#A106 INTEGERS 25 (2025)
ON p-FROBENIUS NUMBERS FOR THE NUMERICAL
SEMIGROUPS GENERATED BY THREE CONSECUTIVE STAR
NUMBERS
Takao Koma su
Ins i u e o Ma hema ics, Henan Academy o Sciences, Zhengzhou, China;
Depa men o Ma hema ics, Ins i u e o Science Tokyo, 2-12-1 Ookayama,
Megu o-ku, Tokyo, Japan
[email p o ec ed]; [email p o ec ed]
Tapas Cha e jee
Depa men o Ma hema ics, Indian Ins i u e o Technology Ropa , Punjab, India
[email p o ec ed]
Palak Na ula
Depa men o Ma hema ics, Indian Ins i u e o Technology Ropa , Punjab, India
[email p o ec ed]
Recei ed: 12/30/24, Re ised: 5/22/25, Accep ed: 11/3/25, Published: 11/25/25
Abs ac
The F obenius coin p oblem in ol es compu ing he la ges in ege , known as he
F obenius numbe , ha canno be exp essed as a non-nega i e in eg al linea com-
bina ion o gi en ela i ely p ime posi i e in ege s. A mo e gene alized e sion o
his p oblem, e med as he p-F obenius numbe , aims o ind he la ges in ege
ha has a mos p ep esen a ions in e ms o linea combina ions, whe e pis any
non-nega i e in ege . In his a icle, we gi e he closed- o m exp essions o he p-
F obenius numbe s o he nume ical semig oups gene a ed by he iple s o he
consecu i e s a numbe s o he cases p= 0,1,and 2.Also, we p esen explici
o mulas o hei p-Syl es e numbe s (which coun he posi i e in ege s ha ing no
mo e han p ep esen a ions) whe e p= 0,1,and 2.
1. In oduc ion
The S a numbe s, deno ed by Sn,a e commonly e e ed o as he cen e ed 12-
gonal numbe s o cen e ed dodecagonal numbe s. The o mula exp essing he n h
s a numbe is gi en by Sn= 6n(n−1)+1,whe e n≥1. The i s ew s a numbe s
DOI: 10.5281/zenodo.17711576
INTEGERS: 25 (2025) 2
[21, A.131] a e as ollows
{Sn}n≥1= 1,13,37,73,121,181,253,337,433,541,661,793,937,1093,
1261,1441,1633,1837,2053, . . . .
These numbe s appea in many numbe heo e ic p oblems. Some well-known o -
mulas include [27, OEIS A306980]
∞
X
n=1
1
Sn
=π an(π/2√3)
2√3,
∞
X
n=0
Sn
n!= 7e, and
∞
X
n=1
Sn
2n= 25.
One can also easily p o e he i s iden i y abo e using he Cauchy’s esidue heo em.
Geome ically, he n h s a numbe consis s o a cen al poin along wi h 12 copies
o he (n−1) h iangula numbe n−1.A no able obse a ion is ha in ini ely many
s a numbe s a e also iangula numbe s and among he ini ial ins ances, we ha e
S1=1= 1,S7= 253 = 22,S91 = 49141 = 313, and S1261 = 9533161 = 4366 on
he OEIS en y A156712. The s a numbe s a e used o a new se o ec o - alued
Teichm¨ulle modula o ms, de ined on he Teichm¨ulle space, s ic ly ela ed o he
Mum o d o ms, which a e holomo phic global sec ions o he ec o bundle [19].
Le A={a1, a2, . . . , ak}be a se o ela i ely p ime posi i e in ege s, whe e
k≥2, and le pbe any non-nega i e in ege . The p-nume ical semig oup Sp(A)
is de ined as he se o in ege s whose non-nega i e in eg al linea combina ions
in e ms o gi en posi i e in ege s a1, a2, . . . , akcan be exp essed in mo e han
pways [18]. Fo some backg ound on he numbe o ep esen a ions, e e , e.g.,
[4,6,10,28]. Fo he se o non-nega i e in ege s N0, he se Gp(A):=N0 Sp(A)
is ini e i and only i gcd (a1, a2, . . . , ak) = 1. Then, he maximum elemen o he
se Gp(A),deno ed by gp(A) is called he p-F obenius numbe . The ca dinali y o
he se Gp(A) is called he p-Syl es e numbe (o he p-genus) and is deno ed by
np(A).This kind o concep is a gene aliza ion o he amous Diophan ine p oblem
o F obenius [2,25], since p= 0 is he classical case, whe e he o iginal F obenius
numbe is deno ed by g(A) = g0(A) and he genus as n(A) = n0(A).He e, he se
Ais called he sys em o gene a o s o he p-nume ical semig oup Sp(A).
When k= 2, he e exis s an explici closed o mula o he p-F obenius numbe
o any non-nega i e in ege p[3]. Howe e , o k= 3, he p-F obenius numbe
canno be gi en by any se o closed o mulas ha can be educed o a ini e se o
ce ain polynomials [9]. Since i is e y di icul o gi e a closed explici o mula o
any gene al sequence o h ee o mo e a iables, many esea che s ha e ied o ind
he F obenius numbe s o some special cases (see, e.g., [14,22,23,24] o mo e
de ails). Recen ly, in [7,8], he F obenius numbe s o he iple s o successi e
cen e ed iangula , cen e ed squa e, cen e ed pen agonal, and cen e ed hexagonal
numbe s we e s udied. Though i is e en mo e di icul when p>0 (see, e.g., [12,15,
16,17]), in [11], he p-F obenius numbe s o h ee consecu i e iangula numbe s
INTEGERS: 25 (2025) 3
we e s udied. In his pape , he p-F obenius numbe s o he h ee consecu i e s a
numbe s a e examined. Ini ially, he ocus is on unde s anding he s uc u e o
he p-Ap´e y se o he classical case, ollowed by o he posi i e in eg al alues o
p= 1,2. Addi ionally, we p esen explici o mulas o hei p-Syl es e numbe s
o p= 0,1,2.
2. P elimina ies
In his sec ion, we ecall he no ion o he p-Ap´e y se [1] and some esul s o
compu e he p-F obenius numbe and p-Syl es e numbe which will play a c ucial
ole in p o ing he main heo em. Le us de ine he p-Ap´e y se .
De ini ion 1. Conside a se o posi i e in ege s A={a1, a2, . . . , ak}(k≥2) wi h
gcd(A) = 1. Wi hou loss o gene ali y, assume ha a1= min(A) and le pbe any
non-nega i e in ege . Then, he p-Ap´e y se o A, deno ed by App(A),is de ined as
App(A) = Ap(a1, a2, . . . , ak)={m(p)
0, m(p)
1, . . . , m(p)
a1−1},
whe e m(p)
iis he leas posi i e in ege o Sp(A),and sa is ies he cong uence m(p)
i≡
i(mod a1), o 0 ≤i≤a1−1. This de ini ion is equi alen o saying ha m(p)
i∈
Sp(A), m(p)
i−a1∈ Sp(A),and m(p)
i≡i(mod a1). No e ha m(0)
0is de ined o be
0. I ollows ha o gi en p,
App(A)≡ {0,1, . . . , a1−1}(mod a1).
In o he wo ds, App(A) o ms a comple e esidue sys em modulo a1.
One o he con enien o mulas o ob ain he p-F obenius numbe and he p-
Syl es e numbe (o p-genus) is ia he elemen s in he co esponding p-Ap´e y se .
The lemma gi en below desc ibes he ela ionship be ween he F obenius numbe
and he Syl es e numbe wi h he associa ed Ap´e y se [13].
Lemma 1. Le gcd(a1, . . . , ak)=1wi h a1= min{a1, . . . , ak}. Then, we ha e
gp(a1, . . . , ak) = max
0≤j≤a1−1m(p)
j−a1,
np(a1, . . . , ak) = 1
a1
a1−1
X
j=0
m(p)
j
−a1−1
2.
Rema k 1. When p= 0 (classical case), he F obenius numbe is essen ially due
o B aue and Shockley [5], and he classical Syl es e numbe is due o Selme
[26]. Mo e gene al o mulas, including he p-powe sum and he p-weigh ed sum,
can also be seen in [13].
INTEGERS: 25 (2025) 4
In o de o discuss he F obenius numbe o iples o successi e s a numbe s, we
mus ensu e ha hey a e ela i ely p ime. He e, we gi e a lemma ha a icula es
his equi emen .
Lemma 2. Fo any h ee consecu i e s a numbe s Sn, Sn+1, and Sn+2,
gcd(Sn, Sn+1, Sn+2)=1.
P oo . We know ha
gcd(Sn, Sn+1, Sn+2) = gcd(gcd(Sn, Sn+1),gcd(Sn+1, Sn+2)).
Le dn= gcd(Sn, Sn+1) and obse e ha (n+ 1)Sn−(n−1)Sn+1 = 2. This
implies dn|2 and since s a numbe s a e odd, we conclude ha dn= 1. Thus,
gcd(Sn, Sn+1) = 1 and hence, gcd(Sn, Sn+1, Sn+2) = 1.
In addi ion, in a la e sec ion we use a classical iden i y known as B´ezou ’s
Lemma, which is s a ed as ollows.
Lemma 3 ([20]).Le aand bbe in ege s wi h g ea es common di iso d. Then
he e exis in ege s xand ysuch ha ax +by =d. Mo eo e , he in ege s o he
o m az +b a e exac ly he mul iples o d.
3. Main Resul s
Now, we de i e he explici exp essions o he p-F obenius numbe s and he p-
Syl es e numbe s o he iples consis ing o successi e s a numbe s, discussed in
Sec ion 3.1 and Sec ion 3.2, espec i ely.
3.1. p-F obenius Numbe s
The p-F obenius numbe s o he nume ical semig oups gene a ed by h ee consecu-
i e s a numbe s a e gi en as ollows.
Theo em 1. Fo p= 0,1,2, we ha e
gp(Sn, Sn+1, Sn+2) = (2nSn+1 + (p+ 2)nSn+2 −Sn,i 6≤n≤9;
(2n−11)Sn+1 + (p+ 3)nSn+2 −Sn,i n≥10.
Rema k 2. Mo e explici ly, we can w i e
gp(Sn, Sn+1, Sn+2) = (24n3+ 42n2+ 34n−1 + (6n3+ 18n2+ 13)p, i 6 ≤n≤9;
30n3−6n2−19n−12 + (6n3+ 18n2+ 13)p, i n≥10.
When p≥3, he si ua ion becomes complex, and no explici o mula has been
de i ed so a . The cases whe e n= 2,3,4,5 a e discussed la e .
INTEGERS: 25 (2025) 5
P oo . Ou main goal is o ind he p-Ap´e y se and es ablish he alidi y o he
heo em case by case o n≥6.
Case 1: p= 0. Fo con enience, subs i u e y,z := ySn+1 +zSn+2 o non-nega i e
in ege s yand z. Then, we can show ha he elemen s o he 0-Ap´e y se a e gi en
as in Table 1.
0,0··· 2n−11,0 2n−10,0··· ··· 2n,0
.
.
..
.
..
.
..
.
.
0,2n··· 2n−11,2n 2n−10,2n··· ··· 2n,2n
0,2n+1 ··· 2n−11,2n+1
.
.
..
.
.
.
.
..
.
.
0,3n··· 2n−11,3n
Table 1: Ap0(Sn, Sn+1, Sn+2)
Fi s ly, we p o e ha he elemen s y,z in Table 1 o m a comple e esidue sys em
modulo Sn.To p o e his we show ha any such wo elemen s in he able a e
incong uen modulo Sn.Assume o a con adic ion ha he e exis o de ed pai s
(y1, z1) and (y2, z2) such ha y1,z1= y2,z2in Table 1wi h
y1,z1≡ y2,z2(mod Sn).(1)
Subs i u ing he alues o y,z in Equa ion (1), we ha e
y1Sn+1 +z1Sn+2 ≡y2Sn+1 +z2Sn+2 (mod Sn).
Re-a anging he abo e equa ion, we ge
(y1−y2)Sn+1 + (z1−z2)Sn+2 ≡0 (mod Sn).
This implies
(y1−y2)(Sn+1 −Sn)+(z1−z2)(Sn+2 −Sn)≡0 (mod Sn).
Consequen ly,
12n(y1−y2) + 12(2n+ 1)(z1−z2)≡0 (mod Sn).
Fu he mo e, gcd(12, Sn)=1,and hence
n(y1−y2) + (2n+ 1)(z1−z2)≡0 (mod Sn).(2)
Fo y1, y2, z1, z2as in Table 1, we ha e
|y1−y2| ≤ 2nand |z1−z2|≤3n. (3)
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F om Equa ion (2), n(y1−y2) + (2n+ 1)(z1−z2) is a mul iple o Sn, and unde
he cons ain s o Equa ion (3), he only possible alues a e 0, Sn,o −Sn. I
n(y1−y2) + (2n+ 1)(z1−z2) = 0, i ollows ha y1,z1= y2,z2which con adic s
ou ini ial assump ion. Now, conside he case when
n(y1−y2) + (2n+ 1)(z1−z2)=Sn= 6n2−6n+ 1.(4)
Using B´ezou ’s lemma, he ex ended Euclidean algo i hm, and he inequali ies in
Equa ion (3), he only in eg al solu ion o Equa ion (4) is
(y1−y2, z1−z2) = (2n−10,2n+ 1).
As a esul ,
y1= 2n−10 + y2,
z1= 2n+1+z2.
Since y2and z2a e non-nega i e in ege s, i ollows ha y1,z1lies ou side Table
1. Hence, ou ini ial assump ion was w ong. On a simila line o easoning, we
can a gue ha n(y1−y2) + (2n+ 1)(z1−z2)=−Sn.Thus, no wo elemen s in
Table 1a e cong uen modulo Sn. Addi ionally, no e ha he coun o elemen s in
Table 1is equal o (2n+ 1)2+n(2n−10) = Sn.The e o e, we ob ain ha he se
{ y,z :y, z ∈Table 1}cons i u es a comple e esidue sys em modulo Sn.In o he
wo ds, we show ha o each i∈ {0,1,...,Sn−1}, he e exis s a unique elemen y,z
in Table 1such ha y,z ≡i(mod Sn).
We now p oceed o p o e ha he elemen s in Table 1a e indeed he smalles
ones in hei co esponding esidue classes. Obse e ha
(2n−10)Sn+1 + (2n+ 1)Sn+2 = (4n+ 3)Sn,(5)
(2n+ 1)Sn+1 −nSn+2 = (n+ 1)Sn,(6)
(3n+ 1)Sn+2 −11Sn+1 = (3n+ 2)Sn.(7)
Consequen ly,
2n−10+y,z ≡ y,z−2n−1(mod Sn) and 2n−10+y,z > y,z−2n−1
(2n+ 1 ≤z≤3n, 0≤y≤10),
2n+1+y,z ≡ y,n+z(mod Sn) and 2n+1+y,z > y,n+z
(0 ≤y≤2n−11,0≤z≤2n),
4n−9+y,z ≡ 2n−10+y,n+z(mod Sn) and 4n−9+y,z > 2n−10+y,n+z
(0 ≤y≤10,0≤z≤n),
y,z ≡ y−11,z+3n+1 (mod Sn) and y,z > y−11,z+3n+1
(0 ≤z≤n−1,11 ≤y≤2n).
INTEGERS: 25 (2025) 7
The e o e, he elemen s in Table 1 o m he 0-Ap´e y se Ap0(Sn, Sn+1, Sn+2).
Now, om Table 1, he e a e wo possibili ies o he la ges alue o he se
Ap0(Sn, Sn+1, Sn+2): 2n,2no 2n−11,3n.No e ha , 2n,2n< 2n−11,3ni and only
i nSn+2 >11Sn+1,which is equi alen o 6n3−48n2−53n−11 >0. Obse e ha
he oo s o he equa ion 6n3−48n2−53n−11 = 0 a e −0.7214,−0.2822,and
9.0036. The e o e, o 6 ≤n≤9, 2n,2nis he la ges elemen o he Ap´e y se , so
by Lemma 1, we ha e ha
g0(Sn, Sn+1, Sn+2)=2nSn+1 + 2nSn+2 −Sn.
While o n≥10, we ha e 2n−11,3nis he maximum among all he elemen s, so
using Lemma 1
g0(Sn, Sn+1, Sn+2) = (2n−11)Sn+1 + 3nSn+2 −Sn.
Case 2: p= 1. The elemen s o he 1-Ap´e y se can be de e mined om hose o
he 0-Ap´e y se as ollows.
2n+1,0··· ··· 4n+1,0
.
.
..
.
.
.
.
..
.
.
2n+1,n ··· ··· 4n+1,n
2n+1,n+1 ··· 4n−10,n+1
.
.
..
.
.
2n+1,2n··· 4n−10,2n
2n−10,2n+1 ··· ··· 2n,2n+1
.
.
..
.
.
2n−10,3n··· ··· 2n,3n
0,3n+1 ··· 2n−11,3n+1
.
.
..
.
.
0,4n··· 2n−11,4n
Table 2: Ap1(Sn, Sn+1, Sn+2)
F om he se o Equa ions (5), (6), and (7) and he ela ion
(2n+ 12)Sn+1 + (2n+ 1)Sn= (4n+ 1)Sn+2,
we ha e he ollowing one- o-one co espondence be ween he elemen s o he 0-
Ap´e y se (on he le -hand side o cong uences) and ha o he 1-Ap´e y se (on
he igh -hand side o cong uences):
y,z ≡ y+2n+1,z−n(mod Sn)
(0 ≤y≤2n, n ≤z≤2n; 0 ≤y≤2n−11,2n+ 1 ≤z≤3n),
y,z ≡ y+2n−10,z+2n+1 (mod Sn) (0 ≤y≤10,0≤z≤n−1),
y,z ≡ y−11,z+3n+1 (mod Sn) (11 ≤y≤2n, 0≤z≤n−1).
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The elemen s in he i s n ows o Table 1a e di ided in o wo pa s. One pa is
simply mo ed below he 0-Ap´e y se o ill in he gap as shown in Table 2. Howe e ,
he emaining po ion is mo ed o he lowe le o he 0-Ap´e y se . Elemen s o he
han he i s n ows o Table 1a e shi ed o he igh side o he 0-Ap´e y se by
shi ing up by n ows.
Se τx,y,z := xSn+ySn+1 +zSn+2. We show ha all he elemen s o Table 2
ha e a leas wo di e en ep esen a ions. In ac , o 0 ≤y≤2n, n ≤z≤2nand
0≤y≤2n−11,2n+ 1 ≤z≤3n, we ha e
τn+1,y,z =τ0,y+2n+1,z−n.
Simila ly, we ge
τ4n+3,y,z =τ0,y+2n−10,z+2n+1 ( o 0 ≤y≤10,0≤z≤n−1),
τ3n+2,y,z =τ0,y−11,z+3n+1 ( o 11 ≤y≤2n,0≤z≤n−1).
F om Table 2, he e a e ou possible choices o he la ges alue o Ap1(Sn, Sn+1, Sn+2):
4n+1,n, 4n−10,2n, 2n,3n,and 2n−11,4n.
Since 18n2+ 18n−1>0,we ha e 4n+1,n < 2n,3nand 4n−10,2n< 2n−11,4n.
Analogous o he case p= 0,we ob ain 2n,3n< 2n−11,4ni and only i n>9.
The e o e, o 6 ≤n≤9, he maximum elemen is 2n,3n, and by Lemma 1,
g1(Sn, Sn+1, Sn+2) = 2nSn+1 + 3nSn+2 −Sn.
When n≥10, 2n−11,4nis he la ges o all he elemen s, and hence using Lemma
1,
g1(Sn, Sn+1, Sn+2) = (2n−11)Sn+1 + 4nSn+2 −Sn.
Case 3: p= 2. The elemen s o he 2-Ap´e y se can be de i ed om hose o he
1-Ap´e y se in he ollowing manne (see Table 3).
4n+2,0··· 6n−9,0 6n−8,0··· 6n+2,0
4n+2,1··· 6n−9,1
.
.
..
.
.
4n+2,n ··· 6n−9,n
4n−9,n+1 ··· 4n+1,n+1
.
.
..
.
.
4n−9,2n··· 4n+1,2n
2n+1,2n+1 ··· 4n−10,2n+1
.
.
..
.
.
2n+1,3n··· 4n−10,3n
2n−10,3n+1 ··· 2n,3n+1
.
.
..
.
.
2n−10,4n··· 2n,4n
0,4n+1 ··· 2n−11,4n+1
.
.
..
.
.
0,5n··· 2n−11,5n
Table 3: Ap2(Sn, Sn+1, Sn+2)
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Simila o he case when p= 1, he elemen s in he i s n ows o he main pa o
he 1-Ap´e y se a e subdi ided in o wo pa s and hen eloca ed benea h he wo
s ai case sec ions o he 1-Ap´e y se . Elemen s o he han he i s n ows o he
main po ion unde go a shi o he igh side o he 1-Ap´e y se which is achie ed
by mo ing up by n ows. The o he wo s ai case pa s a e shi ed o he igh by
(2n+ 1) s eps and upwa d by ns eps. Mo e p ecisely,
y,z ≡ y+2n+1,z−n(mod Sn)
(2n+ 1 ≤y≤4n+ 1, z =n; 2n+ 1 ≤y≤4n−10, n + 1 ≤z≤2n),
y,z ≡ y−11,z+3n+1 (mod Sn) (2n+ 1 ≤y≤2n+ 11,0≤z≤n−1),
y,z ≡ y−2n−12,z+4n+1 (mod Sn) (2n+ 12 ≤y≤4n+ 1,0≤z≤n−1),
y,z ≡ y+2n+1,z−n(mod Sn)
(0 ≤y≤2n−11,3n+ 1 ≤z≤4n; 2n−10 ≤y≤2n, 2n+ 1 ≤z≤3n).
Using Equa ions (5), (6), and (7), we can demons a e ha he elemen s o he
2-Ap´e y se possess a leas h ee dis inc ep esen a ions. Fo 0 ≤y≤2n, z = 2n
and 0 ≤y≤2n−11,2n+ 1 ≤z≤3n, we ha e
τ2n+2,y,z =τn+1,y+2n+1,z−n=τ0,y+4n+2,z−2n.
Simila ly, we ha e
τ4n+3,y,z =τ3n+2,y+2n+1,z−n=τ0,y+2n−10,z+2n+1
(0 ≤y≤10, n ≤z≤2n−1),
τ3n+2,y,z =τ2n+1,y+2n+1,z−n=τ0,y−11,z+3n+1
(11 ≤y≤2n, n ≤z≤2n−1),
τ5n+4,y,z =τn+1,y+2n−10,z+2n+1 =τ0,y+4n−9,z+n+1
(0 ≤y≤10,0≤z≤n−1),
τ4n+3,y,z =τn+1,y−11,z+3n+1 =τ0,y+2n−10,z+2n+1
(11 ≤y≤2n, 0≤z≤n−1) .
In Table 3, by compa ing he six candida es
2n−11,5n, 2n,4n, 4n−10,3n, 4n+1,2n, 6n−9,n,and 6n+2,0,
we ind ha 2n,4nis he la ges elemen o he 2-Ap´e y se , when 6 ≤n≤9,and
2n−11,5nis he la ges elemen o he 2-Ap´e y se , when n≥10. The e o e,
g2(Sn, Sn+1, Sn+2) = (2nSn+1 + 4nSn+2 −Sn,i 6 ≤n≤9;
(2n−11)Sn+1 + 5nSn+2 −Sn,i n≥10.
This comple es he p oo .
