A Frobenius problem suggested by prime k-tuplets

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Contents lists available atScienceDirect

Discrete Mathematics

journalhomepage:www.elsevier.com/locate/disc

A Frobenius problem suggested by prime k-tuplets

Aureliano M. Robles-Pérez

^a

^,∗ , José Carlos Rosales

^b

aDepartamentodeMatemáticaAplicada&InstitutodeMatemáticas(IMAG),UniversidaddeGranada,18071-Granada,Spain bDepartamentodeÁlgebra&InstitutodeMatemáticas(IMAG),UniversidaddeGranada,18071-Granada,Spain

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received18March2022

Receivedinrevisedform14February2023 Accepted15February2023

Availableonlinexxxx

Keywords:

k-tuplets

Admissibleconstellation Frobeniusnumber Numericalsemigroup Apéryset

We study the Frobenius problem for certain k-tuplets, which include primek-tuplets.

Moreover,weanalyzesomepropertiesofthenumericalsemigroupsassociatedwiththese tuplets.

©²⁰²³^TheÂuthor(s).^Published^byÊlsevier^B.V.^Thisîsânôpenâccessârticleûnder^the CCBYlicense(http://creativecommons.org/licenses/by/4.0/).

1. Introduction

Aprimetriplet isasequenceofthreeprimenumbers

(

p1

,

p2

,

p3

)

suchthat p1

<

p2

<

p3 andp3

−

^p1

=

^6.^Inparticular, thesequencesmustbeoftheform

(

p

,

p

+

²

,

p

+

⁶

)

or

(

p

,

p

+

⁴

,

p

+

⁶

)

.Sinceoneofeverythreeconsecutiveoddnumbersis amultipleofthree,andthereforenotprime(exceptforthreeitself),wehavethattheprimetripletsaretheclosestpossible groupingsofthreeprimenumberswiththeexceptionsof

(

2

,

3

,

5

)

and

(

3

,

5

,

7

)

.

Analogously,aprimequadruplet isasequence

(

p1

,

p2

,

p3

,

p4

)

,offourprimenumbers,suchthat p1

<

p2

<

p3

<

p4 and p4

−

^p1

=

^8.^In^this^case,^the^sequences^must^be^of^the^form

(

p

,

p

+

²

,

p

+

⁶

,

p

+

⁸

)

andaretheclosestpossiblegroupings offourprimenumberswiththeexceptionsof

(

2

,

3

,

5

,

7

)

and

(

3

,

5

,

7

,

11

)

.

Toformalizetheconceptofk-tuplet,letusfollow[6]:ifk isanintegergreaterthanone,thens

(

k

)

isthesmallestnumber forwhichthereexists asetofk orderedintegers B_k

= {

^b1

, . . . ,

b_k

}

^such ^that^b1

=

^0,^bk

=

^s

(

k

)

,and

{

^b1mod q

, . . . ,

b_kmod q

} = {

⁰

,

1

, . . . ,

q

−

¹

}

^(that^is,

{

^b1mod q

, . . . ,

b_kmod q

}

îs^not â^complete^setôf^residues^modulo^q) ^forâny^prime^number q.Fromhere, aprimek-tuplet isasequenceofconsecutiveprimenumbers,P_k

= (

^p1

, . . . ,

p_k

)

,suchthat p_i

−

^p1

=

^bi forall i

∈ {

¹

, . . . ,

k

}

^(observe^that ^pk

−

^p1

=

^s

(

k

)

).Thus,roughly speaking,aprimek-tuplet isasequenceofk consecutiveprime numberssuchthatthedifferencebetweentheﬁrstandthelast(thatis,s

(

k

)

)isassmallaspossible.SometimesB_k iscalled anadmissibleset andP_kisnamedanadmissibleconstellation(ofconsecutiveprimenumbers) (see[5]).

Letusobservethat ifwe remove thecondition onthefunction s

(

k

)

,we willhaveother admissibleconstellations. For example,

(

p

,

p

+

⁴

)

(cousinprimes),

(

p

,

p

+

⁶

)

(sexyprimepairs),

(

p

,

p

+

⁶

,

p

+

¹²

)

(sexyprimetriplets),etcetera.

Letusalso note thatfrom thedeﬁnition,it iseasy tosee that we canonly obtaina ﬁnitenumber ofsequences ofk consecutiveprimenumbers

(

p1

, . . . ,

pk

)

suchthat pk

−

^p1

<

s

(

k

)

.Incontrast,thegeneralizedHardy-Littlewoodconjecture

*

Correspondingauthor.

E-mailaddresses:[email protected](A.M. Robles-Pérez),[email protected](J.C. Rosales).

https://doi.org/10.1016/j.disc.2023.113394

0012-365X/©²⁰²³^TheÂuthor(s).^Published^byÊlsevier^B.V.^Thisîsânôpenâccessârticleûnder^the^CC^BY^license^(http://

creativecommons.org/licenses/by/4.0/).

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(see [8,9])impliesthat,foreveryadmissibleset Bk,wecanobtaininﬁnitelymanyprimek-tuplets.Infact, allthecurrent evidenceconﬁrmtheconjectureforanyadmissibleconstellation(whetherornotconsideringtheconditionons

(

k

)

).

Let

(

a1

, . . . ,

ae

)

be a sequence of positive integers such that gcd

(

a1

, . . . ,

ae

) =

^1. ^Then ^a ^classical ^problemⁱⁿ ^additive numbertheoryistheFrobeniusproblem:whatisthegreatestintegerF

(

a1

, . . . ,

ae

)

thatisnotanelementoftheseta1

N + . . . +

^ae

N

?Althoughthereexistsasolutionforthisproblemwhene

=

^{2 (see}^[18]),îtîs^well-known^thatîtîs^not^possible tofindapolynomialformulainordertocomputeF

(

a₁

, . . . ,

a_e

)

ife

≥

^{3 (see}^[4]).^Therefore,^many^efforts^have^been^devoted toobtainingpartialresultsordevelopingalgorithmstogettheanswertothisquestion(see[13]).

Among others, the main objective of this work is to ﬁnd the solution to the Frobenius problem for (prime) triplets (Section 3) and(prime) quadruplets(Section 4). We alsocomment onsome results fork-tuplets withk

≥

^{5 (Section}^5).

Moreover,fromthecomputations,weconjecturethatpolynomialsofdegreetwoallowustocomputetheFrobeniusnumber ofk-tuplets.

Toachieveourpurposes,weusethetheoryofnumericalsemigroups(closelyrelatedtotheFrobeniusproblem)and,in particular,theApérysetofanumericalsemigroup(Section2).

It is worth notingthat the restriction on prime numbers is not essential to get the proposed results. Infact, inthe statementsonApérysets,Frobeniusandpseudo-Frobeniusnumbers,andgenus,wewillconsiderk-tuplesthat onlyverify thecondition oftheadmissibleconstellation;that is,if

{

^p1

,

p2

, . . . ,

p_k

}

^is^a^k-tuplet, ^then

{

^p1

−

^p1

,

p2

−

^p1

, . . . ,

p_k

−

^p1

}

mustbeanadmissibleset.

2. Preliminaries(onnumericalsemigroups)

Let

Z

be thesetofintegersand

N = {

^z

∈ Z |

^z

≥

⁰

}

^.^A^submonoid ^of

(N, +)

^is^a^subset ^{M of}

N

that isclosedunder additionandcontains thezeroelement.Anumericalsemigroup isasubmonoidof

(N, +)

^such^that

N \

^S

= {

ⁿ

∈ N |

ⁿ

∈ /

^S

}

isﬁnite.

Let S bea numericalsemigroup. Since

N \

^{S is} â^finite ^set,^we ^can^define ^two învariantsôf ^S.^Namely, ^the^Frobenius numberofS isthegreatestintegerthatdoesnotbelongtoS,denotedbyF

(

S

)

,andthegenusofS isthecardinalityof

N \

^S, denoted by g

(

S

)

.Letusnote that, innumbertheory,it iscommonto usethe termSylvesternumber insteadofthe term genus(see[10]).

If X isanon-emptysubsetof

N

,thenwedenoteby

^X

^the^submonoid^of

(N, +)

^generated^by ^{X ,}^that^is,

^X

=

λ

1x₁

+ · · · + λ

nx_n

|

ⁿ

∈ N \ {

⁰

},

^x1

, . . . ,

x_n

∈

^X

, λ

1

, . . . , λ

n

∈ N .

Itiswell-known(seeLemma 2.1of[16])that

^X

îsâ^numerical^semigroupîfândônlyîf^gcd

(

X

) =

^1.

If S is a numericalsemigroupand S

=

^X

^,^then ^we^say ^that ^{X is}â ^systemôf^generatorsôf^S. ^Moreover,îf ^S

=

^Y

^for anysubsetY

^{X ,}^then^we^say^that ^{X is}â^minimal^systemôfgenerators of S.InTheorem 2.7of[16],itisshownthateach numericalsemigroupadmitsauniqueminimalsystemofgeneratorsandthatsuchasystemisfinite.Wedenotebymsg

(

S

)

theminimalsystemofgeneratorsofS.Thecardinalityofmsg

(

S

)

,denotedbye

(

S

)

,istheembeddingdimension ofS.

The (extended)Frobeniusproblemforanumericalsemigroup S consistsofﬁnding formulasthatallow ustocompute F

(

S

)

andg

(

S

)

intermsofmsg

(

S

)

.As inthecaseoftheFrobeniusproblemforsequences, suchformulas arewell-known fore

(

S

) =

^2,^butîtîs^not^possible ^to^find^polynomial^formulas ^whenê

(

S

) ≥

^3,^except^for^particular^families ^of^numerical semigroups.

Now,letusdeﬁneausefultooltodescribeanumericalsemigroupS.Ifn

∈

^S

\ {

⁰

}

^,^then^the^Apéry^set^of^{n in}^{S (named}ⁱⁿ honourof[1])isAp

(

S

,

n

) = {

^s

∈

^S

|

^s

−

ⁿ

∈ /

^S

}

^.

ThefollowingresultisLemma 2.4of[16].

Proposition2.1.LetS beanumericalsemigroupandn

∈

^S

\ {

⁰

}

^.^Then^thecardinalityofAp

(

S

,

n

)

isn.Moreover,

Ap

(

S

,

n

) = {

^w

(

0

) =

⁰

,

w

(

1

), . . . ,

w

(

n

−

¹

) },

wherew

(

i

)

istheleastelementofS congruentwithi modulon.

Theknowledge ofAp

(

S

,

n

)

allowsustosolvetheproblemofmembershipofanintegertothenumericalsemigroup S.

Infact,ifx

∈ Z

^,^then ^x

∈

^{S if}ândônlyîf^x

≥

^w

(

xmod n

)

.Moreover,we havethefollowingresultfrom[3] (ﬁrstformula) and[17] (secondone).

Proposition2.2.LetS beanumericalsemigroupandletn

∈

^S

\ {

⁰

}

^.^Then

1. F

(

S

) =

^max

(

Ap

(

S

,

n

)) −

^n, 2. g

(

S

) =

¹_n

(

w∈^Ap(S,n)w

) −

ⁿ⁻₂¹^.

Fromthisproposition,wehavethesolutiontotheFrobeniusproblemforS ifweknowanexplicitdescriptionofAp

(

S

,

n

)

.

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Remark2.3.In[10,11],theauthorusesApérysetstocomputetheFrobeniusnumberandtheSylvesternumber(thatis,the genus)ofnumericalsemigroupsgenerated byarithmetic progressionsandbyarithmetic progressionswithinitialgaps.In particular,ourcorrespondingresultsinCorollary3.13canbededucedfrom[11] (seeRemarks3.16and3.17).

Remark2.4.We can consider differentgeneralizations of the Frobenius number. Forexample, from [12] (and other ref- erences therein), it is possible to deﬁne the p-Frobeniusnumber of a numerical semigroup S

=

^a1

,

a₂

, . . . ,

a_e

^as ^the greatest integer that can be represented atmost p waysby a linear combinationwith non-negativeinteger coeﬃcients ofa1

,

a2

, . . . ,

ae.Thus, the0-FrobeniusnumberistheclassicalFrobeniusnumber.Sincein[12] theauthoruses Apérysets toobtainhisresults,apossiblefutureworkwouldbetocomputethep-Frobeniusnumberfork-tuplets.

LetS beanumericalsemigroup.Followingthenotationintroducedin[15],wesaythatanintegerx isapseudo-Frobenius numberofS ifx

∈ Z \

^{S and}^x

+

^s

∈

^{S for}^all^s

∈

^S

\ {

⁰

}

^.^We^denote^by^PF

(

S

)

thesetofallthepseudo-Frobeniusnumbersof S.ThecardinalityofPF

(

S

)

isanotableinvariantofS (see[2]),theso-calledtype of S,denotedbyt

(

S

)

.

LetS beanumericalsemigroup.Wedeﬁneover

Z

thefollowingbinaryrelation:a

≤

Sb ifb

−

^a

∈

^S.^As^statedⁱⁿ^[16],^it isclearthat

≤

S isanon-strictpartialorderrelation(thatis,reﬂexive,transitive,andanti-symmetric).

The followingresultis Proposition 7of[7] (seealso Proposition 2.20 of[16])andcharacterizes the pseudo-Frobenius numbersintermsofthemaximalelementsofAp

(

S

,

n

)

withrespecttotherelation

≤

S.

Proposition2.5.LetS beanumericalsemigroupandn

∈

^S

\ {

⁰

}

^.^Then

PF

(

S

) = {

w

−

n

|

w

∈

Maximals_≤_S

(

Ap

(

S

,

n

))}.

3. Tripletsintheformofprimetriplets

To have a well-posed Frobenius problem, we can consider six possibilities of tripletssatisfying the condition of the admissibleconstellation(see [14]).However, sincereasoningsandcalculationsare similar,we willstudyindetailthetwo casesconcerningprimetriplets.

Letusrecallthataprimetripletisoftheform

(

p

,

p

+

²

,

p

+

⁶

)

oroftheform

(

p

,

p

+

⁴

,

p

+

^6).^We^improve^this^factⁱⁿ thefollowingproposition.

Proposition3.1.Wehavethat:

1. If

(

p

,

p

+

²

,

p

+

⁶

)

isaprimetriplet,thenp

=

^6k

+

^5,^with^k

∈ N

^. 2. If

(

p

,

p

+

⁴

,

p

+

⁶

)

isaprimetriplet,thenp

=

^6k

+

^7,^with^k

∈ N

^.

Proof. It is clear that if p

∈ N \ {

¹

}

îs ^not ^divisible ^by ^two ôr ^three, ^then ^there êxists ^k

∈ N

^such ^that ^p

=

^6k

+

^{5 or} p

=

^6k

+

^7.

1. Ifp

=

^6k

+

^7,^then ^p

+

²

=

^6k

+

^{9 is}^multipleôf^threeând^can^not^beâ^prime^number.Consequently,if

(

p

,

p

+

²

,

p

+

⁶

)

isaprimetriplet,then p

=

^6k

+

^5.

2. Thereasoningissimilartotheabovecase.

Asaconsequenceofthepreviousproposition,aprimetripletisofoneofthefollowingtwoforms.

1.

(

6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

)

,withk

∈ N

^. 2.

(

6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

)

,withk

∈ N

^.

Since gcd

(

6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

) =

^gcd

(

6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

) =

^1, ^we ^can ^deﬁne ^two ^families ^of ^numerical ^semigroups(

T

1and

T

2).

•

^S

∈ T

1 ifS

=

^6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

^,^with^k

∈ N

^.

•

^S

∈ T

2 ifS

=

^6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

^,^with^k

∈ N

^.

ItiseasytocheckthatifS

∈ T

1

∪ T

2,thene

(

S

) =

^3.^Moreover,^from^[7],^we^deduce^the^following^result.

Proposition3.2.LetS bea numericalsemigroupsuchthate

(

S

) =

^3.^Then^t

(

S

) ∈ {

¹

,

2

}

^.Înâddition,îf ^S

=

ⁿ1

,

n2

,

n3

^,^where n1

,

n2

,

n3arepairwiserelativelyprimenumbers,thent

(

S

) =

^2.

Animmediateconsequenceoftheabovecommentsandresultsisthefollowingproposition.

Proposition3.3.IfS

∈ T

1

∪ T

2,thene

(

S

) =

^{3 and}^t

(

S

) =

^2.

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3.1. Firstcase(family

T

1)

In thissubsection,we are interested instudying thenumerical semigroupsof the form S

=

^6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

^, wherek

∈ N

^.

Straightforwardcomputationsleadtothefollowingresult.

Lemma3.4.Ifk

∈ N

^,^then^we^have^the^equalities

1. 3

(

6k

+

⁷

) =

²

(

6k

+

⁵

) +

¹

(

6k

+

¹¹

)

;

2.

(

2k

+

²

)(

6k

+

¹¹

) = (

^2k

+

³

)(

6k

+

⁵

) +

¹

(

6k

+

⁷

)

; 3. 2

(

6k

+

⁷

) + (

^2k

+

¹

)(

6k

+

¹¹

) = (

^2k

+

⁵

)(

6k

+

⁵

)

.

Letusnowseethekeytothissubsection.

Theorem3.5.Ifk

∈ N

^and^S

=

^6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

^,^then Ap

(

S

,

6k

+

⁵

) = {

^a

(

6k

+

⁷

) +

^b

(

6k

+

¹¹

) | (

^a

,

b

) ∈

^C

} ,

whereC

=

{

⁰

,

1

,

2

} × {

⁰

,

1

, . . . ,

2k

+

¹

}

\ {(

²

,

2k

+

¹

)}

^.

Proof. FromLemma3.4,we easilydeducethatAp

(

S

,

6k

+

⁵

)

isasubset of

{

^a

(

6k

+

⁷

) +

^b

(

6k

+

¹¹

) | (

^a

,

b

) ∈

^C

}

^.^Now ^then, sincethecardinalityofC islessthanorequalto3

(

2k

+

²

) −

¹

=

^6k

+

^5,^and^taking^intoconsiderationProposition2.1,we getthatAp

(

S

,

6k

+

⁵

) = {

^a

(

6k

+

⁷

) +

^b

(

6k

+

¹¹

) | (

^a

,

b

) ∈

^C

}

^.

LetusobservethatwecanrewriteseveralelementsofAp

(

S

,

6k

+

⁵

)

asfollows.

• (

^6k

+

⁷

) +

^b

(

6k

+

¹¹

) = (

^b

+

¹

)(

6k

+

¹¹

) −

^4,^for^all^b

∈ {

⁰

,

1

, . . . ,

2k

+

¹

}

^.

•

²

(

6k

+

⁷

) +

^b

(

6k

+

¹¹

) = (

^b

+

²

)(

6k

+

¹¹

) −

^8,^for^all^b

∈ {

⁰

,

1

, . . . ,

2k

}

^.

Thus,wecaneasilydescribetheApérysetbyarrangingitselementsinincreasingorder.

Corollary3.6.Ifk

∈ N

^and^S

=

^6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

^,^then

Ap

(

S

,

6k

+

⁵

) = {

⁰

; (

^6k

+

¹¹

) −

⁴

,

6k

+

¹¹

;

²

(

6k

+

¹¹

) −

⁸

,

2

(

6k

+

¹¹

) −

⁴

,

2

(

6k

+

¹¹

) ; . . . ;

(

2k

+

¹

)(

6k

+

¹¹

) −

⁸

, (

2k

+

¹

)(

6k

+

¹¹

) −

⁴

, (

2k

+

¹

)(

6k

+

¹¹

) ; (

^2k

+

²

)(

6k

+

¹¹

) −

⁸

, (

2k

+

²

)(

6k

+

¹¹

) −

⁴

}.

Inthisway,wecanidentifyapattern.Infact,inthepreviouscorollary,wehaveused“;”toseparatesuitablegroupsof numbers.

AsmentionedinSection2,knowledgeoftheApérysetallowsustoobtaininformationaboutthenumericalsemigroup.

Thus,inthecurrentcase,wehavethefollowingresult.(Letusobservethat,fromCorollary3.7,werecoverProposition3.3.) Corollary3.7.Ifk

∈ N

^and^S

=

^6k

+

⁵

,

6k

+

⁷

,

6k

+

¹¹

^,^then

1. PF

(

S

) = {

^12k²

+

^28k

+

⁹

,

12k²

+

^28k

+

¹³

}

^; 2. F

(

S

) =

^12k²

+

^28k

+

^13;

3. g

(

S

) =

6k²

+

16k

+

8.

Proof. 1. ByTheorem3.5,wehavethatMaximals_≤_S

(

Ap

(

S

,

6k

+

⁵

)) = {(

^6k

+

⁷

) +(

^2k

+

¹

)(

6k

+

¹¹

),

2

(

6k

+

⁷

) +

^2k

(

6k

+

¹¹

) }

^. Thereby,fromProposition2.5,wecanassertthatPF

(

S

) = {(

^6k

+

⁷

) + (

^2k

+

¹

)(

6k

+

¹¹

) − (

^6k

+

⁵

),

2

(

6k

+

⁷

) +

^2k

(

6k

+

11

) − (

^6k

+

⁵

)} = {

^12k²

+

^28k

+

¹³

,

12k²

+

^28k

+

⁹

}

^.

2. ItisclearthatF

(

S

) =

^max

(

PF

(

S

))

and,therefore,F

(

S

) =

^12k²

+

^28k

+

^13.

3. FromTheorem3.5,wehavethat

Ap

(

S

,

6k

+

⁵

) = {

⁰

,

6k

+

¹¹

, . . . , (

1

+

^2k

)(

6k

+

¹¹

),

6k

+

⁷

, (

6k

+

⁷

) +(

^6k

+

¹¹

), . . . , (

6k

+

⁷

) +(

¹

+

^2k

)(

6k

+

¹¹

),

2

(

6k

+

⁷

),

2

(

6k

+

⁷

) + (

^6k

+

¹¹

), . . . ,

2

(

6k

+

⁷

) +

^2k

(

6k

+

¹¹

) }.

Now,byProposition2.2,wegetthat g

(

S

) =

¹

6k

+

⁵

6k

+

¹¹

+ · · · + (

¹

+

^2k

)(

6k

+

¹¹

) +

^6k

+

⁷

+ (

^6k

+

⁷

) + (

^6k

+

¹¹

) +

(5)

· · · + (

^6k

+

⁷

) + (

¹

+

^2k

)(

6k

+

¹¹

) +

²

(

6k

+

⁷

) +

²

(

6k

+

⁷

) + (

^6k

+

¹¹

) +

· · · +

²

(

6k

+

⁷

) +

^2k

(

6k

+

¹¹

)

− (

6k

+

⁵

) −

¹

2

=

^6k²

+

^16k

+

⁸

.

Remark3.8.FromCorollary3.7,wededucethatif p

=

^6k

+

^{5 with}^k

∈ N

^,^then^F

(

p

,

p

+

²

,

p

+

⁶

) =

^p²⁺^4p₃ ⁻⁶^. Weﬁnishwithanillustrativeexample.

Example3.9.Ifk

=

^1,^then ^S

=

¹¹

,

13

,

17

^, ^PF

(

S

) = {

⁴⁹

,

53

}

^,^F

(

S

) =

^53, ^and^g

(

S

) =

^{30 (by} ^Corollary^3.7). ^Moreover,^from Corollary3.6,weknowthatAp

(

S

,

11

) = {

⁰

,

13

,

17

,

26

,

30

,

34

,

43

,

47

,

51

,

60

,

64

}

^.

3.2. Secondcase(family

T

2)

Nowwestudynumericalsemigroupsoftheform S

=

^6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

^,^where^k

∈ N

^.^The^results^are^similar^to thoseofSubsection3.1.Therefore,weomittheproofs.

Lemma3.10.Ifk

∈ N

1. 3

(

6k

+

¹¹

) =

¹

(

6k

+

⁷

) +

²

(

6k

+

¹³

)

;

2.

(

2k

+

³

)(

6k

+

¹³

) = (

^2k

+

⁴

)(

6k

+

⁷

) +

¹

(

6k

+

¹¹

)

; 3. 2

(

6k

+

¹¹

) + (

^2k

+

¹

)(

6k

+

¹³

) = (

^2k

+

⁵

)(

6k

+

⁷

)

.

∈ N

^and^S

=

^6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

^,^then

Ap

(

S

,

6k

+

7

) = {

a

(

6k

+

11

) +

b

(

6k

+

13

) | (

a

,

b

) ∈

C

} ,

whereC

=

{

⁰

,

1

,

2

} × {

⁰

,

1

, . . . ,

2k

+

²

}

\ {(

²

,

2k

+

¹

), (

2

,

2k

+

²

)}

^. Letusobservethat

• (

^6k

+

¹¹

) +

^b

(

6k

+

¹³

) = (

^b

+

¹

)(

6k

+

¹³

) −

^2,^for^all^b

∈ {

⁰

,

1

, . . . ,

2k

+

²

}

^;

•

²

(

6k

+

¹¹

) +

^b

(

6k

+

¹³

) = (

^b

+

²

)(

6k

+

¹¹

) −

^4,^for^all^b

∈ {

⁰

,

1

, . . . ,

2k

}

^. Thus,wedescribetheApérysetbyarrangingitselementsinincreasingorder.

∈ N

^and^S

=

^6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

^,^then

Ap

(

S

,

6k

+

⁷

) = {

⁰

; (

^6k

+

¹³

) −

²

,

6k

+

¹³

;

²

(

6k

+

¹³

) −

⁴

,

2

(

6k

+

¹³

) −

²

,

2

(

6k

+

¹³

) ; . . . ; (

2k

+

²

)(

6k

+

¹³

) −

⁴

, (

2k

+

²

)(

6k

+

¹³

) −

²

, (

2k

+

²

)(

6k

+

¹³

); (

^2k

+

³

)(

6k

+

¹³

) −

²

}.

∈ N

^and^S

=

^6k

+

⁷

,

6k

+

¹¹

,

6k

+

¹³

^,^then

1. PF

(

S

) = {

^12k²

+

^32k

+

¹⁵

,

12k²

+

^38k

+

³⁰

}

^; 2. F

(

S

) =

^12k²

+

^38k

+

^30;

3. g

(

S

) =

^6k²

+

^20k

+

^16.

Remark3.14.FromCorollary3.13,wehavethatif p

=

^6k

+

^{7 with}^k

∈ N

^,^then^F

(

p

,

p

+

⁴

,

p

+

⁶

) =

^p²⁺^5p₃ ⁺⁶^. Letusseeanexample.

Example3.15.Ifk

=

^0,^then ^S

=

⁷

,

11

,

13

^,^PF

(

S

) = {

¹⁵

,

30

}

^,^F

(

S

) =

^30,^and^g

(

S

) =

^{16 (by}^Corollary^3.13).^Moreover,^from Corollary3.12,wegetthatAp

(

S

,

7

) = {

⁰

,

11

,

13

,

22

,

24

,

26

,

37

}

^.

Remark3.16.The values of F

(

p

,

p

+

⁴

,

p

+

⁶

)

and g

(

p

,

p

+

⁴

,

p

+

⁶

)

, for p

=

^6k

+

^{7 with}^k

∈ N

^, ^can ^be ^obtained ^from Theorems 2 and 3 of[11], respectively. Indeed,it is enough to consider a

=

^p, ^K

=

^1, ^k

=

^3,^d

=

^2, ^and^r

=

^{2 in} ^those theorems.

(6)

Remark3.17.LetusrecallthattheSylvestersum ofanumericalsemigroupS

=

^a1

,

a2

, . . . ,

ae

^is^the^value s

(

S

) =

x∈N\S

x

.

In[11], theauthorcomputestheSylvestersumofnumericalsemigroupsgeneratedbyarithmetic progressionswithinitial gaps.Togettheresult,heagainusestheApérysetAp

(

A

,

a₁

) = {

^w0

=

⁰

,

w₁

, . . . ,

w_a₁₋₁

}

^by^the^formula

s

(

a₁

,

a₂

, . . . ,

a_e

) =

^s

(

S

) =

¹ 2a₁

a

1−1 i=¹

w²_i

−

¹ 2

a

1−1 i=¹

w_i

+

^a²¹

−

¹

12

.

(1)

Asaconsequence,byTheorem 1of[11],wehavethatifp

=

^6k

+

^{7 with}^k

∈ N

^,^then^s

(

p

,

p

+

⁴

,

p

+

⁶

) =

₁₀₈¹

(

2p⁴

+

^24p³

+

93p²

+

^202p

+

⁴³⁵

)

.Ofcourse,wecoulddirectlyapply(1) totheApérysetsofthefamily

T

2 andobtainthesameresult.

Evenmore,we couldconsider(1) forallthefamilieswe analyzeinthiswork,butweprefer toleaveittothereaderasa computationexercise.

4. Quadrupletsintheformofprimequadruplets

As fortriplets, there areseveralcases ofquadruplets fulﬁllingthecondition ofthe admissibleconstellation. Sincethe reasoningsandcomputationsaresimilaronallofthem(see[14]),we willonlystudythosedirectlyassociatedwithprime quadruplets, notforgetting that the resultsare, infact, validfor anyquadrupletssatisfying thecorresponding admissible constellationcondition.

Itiswell-knownthataprimequadrupletisoftheform

(

p

,

p

+

²

,

p

+

⁶

,

p

+

⁸

)

.Inthefollowingresult,weimprovethis expression.

Proposition4.1.Wehavethatif

(

p

,

p

+

²

,

p

+

⁶

,

p

+

⁸

)

isaprimequadruplet,theneitherp

=

^4k

+

^{5 or}^p

=

^4k

+

^{7 for}^some^k

∈ N

^.

Proof. Itisclearthatifp

∈ N \ {

¹

}

^,^then^there^exists^k

∈ N

^such^that^p

=

^4k

+

^i,^withⁱ

∈ {

⁰

,

1

,

2

,

3

}

^.^But,^since^p

,

p

+

²

,

p

+

6

,

p

+

^{8 are}^prime ^numbers,^then ⁱ

=

^{0 and} ⁱ

=

^2.În âddition,^let ûs^note ^that

(

1

,

3

,

7

,

9

)

and

(

3

,

5

,

9

,

11

)

are not prime quadruplets.

Incontrasttothecaseoftriplets,auniqueformdeﬁnestheprimequadruplets.However,weneedtwodifferentexpres- sionstosolvetheFrobeniusproblem.

• (

^4k

+

⁵

,

4k

+

⁷

,

4k

+

¹¹

,

4k

+

¹³

)

,withk

∈ N

^.

• (

^4k

+

⁷

,

4k

+

⁹

,

4k

+

¹³

,

4k

+

¹⁵

)

,withk

∈ N

^.

Thus,since gcd

(

4k

+

⁵

,

4k

+

⁷

) =

^gcd

(

4k

+

⁷

,

4k

+

⁹

) =

^{1 for}^all^k

∈ N

^,^we^have^two^families ^of^numerical^semigroups⁽

Q

1 and

Q

2).

•

^S

∈ Q

1 ifS

=

^4k

+

⁵

,

4k

+

⁷

,

4k

+

¹¹

,

4k

+

¹³

^,^with^k

∈ N

^.

•

^S

∈ Q

2 ifS

=

^4k

+

⁷

,

4k

+

⁹

,

4k

+

¹³

,

4k

+

¹⁵

^,^with^k

∈ N

^.

Let us observe that

⁵

,

7

,

11

,

13

^and

⁷

,

9

,

13

,

15

âre ^numerical ^semigroups ^with êmbedding ^dimension êqual ^to ^four.

Moreover,since4k

+

¹³

<

2

(

4k

+

⁵

)

and4k

+

¹⁵

<

2

(

4k

+

⁷

)

forallk

≥

^1,^we^can^assert^that^e

(

S

) =

^{4 for}^every ^S

∈ Q

1

∪Q

2.

Remark4.2.We have S

=

¹

,

3

,

7

,

9

=

¹

^and ^S

=

³

,

5

,

9

,

11

=

³

,

5

,

11

^. ^Thus, ^e

(

S

) <

4 inboth cases. Thisis another reasonforeliminating

(

1

,

3

,

7

,

9

)

and

(

3

,

5

,

9

,

11

)

aspossibilitiesinProposition4.1.

Remark4.3.ItispossibletoimproveProposition4.1alittlemore.Thus,ifweassumethattheelementsonthequadruplet arenotdivisiblebytwoorthree,thenwehavethateither p

=

^12k

+

^{5 or}^p

=

^12k

+

^{11 for}^some^k

∈ N

^.Êven^more,îf^we considerthat theelements arenot divisiblebytwo, threeorfive,thenall primequadruplets(except

(

5

,

7

,

11

,

13

)

)are of theformp

=

^30k

+

^{11 for}^some^k

∈ N

^.Înâny^case,^weôbtainessentiallythesameconclusionsforthefamilies

Q

1 and

Q

2 associatedwithp

=

^12k

+

^{5 and} ^p

=

^12k

+

^11.

4.1. Firstexpression(family

Q

1)

LetS beanumericalsemigroupoftheformS

=

^4k

+

⁵

,

4k

+

⁷

,

4k

+

¹¹

,

4k

+

¹³

^,^where^k

∈ N

^.^Again^we^omit^the^proofs oftheresultsbecausetheyaresimilartothoseofSubsection3.1.

(7)

Remark4.4.Thenumerical semigroup S

=

⁵

,

7

,

11

,

13

^behaves ^somewhat differentlyfromthe restofthe cases.Infact, Ap

(

S

,

5

) = {

⁰

,

7

,

11

,

13

,

14

}

^and

1. PF

(

S

) = {

⁶

,

8

,

9

}

^; 2. F

(

S

) =

^9;

3. g

(

S

) =

^7;

4. t

(

S

) =

^3.

Itisclearthatexceptthevalueofg

(

S

)

,Corollary4.8doesnotgivetheseresults.

Lemma4.5.Ifk

∈ N \ {

⁰

}

1. 3

(

4k

+

⁷

) =

²

(

4k

+

⁵

) +

¹

(

4k

+

¹¹

)

; 2. 3

(

4k

+

¹¹

) =

¹

(

4k

+

⁷

) +

²

(

4k

+

¹³

)

;

3.

(

k

+

²

)(

4k

+

¹³

) = (

^k

+

³

)(

4k

+

⁵

) +

¹

(

4k

+

¹¹

)

; 4. 1

(

4k

+

⁷

) +

¹

(

4k

+

¹¹

) =

¹

(

4k

+

⁵

) +

¹

(

4k

+

¹³

)

; 5. 1

(

4k

+

⁷

) + (

^k

+

¹

)(

4k

+

¹³

) = (

^k

+

⁴

)(

4k

+

⁵

)

; 6. 2

(

4k

+

⁷

) +

¹

(

4k

+

¹³

) =

¹

(

4k

+

⁵

) +

²

(

4k

+

¹¹

)

;

7. 1

(

4k

+

¹¹

) + (

^k

+

¹

)(

4k

+

¹³

) = (

^k

+

²

)(

4k

+

⁵

) +

²

(

4k

+

⁷

)

; 8. 2

(

4k

+

¹¹

) +

^k

(

4k

+

¹³

) = (

^k

+

³

)(

4k

+

⁵

) +

¹

(

4k

+

⁷

)

.

∈ N \ {

⁰

}

^and^S

=

^4k

+

⁵

,

4k

+

⁷

,

4k

+

¹¹

,

4k

+

¹³

^,^then

Ap

(

S

,

4k

+

⁵

) = {

^a

(

4k

+

⁷

) +

^b

(

4k

+

¹¹

) +

^c

(

4k

+

¹³

) | (

^a

,

b

,

c

) ∈

^C

} ,

whereC

=

^Ca

∪

^Cb

∪

^Cc,withC_a

=

{

¹

} × {

⁰

} × {

⁰

,

1

, . . . ,

k

}

∪ {(

²

,

0

,

0

)}

^,^Cb

= ({

⁰

} × {

¹

,

2

} × {

⁰

,

1

, . . . ,

k

}) \ {(

⁰

,

2

,

k

)}

^,^and Cc

= {

⁰

} × {

⁰

} × {

⁰

,

1

, . . . ,

k

+

¹

}

^.

Letusobservethat

• (

^4k

+

⁷

) +

^c

(

4k

+

¹³

) = (

^c

+

¹

)(

4k

+

¹³

) −

^6,^for^all^c

∈ {

⁰

,

1

, . . . ,

k

}

^;

• (

^4k

+

¹¹

) +

^c

(

4k

+

¹³

) = (

^c

+

¹

)(

4k

+

¹³

) −

^2,^for^all^c

∈ {

⁰

,

1

, . . . ,

k

}

^;

•

²

(

4k

+

¹¹

) +

^c

(

4k

+

¹³

) = (

^c

+

²

)(

4k

+

¹⁵

) −

^4,^for^all^c

∈ {

⁰

,

1

, . . . ,

k

−

¹

}

^. Therefore,wecangivetheApérysetbyarrangingitselementsinincreasingorder.

∈ N \ {

⁰

}

^and^S

=

^4k

+

⁵

,

4k

+

⁷

,

4k

+

¹¹

,

4k

+

¹³

^,^then

Ap

(

S

,

4k

+

⁵

) = {

⁰

; (

^4k

+

¹³

) −

⁶

, (

4k

+

¹³

) −

²

,

4k

+

¹³

;

²

(

4k

+

⁷

) ;

2

(

4k

+

¹³

) −

⁶

,

2

(

4k

+

¹³

) −

⁴

,

2

(

4k

+

¹³

) −

²

,

2

(

4k

+

¹³

) ; . . . ;

(

k

+

¹

)(

4k

+

¹³

) −

⁶

, (

k

+

¹

)(

4k

+

¹³

) −

⁴

, (

k

+

¹

)(

4k

+

¹³

) −

²

, (

k

+

¹

)(

4k

+

¹³

) }.

∈ N \ {

⁰

}

^and^S

=

^4k

+

⁵

,

4k

+

⁷

,

4k

+

¹¹

,

4k

+

¹³

^,^then

1. PF

(

S

) = {

^4k

+

⁹

,

4k²

+

^13k

+

²

,

4k²

+

^13k

+

⁴

,

4k²

+

^13k

+

⁶

,

4k²

+

^13k

+

⁸

}

^; 2. F

(

S

) =

^4k²

+

^13k

+

^8;

3. g

(

S

) =

^2k²

+

^8k

+

^7;

4. t

(

S

) =

^5.

Remark4.9.ByCorollary4.8,ifp

=

^4k

+

^{5 for}^some^k

∈ N \ {

⁰

}

^,^then^F

(

p

,

p

+

²

,

p

+

⁶

,

p

+

⁸

) =

^p²⁺^3p₄ ⁻⁸^. Weﬁnishwithanillustrativeexample.

Example4.10.For k

=

^{24 we} ^have ^S

=

¹⁰¹

,

103

,

107

,

109

^and, ^by ^Corollary ^4.8, ^PF

(

S

) = {

¹⁰⁵

,

2618

,

2620

,

2622

,

2624

}

^, F

(

S

) =

^2624,^and^g

(

S

) =

^1351. ^Moreover,^by ^Corollary ^4.7,^Ap

(

S

,

11

) = {

⁰

,

103

,

107

,

109

,

206

,

212

,

214

,

216

,

218

, . . . ,

2719

,

2721

,

2723

,

2725

}

^.