Invariant relations between binary Goldbach s decompositions numbers coded in a 4 letters language

Transcription

1 Ivariat relatios betwee biary Goldbach s decompositios umbers coded i a letters laguage Deise Vella-Chemla To cite this versio: Deise Vella-Chemla. Ivariat relatios betwee biary Goldbach s decompositios umbers coded i a letters laguage <hal > HAL Id: hal Submitted o Ja 015 HAL is a multi-discipliary ope access archive for the deposit ad dissemiatio of scietific research documets, whether they are published or ot. The documets may come from teachig ad research istitutios i Frace or abroad, or from public or private research ceters. L archive ouverte pluridiscipliaire HAL, est destiée au dépôt et à la diffusio de documets scietifiques de iveau recherche, publiés ou o, émaat des établissemets d eseigemet et de recherche fraçais ou étragers, des laboratoires publics ou privés.

2 Ivariat relatios betwee biary Goldbach s decompositios umbers coded i a letters laguage Deise Vella-Chemla October 01 1 Itroductio Goldbach s cojecture states that each eve iteger except is the sum of two prime umbers. This ote presets a set of ivariat relatios ucovered betwee some Goldbach s decompositios umbers, the decompositios beig coded i a letters laguage. We will try to use those ivariat relatios to attempt to obtai a recurrece demostratio of Goldbach s biary cojecture. I the followig, oe is iterested i decompositios of a eve umber as a sum of two odd itegers p+q with 3 p /, / q 3 ad p q. We call p a s first rage sommat ad q a s secod rage sommat. Notatios : We will desigate by : a : a s decompositio of the form p + q with p ad q primes ; b : a s decompositio of the form p + q with p compoud ad q prime ; c : a s decompositio of the form p + q with p prime ad q compoud ; d : a s decompositio of the form p + q with p ad q compoud umbers. Example : l 0 a c c b a c d a c The mai array ( We desigate by T = (L, C = l,m the array cotaiig l,m elemets that are oe of a, b, c, d letters. belogs to the set of eve itegers greater tha or equal to 6. m, belogig to the set of odd itegers greater tha or equal to 3, is a elemet of list of first rage sommats. Let us cosider g fuctio defied by : g : N N + 1 x x + 1 g(6 = 3, g(8 = 3, g(10 = 5, g(1 = 5, g(1 = 7, g(16 = 7, etc. 1

3 g( fuctio defies the greatest of first rage sommats. As we oly cosider decompositios of the form p+q where p q, i T will oly appear letters l,m such that m +1 i such a way that T array first letters are : l 6,3, l 8,3, l 10,3, l 10,5, l 1,3, l 1,5, l 1,3, l 1,5, l 1,7, etc. Here are first lies of array T. Remarks : C L 6 a 8 a 10 a a 1 c a 1 a c a 16 a a c 18 c a a d 0 a c a b a a c b a c a a d a 6 a c a b c a 8 c a c b a c 30 c c a d a a d 3 a c c b c a b 3 a a c d a c b a 36 c a a d c a d a... Figure 1 : words of eve umbers betwee 6 ad 36 1 words o array s diagoals called diagoal words have their letters either i A ab = {a, b} alphabet or i A cd = {c, d} alphabet. a diagoal word codes decompositios that have the same secod rage sommat. For istace, o Figure, letters aaabaa of the diagoal that begis at letter l 6,3 = a code decompositios 3 + 3, 5 + 3, 7 + 3, 9 + 3, ad let us desigate by l the lie whose elemets are l,m. Lie l cotais elemets. beig fixed, let us call C,3 the colum formed by l k,3 for 6 k. I this colum C,3, let us distiguish two parts, the top part ad the bottom part of the colum. + Let us call H,3 colum s top part, i.e. set of l k,3 where 6 k. + Let us call B,3 colum s bottom part, i.e. set of l k,3 where < k.

4 H 3,3 B 3,3 Z a = 5 Z c = Y a = 5 Y c = 3 6 : 8 : 10 : 1 : 1 : 16 : 18 : 0 : : : 6 : 8 : 30 : 3 : 3 : a a a c a a a c a a a c c a a d a c a b a a c b a c a a d a a c a b c a c a c b a c c c a d a a d a c c b c a b a a c d a c b a Figure : = 3 X a =, X b = 1, X c =, X d = 1 To better uderstad coutigs i ext sectio, we will use projectio P of lie o bottom part of first colum B,3 that associates letters at both extremities of a diagoal (it associates to the letter that codes p + q decompositio the letter that codes 3 + q decompositio. If we cosider applicatio proj such that proj(a = proj(b = a ad proj(c = proj(d = c the, sice 3 is prime, proj(l,k+1 = l k+,3. We ca also uderstad the effect of this projectio (that preserves secod rage sommat by aalyzig decompositios : if p + q is coded by a a or a b letter, it correspods to two possible cases i which q is prime, ad so 3 + q decompositio, cotaiig two prime umbers, will be coded by a a letter ; if p + q is coded by a c or a d letter, it correspods to two possible cases i which q is compoud, ad so 3 + q decompositio, of the form prime + compoud will be coded by a c letter. We will also use i ext sectio projectio P of lie o top part of the first colum H,3 that associates to the letter that codes p + q decompositio the letter that codes 3 + p decompositio (let us ote that first rage sommat becomes secod rage sommat ; let us aalyze how such a projectio will affect decompositios : if p + q is coded by a a or c letter, it correspods to the two possible cases i which p is prime, the 3 + p decompositio, cotaiig two prime umbers will be coded by a a letter ; if p + q is coded by a b or d letter, it correspods to the two possibl cases i which p is compoud, the 3 + p decompositio, of the form prime + compoud will be coded by a c letter. 3 Computatios 1 We ote : X a ( the umber of s decompositios of the form prime + prime ; X b ( the umber of s decompositios of the form compoud + prime ; X c ( the umber of s decompositios of the form prime + compoud ; X d ( the umber of s decompositios of the form compoud + compoud. X a (, X b (, X c (, X d ( variables are coutig logical assertios umbers. X a ( + X b ( + X c ( + X d ( = is the umber of elemets of lie. Example : = 3 : X a (3 = #{3 + 31, 5 + 9, , } = X b (3 = #{ } = 1. X c (3 = #{7 + 7, } = 3

5 X d (3 = #{9 + 5} = 1 Let Y a ( (resp. Y c ( beig the umber of a letters (resp. c that appear i B,3. We recall that there are oly a ad c letters i first colum because it cotais letters associated with decompositios of the form 3 + x ad because 3 is prime. Example : Y a (3 = #{3 + 17, , 3 + 3, 3 + 9, } = 5 Y c (3 = #{3 + 1, 3 + 5, 3 + 7} = 3 3 Because of P projectio that is a bijectio, ad because of a, b, c, d letters defiitios, Y a ( = X a ( + X b ( ad Y c ( = X c ( + X d (. Thus, trivially, Y a ( + Y c ( = X a ( + X b ( + X c ( + X d ( =. Example : Y a (3 = #{3 + 17, , 3 + 3, 3 + 9, } X a (3 = #{3 + 31, 5 + 9, , } X b (3 = #{ } Y c (3 = #{3 + 1, 3 + 5, 3 + 7} X c (3 = #{7 + 7, } X d (3 = #{9 + 5} Let Z a ( (resp. Z c ( beig the umber of a letters (resp. c that appear i H,3. Example : Z a (3 = #{3 + 3, 3 + 5, 3 + 7, , } = 5 Z c (3 = #{3 + 9, } = Z a ( + Z c ( =. Remidig idetified properties Y a ( = X a ( + X b ( (1 Y c ( = X c ( + X d ( ( Y a ( + Y c ( = X a ( + X b ( + X c ( + X d ( = Z a ( + Z c ( = (3 (

6 Let us add four ew properties to those oes : X a ( + X c ( = Z a ( + δ p ( (5 with δ p ( equal to 1 whe is the double of a prime umber ad equal to 0 otherwise. X b ( + X d ( = Z c ( + δ c imp ( (6 with δ odd compoud ( equal to 1 whe is the double of a compoud odd umber ad equal to 0 otherwise (whe there exists k such that = k (doubles of eve umbers or whe is the double of a prime umber. Z c ( Y a ( = Y c ( Z a ( δ k+ ( (7 with δ k+ ( equal to 1 whe is the double of a odd umber ad 0 otherwise. Z c ( Y a ( = X d ( X a ( δ odd compoud ( (8 Properties 1, ad 3 come simply from projectio P s defiitio. Properties 5 ad 6 come simply from projectio P s defiitio. Oe ca otice a certai redudacy betwee the 3 booleas that are itroduced here δ p (, δ odd compoud ( ad δ k+ (. Trivial logical assertio (δ p ( δ odd compoud ( δ k+ ( is always verified. Demostratios.1 Utilitaries Let us demostrate that if is a odd umber double (i.e. of the form k+, the = +1. (k + k Ideed, the left part of the equality is equal to = = k. (k + k The right part of the equality is equal to + 1 = + 1 = (k = k. Let us demostrate that if is a eve umber double (i.e. of the k, the =. k k = k 1 ad = k 1.. Properties 7 ad 8..1 Property 7 Property 7 eociates that Z c ( Y a ( = X d ( X a ( δ odd compoud ( with δ odd compoud ( that equals 1 if is the double of a odd compoud iteger ad equals 0 otherwise. By defiitio, Z c ( couts the umber of decompositios of the form α = 3 + compoud with compoud strictly lesser tha / (let us call E this decompositios set. So Z c ( couts also the umber of s decompositios of the form β = compoud + y by bijectio of secod rage sommat of α decompositio o first rage sommat of β decompositio. But the umber of decompositios of the form compoud + y is equal to X b ( + X d ( by defiitio of those variables. By defiitio, Y a ( couts the umber of decompositios of the form γ = 3 + prime with prime strictly greater tha / (let us call F this decompositios set. So Y a ( couts also the umber of s decompositios of the form η = x + prime by bijectio of secod rage sommat of γ decompositio o secod rage sommat of η decompositio. But the umber of decompositios of the form x + prime is equal to X b ( + X a ( by defiitio of those variables. s decompositios of the form compoud + prime are at the same time i E ad i F. By computig Z c ( Y a (, we also obtai the value of X d ( X a ( by defiitio of what variables Y a (, Z c (, X d ( et X a ( cout. 5

7 .. Property 8 Let us demostrate that Z c ( Y a ( = Y c ( Z a ( δ k+ ( with δ k+ ( that equals 1 if is the double of a odd umber ( k 3, = k + ad 0 otherwise. We use a recurrece reasoig : i We iitialize recurreces accordig to the 3 types of umbers that must be studied : eve doubles de pairs (k, odd doubles (k + with the odd beig prime or compoud. Proprety 8 is true for = 1, 16 ad 18. Let us provide variables values i a array for those 3 eve umbers : Z c ( Y a ( Y c ( Z a ( δ k ii Property 8 is equivalet to Z a ( + Z c ( + δ k+ ( = Y a ( + Y c (. Four cases must be cosidered : two cases wherei is a odd s double (prime or compoud ad + is a eve s double ad two cases wherei is a eve s double ad + is a odd s double (prime or compoud. iia eve s double ad + prime s double : δ p δ odd compoud δ k We make the hypothesis that property 8 is true for, Z a ( + Z c ( + δ k+ ( = Y a ( + Y c ( (H Let us demostrate that it is true for +, Z a ( + + Z c ( + + δ k+ ( + = Y a ( + + Y c ( + (Ccl We have Z a ( + = Z a ( ad Z c ( + = Z c (. Let us remai property 3 cocerig Y s : Y a ( + Y c ( = (3 I (Ccl, we ca, par by our recurrece hypothesis ad by property (3, replace the left member of the equality by Z a ( + Z c ( + 1 ad the by Y a ( + Y c ( + 1 (by (H ad the by + 1 (by (3 that is equal to. But i (Ccl, we ca also replace the right member of the equality by because of property (3. We have for + a equality betwee left ad right sides, i.e. property 8 is verified by +. From the hypothesis that property is true for, we have iferred property is true for +. iib eve s double ad + odd compoud s double : δ p δ odd compoud δ k

8 Let us make the hypothesis that property 8 is true for, Z a ( + Z c ( + δ k+ ( = Y a ( + Y c ( (H Let us demostrate it is true for +, Z a ( + + Z c ( + + δ k+ ( + = Y a ( + + Y c ( + (Ccl We have Z a ( + = Z a ( ad Z c ( + = Z c (. We have Y a ( + = Y a ( + 1 ad Y c ( + = Y c (. Let us recall property 3 cocerig Y s : Y a ( + Y c ( = (3 I (Ccl, oe ca, by recurrece hypothesis ad by property 3, replace equality s left member by Z a ( + Z c ( + 1, the by Y a ( + Y c ( + 1 (because of (H, ad the by + 1 (because of (3 that is equal to. But i (Ccl, oe ca also replace equality s right member by because of Y a ( s ad Y c ( s evolutios. There is also for + equality betwee left ad right members of the equatio, i.e. property 8 is verified by +. From the hypothesis that property is true for, we deduced property is true for +. iic prime s double ad + eve s double : δ p δ odd compoud δ k Let us make the hypothesis that property 8 is true for, Z a ( + Z c ( + δ k+ ( = Y a ( + Y c ( (H Let us demostrate it is true for +, Z a ( + + Z c ( + + δ k+ ( + = Y a ( + + Y c ( + (Ccl We have Z a ( + = Z a ( + 1 et Z c ( + = Z c (. Let us recall property 3 cocerig Y s : Y a ( + Y c ( = (3 I (Ccl, oe ca, by recurrece hypothesis ad by property 3, replace equality s left member by Z a ( + Z c (+1 the by Y a (+Y c ( (because of (H, ad the by (because of (3 that is equal to. But i (Ccl, oe ca also replace equality s right member by because of property (3. There is also for + equality betwee left ad right members of the equatio, i.e. property 8 is verified by +. From the hypothesis that property is true for, we deduced property is true for +. 7

9 iid odd compoud s double ad + eve double : δ p δ odd compoud δ k Let us make the hypothesis that property 8 is true for, Z a ( + Z c ( + δ k+ ( = Y a ( + Y c ( (H Let us demostrate it is true for +, Z a ( + + Z c ( + + δ k+ ( + = Y a ( + + Y c ( + (Ccl We have Z a ( + = Z a ( et Z c ( + = Z c ( + 1. Let us recall property 3 cocerig Y s : Y a ( + Y c ( = (3 I (Ccl, oe ca, by recurrece hypothesis ad by property 3, replace equality s left member by Z a ( + Z c ( + 1 puis par Y a ( + Y c ( (because of (H the by (because of (3 that is equal to. But i (Ccl, oe ca also replace equality s right member by because of property (3. There is also for + equality betwee left ad right members of the equatio, i.e. property 8 is verified by +. From the hypothesis that property is true for, we deduced property is true for +. 5 Variables evolutio 5.1 Coutig decompositios of the forme 3 + p (called bottom-oes 1 Z a (+ = Z a ( ad Z c (+ = Z c ( whe is a eve double. Ideed, Z a ( (resp. Z c ( couts umber of primes (resp. odd compoud umbers strictly lesser tha ad there are as may as primes (resp. odd compoud umbers lesser tha a eve umber tha there are primes (resp. odd compoud umbers lesser tha its immediate successor. Z a ( + = Z a ( + 1 ad Z c ( + = Z c ( whe is a prime double ; 3 Z a ( + = Z a ( ad Z c ( + = Z c ( + 1 whe is a odd compoud umber s double. 5. Coutig decompositios of the form 3 + q (called top-oes I the case i which + is a odd s double, we oly add a umber to H +,3 iterval ; if this umber 1 is prime (resp. compoud, Y a ( + = Y a ( + 1 (resp. Y c ( + = Y c ( + 1. I the case i which + is a eve s double, we oly add a umber to the top extremity of the iterval ad we take off a umber at the bottom extremity of the iterval. From this fact, cases have to be studied. Let us study how the decompositios set H +,3 evoluate. if 1 ad are both primes, we take off at the bottom ad we put o at the top of the iterval H +,3 two letters of same kid, so Y a ( + = Y a ( ; if 1 is prime ad is compoud the Y a( + = Y a ( + 1 ; if 1 is compoud ad is prime the Y a( + = Y a ( 1 ; if 1 ad are both compoud, we take off at the bottom ad we put o at the top of the iterval H +,3 two letters of same kid, the Y a ( + = Y a (. 8

10 6 Usig variables gaps We are goig to show that X a ( ca ever be equal to 0 for C, i.e. that every eve umber C ca be writte as a sum of two primes, i.e. verifies biary Goldbach s cojecture. Let us preset i more details what variables Z a (, Z c (, Y a ( et Y c ( represet. Z a ( couts with a maximum error of the umber of prime umbers that are lesser tha or equal to ; Z a ( = π + δ Za ( (9 with δ Za ( equal to if is a prime s double ad equal to 1 otherwise ; Z c ( couts with a maximum error of 1 the umber of odd compoud umbers that are lesser tha or equal to ; Z c ( = π + δ Zc ( (10 with δ Zc ( equal to 1 if is a prime s double ad equal to 0 otherwise ; Y a ( cout with a maximum error of 1 the umber of prime umbers that are betwee ad ; Y a ( = π( π + δ Ya ( (11 with δ Ya ( equal to 1, 0 or 1 (δ Ya ( is equal to 0 whe 1 ad / are both primes or both compoud, δ Ya ( is equal to 1 whe 1 is prime while / is compoud, ad at last, δ Ya ( is equal to 1 whe 1 is compoud while / is prime ; Y c ( couts with a maximum error of 1 the umber of odd compoud umbers that are betwee ad ; Y c ( = π( + π + δ Yc ( (1 with δ Yc ( equal to 1, 0 or 1. δ Yc ( s values will be provided i a detailled way i sectio demostratig property Demostratios of properties 9, 10, 11 ad Property 9 : ivariat relatio cocerig Z a ( i Iitialisatio : (9 is true for = 1, 16, 18 ad 0. Ideed, we have : Z a ( π δ Za ( ii Let us demostrate by recurrece that if (9 is true for, it is also true for +. Let us make the hypothesis that property 9 is true for, Z a ( = π + δ Za ( (H 9

11 Let us demostrate it is true for +, ( + Z a ( + = π + δ Za ( + (Ccl Let us distiguish cases, as i paragraph... iia eve s double, + prime s double. ( + I that case, Z a ( + = Z a (, π Z a ( + = π + 1 ad δ Za ( = 1. = Z a (( = π + δ Za ( ( + = π 1 + δ Za ( ( + = π 1 1 ( + = π + δ Za ( + with δ Za ( + = as expected sice i that case, + is a prime s double. iib eve s double, + odd compoud s double. ( + I that case, Z a ( + = Z a (, π = π ad δ Za ( = 1. Z a ( + = Z a (( = π + δ Za ( ( + = π 1 ( + = π + δ Za ( + with δ Za ( + = 1 as expected sice i that case, + is a odd compoud s double. iic prime s double, + eve s double. ( + I that case, Z a ( + = Z a ( + 1, π = π ad δ Za ( =. Z a ( + = Z a (( + 1 = π + δ Za ( + 1 ( + = π + 1 ( + = π 1 ( + = π + δ Za ( + with δ Za ( + = 1 as expected sice i that case, + is a eve s double. iid odd compoud s double, + eve s double. ( + I that case, Z a ( + = Z a (, π = π ad δ Za ( = 1. Z a ( + = Z a (( = π + δ Za ( ( + = π 1 ( + = π + δ Za ( + 10

12 with δ Za ( + = 1 as expected sice i that case, + is a eve s double Property 10 : ivariat relatio cocerig Z c ( i Iitialisatio : (10 is true for = 1, 16, 18 ad 0. We have : Z c ( π δ Zc ( ad we verify that (10 is true for those umbers. ii Let us demostrate by recurrece that if (10 is true for, it is also true for +. Let us make the hypothesis that property 10 is true for, Z c ( = π + δ Zc ( (H Let us demostrate it is also true for +, Let us study our cases agai. + Z c ( + = π iia eve s double, + prime s double. + Das ce cas, Z c ( + = Z c (, = I that case, Z c ( + +, π + δ Zc ( + (Ccl ( + = π + 1 et δ Zc ( = 0. = Z c ( = π + δ Zc ( + ( + = π δ Zc ( + ( + = π ( + = π + δ Zc ( + with δ Zc ( + = 1 as expected sice i that case + is a prime s double. iib eve s double, + odd compoud s double. + ( + I that case, Z c ( + = Z c (, =, π Z c ( + = π et δ Zc ( = 0. = Z c ( = π + δ Zc ( + ( + = π + δ Zc ( + ( + = π ( + = π + δ Zc ( + with δ Zc ( + = 0 as expected i that case sice + is a odd compoud s double. 11

13 iic prime s double, + eve s double. + I that case, Z c ( + = Z c (, = Z c ( + + 1, π ( + = π et δ Zc ( = 1. = Z c ( = π + δ Zc ( + ( + = π 1 + δ Zc ( + ( + = π ( + = π + δ Zc ( + with δ Zc ( + = 0 as expected sice i that case + is a eve s double. iid odd compoud s double, + eve s double. + ( + I that case, Z c ( + = Z c ( + 1, = + 1, π Z c ( + = Z c ( = π + δ Zc ( + + = 1 π + ( + = π = π = π ad δ Zc ( = δ Zc ( + δ Zc ( + with δ Zc ( + = 0 as expected sice i that case + is a eve s double Property 11 : ivariat relatio cocerig Y a ( The objective is to demostrate why the ivariat relatio cocerig Y a ( ad that is : Y a ( = π( π + δ Ya ( is always verified. I this aim, 16 cases are to be distiguished, accordig to primality characters of umbers :, +, 1 ad +1, those primality characters havig a ifluece o Y a (, π(, π ad δ Ya ( evolutios. i recurreces iitialisatios : for 5 cases amog the 16 to be evisagd, there is a cotradictio (we called them C1 to C5 i bottom of the array below, cotradictio comig for cases C1 to C from the fact that oe ca t have simmultaeously / ad ( + / that are both prime sice they are cosecutive itegers ; C5 is a cotradictio because if 1 ad + 1 ar twi primes, their father (the eve umber betwee them ca be divided by 3 ad so the half of this father, that is /, ca t be a prime umber too ; this is represeted by crosses i first, third ad fourth colums. For the remaiig 11 cases, we must iitialize recurreces. I the array below, we provide values that permit easily to verify that for small itegers, property 11 is systematically verified. 1

14 / ( + / is prime is prime is prime is prime Y a ( π( π(/ δ Ya ( 1 x x x x x x x x x x x x x x x x x C1 x x x x C x x x C3 x x x C x x C5 x x x ii Let us demostrate by recurrece that if (11 is true for, it is true for + too. Let us make the hypothesis that property 11 is true for, Y a ( = π( π + δ Ya ( (H Let us demostrate it is true for +, ( + Y a ( + = π( + π + δ Ya ( + (Ccl Let us study the 11 cases listed i last sectio. iia + 1 compoud, ( + / compoud, / prime, 1 prime ( + I that case, π( + = π(, π = π, δ Ya ( = 0 ad Y a ( + = Y a (. Y a ( + = Y a ( = π( π + δ Ya ( ( + = π( + π + δ Ya ( + sice δ Ya ( + = 0 i that case. It implies that (Ccl is verified. iib + 1 prime, ( + / compoud, / compoud, 1 compoud ( + I that case, π( + = π( + 1, π = π, δ Ya ( = 0 ad Y a ( + = Y a (. Y a ( + = Y a ( = π( π + δ Ya ( ( + = π( + 1 π + δ Ya ( + sice δ Ya ( + = 1 i that case. This implies that (Ccl is verified. iic + 1 compoud, ( + / compoud, / prime, 1 compoud 13

15 ( + I that case, π( + = π(, π = π, δ Ya ( = 1 ad Y a ( + = Y a ( 1. Y a ( + = Y a ( 1 = π( π + δ Ya ( 1 ( + = π( + π ( + = π( + π + δ Ya ( + sice δ Ya ( + = 0 i that case. This implies that (Ccl is verified. iid + 1 prime, ( + / prime, / compoud, 1 prime ( + I that case, π( + = π( + 1, π = π + 1, δ Ya ( = 1 ad Y a ( + = Y a ( + 1. Y a ( + = Y a ( + 1 = π( π + δ Ya ( = π( + 1 π ( + = π( + π sice δ Ya ( + = 0 i that case. This implies that (Ccl is verified δ Ya ( + iie + 1 compoud, ( + / prime, / compoud, 1 prime ( + I that case, π( + = π(, π = π + 1, δ Ya ( = 1 ad Y a ( + = Y a ( + 1. Y a ( + = Y a ( + 1 = π( π + δ Ya ( + 1 ( + = π( + π ( + = π( + π + δ Ya ( + sice δ Ya ( + = 1 i that case. This implies that (Ccl is verified. iif + 1 prime, ( + / prime, / compoud, 1 compoud ( + I that case, π( + = π( + 1, π = π + 1, δ Ya ( = 0 ad Y a ( + = Y a (. Y a ( + = Y a ( = π( π + δ Ya ( + = π( + 1 π ( + = π( + π sice δ Ya ( + = 0 i that case. This implies that (Ccl is verified δ Ya ( + iig + 1 compoud, ( + / prime, / compoud, 1 compoud ( + I that case, π( + = π(, π = π + 1, δ Ya ( = 0 ad Y a ( + = Y a (. 1

16 Y a ( + = Y a ( = π( π + δ Ya ( ( + = π( + π ( + = π( + π + δ Ya ( + sice δ Ya ( + = 1 i that case. This implies that (Ccl is verified. iih + 1 prime, ( + / compoud, / compoud, 1 prime ( + I that case, π( + = π( + 1, π = π, δ Ya ( = 1 ad Y a ( + = Y a ( + 1. Y a ( + = Y a ( + 1 = π( π + δ Ya ( = π( + 1 π ( + = π( + π sice δ Ya ( + = 1 i that case. This implies that (Ccl is verified δ Ya ( + iii + 1 compoud, ( + / compoud, / compoud, 1 prime ( + I that case, π( + = π(, π = π, δ Ya ( = 1 ad Y a ( + = Y a ( + 1. Y a ( + = Y a ( + 1 = π( π + δ Ya ( + 1 ( + = π( + π ( + = π( + π + δ Ya ( + sice δ Ya ( + = 0 i that case. It implies that (Ccl is verified. iij + 1 prime, ( + / compoud, / prime, 1 compoud ( + I that case, π( + = π( + 1, π = π, δ Ya ( = 1 ad Y a ( + = Y a ( 1. Y a ( + = Y a ( 1 = π( π + δ Ya ( 1 + = π( + 1 π ( + = π( + π sice δ Ya ( + = 1 i that case. This implies that (Ccl is verified δ Ya ( + iik + 1 compoud, ( + / compoud, / compoud, 1 compoud ( + I that case, π( + = π(, π = π, δ Ya ( = 0 ad Y a ( + = Y a (. Y a ( + = Y a ( = π( π + δ Ya ( ( + = π( + π + δ Ya ( + sice δ Ya ( + = 0 i that case. This implies that (Ccl is verified. 15

17 6.1. Property 1 : ivariat relatio cocerig Y c ( The objective is to demostrate why the ivariat relatio cocerig Y c ( ad that is : Y c ( = π( + π + δ Yc ( is always verified. i recurreces iitializatios : / ( + / Y is prime is prime is prime is prime c ( π( π(/ δ Yc ( 1 x x x x x x x x x x x x x x x x x It is easy to verify that for small itegers, property 1 is well systematically verified. ii Let us demostrate by recurrece that if (1 is true for, it is true for +. Let us make the hypothesis that property 1 is true for, Y c ( = π( + π + δ Yc ( (H Let us demostrate it is true for +, + Y c ( + = π( + + π + + δ Yc ( + (Ccl We fid here aother time the 11 cases idetified for demostratig property 11 ; sometimes, there are two subcases, whe / is compoud, accordig to the fact that is a eve s double i which case + + = or that is a odd s double = + 1. I the array below are provided δ Yc ( s values accordig to the booleas that must be cosidered (i first colum, we idicate a umber of lie l i that will be useful i the followig ; a example of a small iteger is provided i peultimate colum, to fix ideas ; the letter correspodig to the recurrece type iterveig (cf after the array is provided i last colum : 16

18 1 + 1 / ( + / is prime is prime is prime is prime δ k+ ( δ Yc ( ex case l 1 x x x 0 1 a l x x 0 0 e l 3 x x 0 5 i l x 1 8 i l 5 x x x 1 b l 6 x x 1 36 f l 7 x x 0 66 j l 8 x 1 16 j l 9 x x 1 6 c l 10 x 1 56 g l 11 x 0 50 k l k l 13 x x x 0 60 d l 1 x x x 1 18 h l 15 x x h iia + 1 compoud, ( + / compoud, / prime, 1 prime ( + I that case, π( + = π(, π = π, δ Yc ( = 0 ad Y c ( + = Y c (. Y c ( + = Y c ( = π( + π + δ Yc ( + + = 1 π( + + π + + = π( + + π δ Yc ( + sice δ Yc ( + = 1 i that case ( + correspods tha to oe of the cases of lies l 6, l 8, l 10 or l 1 because beig a odd s double, + ca t be oe. This implies that (Ccl is verified. iib + 1 prime, ( + / compoud, / prime, 1 compoud ( + I that case, π( + = π( + 1, π = π, δ Yc ( = 1 ad Y c ( + = Y c ( + 1. Y c ( + = Y c ( + 1 = π( + π + δ Yc ( = 1 π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to oe of the cases of lies l, l 13 or l 15 because beig a odd s double, + ca t be oe. This implies that (Ccl is verified. iic + 1 compoud, ( + / compoud, / prime, 1 compoud ( + I that case, π( + = π(, π = π, δ Yc ( = 1 ad Y c ( + = Y c ( + 1. Y c ( + = Y c ( + 1 = π( + π + δ Yc ( = 1 π( + + π + + = π( + + π δ Yc ( +

19 sice δ Yc ( + = 1 i that case ( + correspods the to oe of the cases of lies l 6, l 8, l 10 or l 1 because beig a odd s double, + ca t be oe. This implies that (Ccl is verified. iid + 1 prime, ( + / prime, / compoud, 1 prime ( + I that case, π( + = π( + 1, π = π + 1, δ Yc ( = 0 ad Y c ( + = Y c (. is madatory a eve s double. Y c ( + = Y c ( = π( + π + δ Yc ( + + = π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to the case of lie l 1. This implies that (Ccl is verified. iie + 1 compoud, ( + / prime, / compoud, 1 prime ( + I that case, π( + = π(, π = π + 1, δ Yc ( = 0 ad Y c ( + = Y c (. is madatory a eve s double. Y c ( + = Y c ( = π( + π + δ Yc ( + ( + = π( + + π ( + = π( + + π + δ Yc ( + sice δ Yc (+ = 1 i that case (+ correspods the to oe of the cases of lies l 5 or l 9. This implies that (Ccl is verified. iif + 1 prime, ( + / prime, / compoud, 1 compoud ( + I that case, π( + = π( + 1, π = π + 1, δ Yc ( = 1 ad Y c ( + = Y c ( + 1. is madatory a eve s double. Y c ( + = Y c ( + 1 = π( + π + δ Yc ( = π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to the case of lie l 1. This implies that (Ccl is verified. iig + 1 compoud, ( + / prime, / compoud, 1 compoud ( + I that case, π( + = π(, π = π + 1, δ Yc ( = 1 ad Y c ( + = Y c ( + 1. is madatory a eve s double. 18

20 Y c ( + = Y c ( + 1 = π( + π + δ Yc ( ( + = π( + + π ( + = π( + + π + δ Yc ( + sice δ Yc (+ = 1 i that case (+ correspods the to oe of the cases of lies l 5 or l 9. This implies that (Ccl is verified. iih + 1 prime, ( + / compoud, / compoud, 1 prime ( + I that case, π( + = π( + 1, π = π. iih1 eve s double, δ Yc ( = 0 ad Y c ( + = Y c (. Y c ( + = Y c ( = π( + π + δ Yc ( + + = π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 1 i that case ( + correspods the to oe of the cases of lies l 3 or l 1. This implies that (Ccl is verified. iih odd s double, δ Yc ( = 1 ad Y c ( + = Y c ( 1. Y c ( + = Y c ( 1 = π( + π + δ Yc ( = 1 π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to oe of the cases of lies l, l, l 13 or l 15. This implies that (Ccl is verified. iii + 1 compoud, ( + / compoud, / compoud, 1 prime ( + I that case, π( + = π(, π = π. iii1 eve s double, δ Yc ( = 0 ad Y c ( + = Y c (. Y c ( + = Y c ( = π( + π + δ Yc ( + ( + = π( + + π ( + = π( + + π + δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to oe of the cases of lies l 7 or l 11. This implies that (Ccl is verified. iii odd s double, δ Yc ( = 1 ad Y c ( + = Y c ( 1. 19

21 Y c ( + = Y c ( 1 = π( + π + δ Yc ( = 1 π( + + π + + = π( + + π δ Yc ( + sice δ Yc ( + = 1 i that case ( + correspods the to oe of the cases of lies l 6, l 8, l 10 or l 1. This implies that (Ccl is verified. iij + 1 prime, ( + / compoud, / compoud, 1 compoud ( + I that case, π( + = π( + 1, π = π. iij1 eve s double, δ Yc ( = 1 ad Y c ( + = Y c ( + 1. Y c ( + = Y c ( + 1 = π( + π + δ Yc ( = π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 1 i that case ( + correspods the to oe of the cases of lies l 3 or l 1. This implies that (Ccl is verified. iij odd s double, δ Yc ( = 0 ad Y c ( + = Y c (. Y c ( + = Y c ( = π( + π + δ Yc ( + + = 1 π( π + + = π( + + π δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to oe of the cases of lies l, l, l 13 or l 15. This implies that (Ccl is verified. iik + 1 compoud, ( + / compoud, / compoud, 1 compoud ( + I that case, π( + = π(, π = π. iik1 eve s double, δ Yc ( = 1 ad Y c ( + = Y c ( + 1. Y c ( + = Y c ( + 1 = π( + π + δ Yc ( ( + = π( + + π ( + = π( + + π + δ Yc ( + sice δ Yc ( + = 0 i that case ( + correspods the to oe of the cases of lies l 7 or l 11. This implies that (Ccl is verified. iik odd compoud s double, δ Yc ( = 0 ad Y c ( + = Y c ( 0

22 Y c ( + = Y c ( = π( + π + δ Yc ( + + = 1 π( + + π + + = π( + + π δ Yc ( + sice δ Yc ( + = 1 i that case ( + correspods to cases i lies l 6, l 8, l 10 or l 1. This implies that (Ccl is verified. 6. Order relatios betwee variables 6..1 Study of iequalities Z c ( > Z a (, Z a ( > Y a ( ad Y c ( > Z c ( From properties 9 ad 10, we deduce that Z c ( Z a ( = equal to 1, or 3. π + δ Zc Z a ( with δ Zc Z a ( Z c ( Z a ( seems globally icreasig with little variatios. Oe observes that Z c ( > Z a ( for 0. For greater values, strict iequality will be always verified because is icreased much more ( ofte tha π. From properties 9 ad 11, we deduce that Z a ( Y a ( = π π( + δ Za Y a ( with δ Za Y a ( equal to 3,, 1 or 0. Z a ( Y a ( seems globally icreasig with little variatios. Oe observes that Z a ( > Y a ( for 36. From properties 10 ad 1, we deduce that Y c ( Z c ( = π π( + δ Yc Z c ( with δ Yc Z c ( equal to, 1, 0 or 1. Y c ( Z c ( is a icreasig fuctio of. Oe observes that Y c ( > Z c ( for. 6.. Strict order betwee variables Y a (, Y c (, Z a ( ad Z c ( Variables Y a (, Z a (, Z c ( et Y c ( are strictly ordered i the followig way : Y a ( < Z a ( < Z c ( < Y c ( for all 0. A picture showig variables gaps is provided below, that illustrates their itricatio : π( π(/ π(/ / π(/ / π( + π(/ Y a < Za < Zc < Y c Xa Xd From properties 9, 10, 11 ad 1, we deduce : Z c ( Y a ( = π( + δ Zc Y a ( with δ Zc Y a ( equal to 1, 0, 1 or ; 1

23 Y c ( Z a ( = π( + δ Yc Z a ( with δ Yc Z a ( equal to 0, 1, or 3 ; X d ( X a ( = Z c ( Y a ( + δ odd compoud ( = π( + δ Xd X a ( with δ Xd X a ( equal to 1, 0, 1 or. Oe ca also observe by program ( that : X c ( X b ( = π π( + δ Xc X b ( with δ Xc X b ( equal to, 1, 0 or 1. Also, oe ca deduce from equalities ( yet provided that : X a ( + X b ( = π( π + δ Xc X b ( with δ Xa+X b ( very small. The fuctio π( that appear i right members of three first equalities above is globally icreasig with little variatios. It s equal to 0 for = 1. + Sice = + 1 whe is a eve s double (i.e. for oe eve umber each two while π( + = π( + 1 whe + 1 is prime (far less ofte tha for oe eve umber each two, for all greater tha a small value (i.e. > 1, we will always have π( > Study of iequality X a ( > 0 To be sure that X a ( would ever be equal to 0, we should have to show that sice a certai value of, the iequality X d ( > π( + is always verified. Ideed, this iequality, combied with the ivariat X d ( X a ( = π( + δ Xd X a (, would have as cosequece that X a ( would be always strictly positive. We observe that it seems that for 30. X d ( > π( + Ituitively, oe uderstads what process is at work : the umber of compoud umbers becomes so big comparatively to the umber of prime umbers that each compoud umber has may more chaces to be the complemetary to of aother compoud umber rather tha a prime oe. It is thus oe ca observe X d ( becomes quickly greater tha the sum X b ( + X c (, i which each term is greater tha X a (. 6. Subsidiary exercize Let us propose as a exercize that we must match (with a bijectio umbers of two sets ; i the first set cotaiig k umbers, A are compoud ad k A are prime while i the secod set, cotaiig also k umbers, B umbers are compoud while k B are prime. If A k/ ad B k/, the matchig ca put i bijectio two umbers that are ot madatory compoud. But as soo as A > k/ ad B > k/, some matchig must ecessarily put i bijectio a compoud umber from the first set with a compoud umber of the secod set. We ca easily uderstad thaks to the drawig below that the umber of matchigs associatig bijectively two compoud umbers is ecessarily greater tha A + B k. A B

24 6.5 X a ( costraied to be strictly positive Amog odd umbers that are betwee 3 ad /, very quickly, more tha a half are compoud. O the same maer, amog odd umbers that are betwee / ad 3, very quickly, more tha a half are compoud. By this way, there are rather quickly both more tha a half of umbers that ca be a s decompositio first rage sommat that are compoud ad more tha a half of umbers that ca be a s decompositio secod rage sommat that are compoud too (i.e. Y c ( > ad Z c ( > as soo 8 8 as. Ideed, for, π( + π > ad π >. 8 8 X d ( coutig s decompositios of the form compoud+compoud is the systematically greater tha Y c ( + Z c ( (cf subsidiary exercize of precedet sectio ad will remai greater tha it defiitively, which esures that X a ( will be always strictly positive, which proves biary Goldbach s cojecture. 3