same factor sequence as the number you started with.

This method may be written as the formulas

G = 2*S = a*Pa + b* Pb
  • Pa & pb = partitioning of critical primes
  • a,b = arbitrary integers to make partition add to G
S1eq = (a*Pa - b*Pb) / 2
  • S1eq = the symmetry point of equivalent Goldbach sequences
Example 1: For ease, we pick G = 1 and Critical primes = 2,3,5,and 7.
a1 Factors
2,3,5,7,11
2,3,5,7
2,3,5,11
2,3,7,11
2,5,7,11
2,3,5
2,3,7
2,3,11
a2 factors
2
2,11
2,7
2,5
2,3
2,7,11
2,5,11
2,5,7
*
a1   0 -2308 420 1892 660 1652 462 1850 770 1542 1080 1232 -1428 -880 1122 1190
a2   2 2310 -418 -1890 -658 -1650 -460 -1848 -768 -1540 -1078 -1230 1430 882 -1120 -1188
*
switched * 4618 838 3782 1318 3302 922 3698 1538 3082 2158 2462 2858 1762 2242 2378
G's 2 4618 838 3782 1318 3302 922 3698 1538 3082 2158 2462 548 1762 2242 2378
S1eq 1 2309 419 1891 659 1651 461 1849 769 1541 1079 1231 274 881 1121 1189
  • We arranged all the Primes from 2 to 11 in the 8 possible combinations where both groups contain 2 (since G is always even)
  • We found integer multiples of these factor combinations that add together to make G=2. Due to pattern symmetry , there are two numbers for each combination. One with the pattern approach zero, one with the pattern approaching P(p).
  • We Switched the sign and added. This gave the Goldbach number where the pattern recurs.
  • We divided by 2 to find the symmetry points for each sequence.

Observations:

  • The higher the number, the more critical primes, the larger the number of evens with recurrent factor patterns. [2^n]
  • However, the larger the number, the wider the spread of the recurrent numbers [P(p)]
  • To Prove: If for all large numbers, we can show the existence of a smaller number with pattern recursion, then Goldbach's Conjecture is proved.

Parts to this discussion:

  1. Reviewing Pattern Recurrence & Introductory Definitions
  2. Complete Pattern Recurrence
  3. Sufficient Pattern Recurrence on Scale
  4. Incomplete Pattern Recurrence on Scale
  5. Summation

Related pages at this site

 

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Part 2: Scaled Pattern Recurrence: Sufficient Partial Pattern (SPR)

scaled by 3

Every Goldbach sequence may be viewed at multiple scales, or zooms, Dn. Each scaled sequence will have pattern recursion with other Goldbach sequences at normal zoom, D1. This is analogous to symmetry of scale found in fractal geometry. Each scaled sequence will receive sufficient pattern information from the D1 sequences it shares recursion with. This will determine where its holes must occur. If a hole occurs at a specific point in the D1 recurrent, then a hole must occur in the same location in the Dn scaled sequence.

To the left, at the top, we see G=2 scaled by step size D = 3, with its critical factor pattern below. Below that, we see three recurrents with step size D = 1 which insert 3 back into the factor pattern. Factor pattern holes are marked in pink.

Each place a hole appears in one of the D1 recurrents, a hole also appears in the D=3 scaling of G = 2 at the top. The S1 sequences sufficiently masks the S3 sequence. That means, where ever a hole appears in any of the S1 sequences, a hole must appear in the S3 sequence also.

Definitions

  • Dn: the size of the increment in a scaled Goldbach sequence.
  • Sn: the symmetry point of the scaled Goldbach sequence
  • D1 & S1: represent a Goldbach sequence with an increment of 1. (no scaling)

Method for finding Sufficient Recurrents
Formula:  S1 = (P(p)*a/Dp + Sn)/Dn

  • Dn: scale of zoom - a counting number
  • Sn: Symmetry point of scaled Goldbach sequence
  • P(p): product of all critical primes
  • DP: multiple of all critical factors of d (eg: if d=24, critical factors are 2&3, D=2*3=6)
  • a: an arbitrary integer to ensure an inter result to the division
  • any combination of a, D, and d with Sn and P(p) that produces and integer S1 in this formula generates a new symmetry point, S1, which has scaled pattern recursion which sufficiently masks the Sn sequence.

Observations:

  • To Prove Goldbach's Conjecture: show that for any Sn, there is a smaller S1, sharing pattern recursion with Sn, for reasonable choices of Dn.
  • Reasonable choices of Dn appear to be on the scale of all counting numbers smaller than sqrt(Sn).
  • For every d, there should be 2D recurrents in range that fit the formula above
    • Scale: Dn Prime Factors: D Recurrents
      2 2 2*2 = 4
      3 3 2*3 = 6
      4 2 2*2 = 4
      5 5 2*5 = 10
      8 2 2*2 = 4
      9 3 2*3 = 6
      12 2,3 2*6 = 12
  • For each prime factor product, DP, two S1 recurrents will exist in each P(p) / Dn portion of the range.
  • Every S1 recurrent will have 2^n D1 recurrents of its own (see part 1). The list may include redundancies.
  • We still need to determine how many non-redundant recurrents there are,or how many will be small numbers to use this method to solve Goldbach.
See demo in PDF
 

Part 3: Scaled Pattern: Incomplete Pattern Recurrence (IPR)

Easier to work with, but not as solid for proofs, is the recognition that each Goldbach sequence has incomplete recurrence with a set of scaled Goldbach sequences. Places where holes occur in the scaled sequences, holes have high odds of appearing in the original sequence. The scaling constant will mask some of the holes in the S1 sequences. Thus, the hole pattern is incompletely received from the scaled sequences.

Method for finding Incomplete Scaled Recurrents

This formula will determine a set of numbers providing incomplete pattern recurrence for a starting number, S1.

Sn = P(p)* a/DP+ S1*Dn

Again, the arbitrary integer a is picked to ensure that Sn will be an integer.

 

 
 

Summation and Observations

The observations above make Goldbach's Conjecture probably true. They also suggest that the distribution of primes is rather smooth as predicted by Reimann's Hypothesis.

  • Every Goldbach sequence will have 2^n complete recurrents (including itself)
  • Every Goldbach sequence will have at least n sufficient scaled recurrents
    • each sufficient scaled recurrent will have 2^n complete recurrents
  • Every Goldbach sequence will have at least n incomplete recurrents
    • each incomplete recurrent will have 2^n complete recurrents
  • Every sufficient recurrent may be associated with more incomplete recurrents

This expanding pattern recurrence with symmetry of scale would seem to demand an smooth distribution of primes, as well as require that each large number has sufficient recurrence with a smaller number. If this is correct then Goldbach's Conjecture must be true, and Reimann's hypothesis is probably true. However, to determine whether pattern recurrence may solve Goldbach's conjecture will require some more work.

 
 

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