On Quotients of a More General Theorem of Wilson

#A93 INTEGERS 25 (2025)
ON QUOTIENTS OF A MORE GENERAL THEOREM OF WILSON
I an V. Mo ozo
The Ci y College o New Yo k, New Yo k, New Yo k
[email p o ec ed]
Recei ed: 11/27/24, Accep ed: 7/1/25, Published: 11/5/25
Abs ac
The basis o his wo k is a simple, ex ended co olla y o Wilson’s heo em. This
co olla y gene a es many mo e quo ien s han hose al eady gene a ed by Wilson’s
heo em, and i is o in e es o de i e how hey ela e o each o he and build on he
es ablished p ope ies o he o iginal quo ien s. The main esul s a e exp essions
o sums o hese quo ien s, modula cong uences ha ex end he esul s o Lehme ,
and gene a ing unc ions.
1. In oduc ion
In o de o e icien ly desc ibe he esul s p esen ed in his wo k, we will de ine
P1:= P∪ {1}, whe e Pis he se o all p imes, as he se o posi i e non-composi e
numbe s. Wi h his said, Wilson’s heo em is a p imali y es by which
W(n) = 1+(n−1)!
n
is an in ege i and only i n≥1 is a posi i e non-composi e. Also, {W(p) : p∈P1}
is he se o Wilson quo ien s. We may, in ac , gene alize his no ion by obse ing
ha (n−1)! = (n−k−1)! Qk
m=1(n−m) implies
W(n) = 1+(n−k−1)! Qk
m=1(n−m)
n.
The Qk
m=1(n−m) e m is a alling ac o ial, so we can implemen an expansion
o his p oduc , he coe icien s o which a e he S i ling numbe s o he i s kind
s(a, b), de ined by (x)n=Qn−1
k=0 (x−k) = Pn
k=0 s(n, k)xk. Thus, Qk
m=1(n−m) =
1
n(n)k+1 and
W(n) = 1+(n−k−1)! 1
n(n)k+1
n=1+(n−k−1)! Pk+1
i=0 s(k+ 1, i)ni−1
n,
DOI: 10.5281/zenodo.17535209
INTEGERS: 25 (2025) 2
and since s(k+ 1,0) = 0 o k≥0, we ha e
W(n) = 1+(n−k−1)! Pk+1
i=0 s(k+ 1, i + 1)ni
n.(1)
Obse ing ha Pk+1
i=1 s(k+ 1, i + 1)ni≡0 (mod n) and s(k+ 1,1) = (−1)kk!,
Wilson’s c i e ion ex ends o a co olla y wi h an addi ional pa ame e .
Co olla y 1. Gi en nonnega i e in ege s n≥1and k < n,
(−1)kk!(n−k−1)! ≡ −1 (mod n)
i and only i nis non-composi e.
Inciden ally, Wilson’s heo em is a co olla y o i s own co olla y when k= 0, and
we will hus e e o he mo e gene al Wilson-like quo ien s as
Mk(n) = 1+(−1)kk!(n−k−1)!
n,
whe e, by Co olla y 1, Mk(n) is an in ege i and only i n∈P1. The alues o
some o hese quo ien s a e lis ed in Table 3 a he end o he pape . No e ha
M0(n) = W(n). These quo ien s yield nonposi i e esul s o odd k, and he e o e
we can ge id o he signs by de ining
M+
k(n) = |Mk(n)|= (−1)kMk(n) = (−1)k+k!(n−k−1)!
n.(2)
We will b oadly e e o he unc ions Mk(n) and M+
k(n) as M-numbe s.
2. Sums o Mk(n) and M+
k(n)
One a ea o in es iga ion is he s udy o ini e sums o alues o he unc ions Mk
and M+
k, which we will b oadly e e o as Z-numbe s. Conside ing he sums
Z(n) =
n−1
X
k=0
Mk(n), Z+(n) =
n−1
X
k=0
M+
k(n),(3)
we can de i e wo heo ems conce ning hem.
Theo em 1. The unc ion Z(n)is an in ege o all n∈N.
P oo . I n∈P1, hen Mk(n)∈Zimplies Z(n)∈Zby Co olla y 1. O he wise,
conside he e m (−1)kk!(n−k−1)!, whe e nis composi e. I i is addi ionally
no equal o 4, we see ha
k!(n−k−1)! = k!(n−(k+ 1))(n−(k+ 2)) . . . (n−(n−1)) ,
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and hence,
k!(n−k−1)! ≡(−1)n−k−1k!(k+1)(k+2) . . . (n−1) ≡(−1)n−k−1(n−1)! (mod n).
Mo eo e , since n= 4, i ollows ha (n−1)! ≡0 (mod n), implying
(−1)kk!(n−k−1)! ≡0 (mod n).
I ollows ha
n−1
X
k=0
1+(−1)kk!(n−k−1)! ≡
n−1
X
k=0
1≡0 (mod n),
which, wi h he ac ha Z(4) = 1, shows ha Z(n)∈Z o all n∈N.
Theo em 2. The unc ion Z+(n)is a na u al numbe i and only i nis e en o
non-composi e.
P oo . I n∈P1, hen M+
k(n)∈Zimplies Z+(n)∈Zby Co olla y 1. O he wise,
conside he e m (−1)kk!(n−k−1)!, whe e nis composi e. I i is addi ionally
no equal o 4, he p eceeding p oo showed ha
k!(n−k−1)! ≡0 (mod n).
This implies ha
n−1
X
k=0
(−1)k+k!(n−k−1)! ≡
n−1
X
k=0
(−1)k≡1+(−1)n−1
2(mod n),
which, wi h he ac ha Z+(4) = 4, shows ha Z+(n)∈Zi and only i nis e en o
non-composi e. In his case, we ac ually ha e ha Z+(n)∈Nsince Z+(n)>0.
Some alues o Z(n) and Z+(n) a e lis ed in Tables 1 and 2 in he las sec ion.
2.1. Fo mula o Z(n)
Fu he mo e, we can de i e a closed o mula o Z(n). Le S=Pn−1
k=0 (−1)kk!(n−
k−1)!. Then
Z(n) = n−1
X
k=0
1
n!+S
n= 1 + S
n.
Now, di iding Sby (n−1)! yields
S
(n−1)! =
n−1
X
k=0
(−1)kk!(n−k−1)!
(n−1)! =
n−1
X
k=0
(−1)k
n−1
k.
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An iden i y o he same sum [1] ells us ha
n
X
k=0
(−1)k
n
k=(1 + (−1)n) (n+ 1)
n+ 2 , n ∈N0,
so
S
(n−1)! =1+(−1)n−1n
n+ 1 ,
and i ollows algeb aically ha
Z(n) = 1 + 1+(−1)n−1(n−1)!
n+ 1 .
We obse e ha Z(2n) = 1 and Z(2n−1) = 1 + (2n−2)!
n. The o me case is
ob iously a na u al numbe . In he la e case, n= 1 is a i ial case, and since
2n−2≥n o n≥2, we ha e ha n|(2n−2)! o all n∈N. This is a p oo o a
s onge e sion o Theo em 1: Z(n)∈N o all n∈N.
2.2. Fo mula o Z+(n)
We can also de i e a o mula o Z+(n) by ini ially se ing S=Pn−1
k=0 k!(n−k−1)!,
which p oduces
Z+(n) = n−1
X
k=0
(−1)k
n!+S
n=1+(−1)n−1
2n+S
n.
Di iding Sby (n−1)! p oduces
S
(n−1)! =
n−1
X
k=0
k!(n−k−1)!
(n−1)! =
n−1
X
k=0
1
n−1
k.
An iden i y om [1] ells us ha
n
X
k=0
1
n
k=n+ 1
2n
n
X
=0
2
+ 1 , n ∈N0,
so
S
(n−1)! =n
2n−1
n−1
X
=0
2
+ 1 ,
and i ollows algeb aically ha
Z+(n) = 1+(−1)n−1
2n+(n−1)!
2n
n
X
=1
2
.
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To deal wi h he emaining pa ial sum, we make use o he Le ch anscenden , a
special unc ion deno ed by Φ and de ined by Φ(z, s, a) = P∞
n=0
zn
(n+a)s. I has he
p ope y ha
Φ(z, s, a) = znΦ(z, s, n +a) +
n−1
X
=0
z
( +a)s
o ℜ(a),ℜ(s)>0, n∈N, and z∈C[2]. We employ his equi alence by se ing
a= 1, s= 1, and z= 2, which yields
Φ(2,1,1) = 2nΦ(2,1, n + 1) +
n−1
X
=0
2
+ 1 .
We also de ine he polyloga i hm o be he unc ion Lis(z), commonly gi en by he
Di ichle se ies Lis(z) = P∞
n=1
zn
ns. Because Φ(z, s, 1) = 1
zLis(z) [3], we ha e
Li1(2) = −iπ = 2n+1Φ(2,1, n + 1) +
n
X
=1
2
,
and hence, n
X
=1
2
=−iπ −2n+1Φ(2,1, n + 1) .
By subs i u ion we de i e ha
Z+(n) = 1+(−1)n−1
2n−(n−1)! 2Φ(2,1, n + 1) + 2−niπ.(4)
3. Modula Cong uences
3.1. Cong uences o Mk(p) and M+
k(p)
Be noulli numbe s Bka e signed a ional numbe s ha a e ubiqui ous in numbe
heo y and analysis. We can de ine hem p ecisely using he exponen ial gene a ing
unc ion x
ex−1=P∞
k=0
Bkxk
k!, in which hey a ise. When pis p ime, i is known ha
M0(p)≡W(p)≡B2(p−1) −Bp−1(mod p),
which is ob ained om he cong uence ela ion
p−1 + p W (p)≡pB (p−1) (mod p2)
by sub ac ion a e subs i u ing = 1 and = 2 [4]. Recalling Equa ion (1), i is
clea ha
(p−k−1)! Pk+1
i=2 s(k+ 1, i + 1)pi
p≡0 (mod p),

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so
M0(p)≡1+(p−k−1)!(s(k+ 1,1) + ps(k+ 1,2))
p
≡Mk(p) + s(k+ 1,2)(p−k−1)! (mod p).
The S i ling numbe s o he i s kind s(k+ 1,2) can be exp essed wi h ha monic
numbe s as s(k+ 1,2) = (−1)k+1k!Hk, which yields
M0(p)≡Mk(p)+(−1)k+1k!Hk(p−k−1)! ≡Mk(p) + Hk(mod p) (5)
by Co olla y 1. By sub ac ion,
Mk(p)≡B2(p−1) −Bp−1−Hk(mod p).(6)
F om Equa ion (5), we cons uc a cong uence analogous o Lehme ’s:
p−1 + p Mk(p)≡pB (p−1) −p Hk(mod p2).
Fo he unsigned M-numbe s, u ilizing Equa ion (2) p oduces he cong uences
M+
k(p)≡(−1)kMk(p)≡(−1)kB2(p−1) −Bp−1−Hk(mod p),(7)
(−1)k(p−1) + p M+
k(p)≡(−1)kpB (p−1) −p Hk(mod p2).
3.2. Cong uences o Z(n) and Z+(n)
Theo em 3. I nis composi e, hen Z(n)≡1 (mod n). I nis non-composi e,
hen Z(n)≡ −1 (mod n).
P oo . Recalling he o mulae Z(2n) = 1 and Z(2n−1) = 1 + (2n−2)!
n o n∈N,
obse e ha i n= 2k−1 is composi e, hen we can apply he same a gumen as
in he p oo o Theo em 1 o show ha 2k−1|(2k−2)!. Thus, 2k−1|(2k−2)!
k
since k∤2k−1, implying Z(n)≡1 (mod n) o n∈ P1. Howe e , i n= 2k−1 is
non-composi e, hen k(Z(2k−1) + 1) = 2k+ (2k−2)! ≡1−1≡0 (mod 2k−1)
by Wilson’s heo em. Thus, since k∤2k−1, i ollows ha Z(2k−1) + 1 ≡0
(mod 2k−1), implying Z(n)≡ −1 (mod n) o n∈P1, p o ing ou heo em.
F om Equa ion (6), we ha e ha
p−1
X
k=1
Hk≡
p−1
X
k=1
(M0(p)−Mk(p)) ≡(p−1)M0(p)−Z(p) + M0(p)
≡pM0(p)−Z(p)≡2+(p−1)! (mod p).
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Since 1 + (p−1)! ≡0 (mod p) by Wilson’s heo em,
p−1
X
k=1
Hk≡1 (mod p)
o pp ime. We can also conside he ecu si e na u e o ha monic numbe s gi en
by Hn= 1+ 1
nPn−1
k=1 Hk, which by ecu sion yields Hn+1 =1
n+1 +Hn. Subs i u ing
his in o he p e ious sum implies he ollowing cong uence o pp ime:
Hp≡1 + 1
p(mod p).
Theo em 4. We ha e he cong uence Z(2n−1) ≡0 (mod n)i and only i nis
non-composi e.
P oo . Since Z(2n−1) = 1 + (2n−2)!
n= 1 + (2n−2)(2n−3) · · · (n+ 1)(n−1)!,
i n∈P1, hen Z(2n−1) ≡1−(2n−2)!
n!(mod n) by Wilson’s heo em. Howe e ,
(2n−2)!
n!≡(n−2)! (mod n), so Z(2n−1) ≡1−(n−2)! (mod n), and hus
Z(2n−1) ≡0 (mod n)
by Co olla y 1.
Fo he o he di ec ion, obse e ha (2n−2)!
n≡(n−1)!(n−2)! ≡ −1 (mod n)
implies ha nis non-composi e by Co olla y 1. This is because i nwe e o he wise
composi e, hen (n−1)!(n−2)! ≡0 (mod n).
Theo em 5. Fo e en n= 2, he unc ion Z+(n)obeys wo cong uences:
Z+(2n)≡0 (mod 2n), n ∈ P1 {2},(8)
Z+(2n)≡n+ 1 (mod 2n), n ∈P {2}.(9)
P oo . Recalling Equa ion (3),
Z+(2n) =
2n−1
X
k=0
(−1)k+k!(2n−k−1)!
2n=1
n
n−1
X
k=0
k!(2n−k−1)!
due o he e en numbe o e ms and consequen ial symme y o he summands.
E iden ly, since n∈ P1 {2}and i n= 2 and n= 4, we ha e n2|(2n−k−1)! o
all 0 ≤k≤n−1. Addi ionally, 2 |k! o 2 ≤k≤n−1 and
(2n−1)! + (2n−2)!
2n2=(2n−2)!
n∈N
o n∈N. Wi h he cases Z+(4) = 4 and Z+(8) = 1536, his p o es Equa ion (8).
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Fo he second cong uence, again conside Z+(2n) = 1
nPn−1
k=0 k!(2n−k−1)!. Le
us selec dis inc 0 ≤k1< k2≤n−1 and ake he sum o wo e ms:
k1!(2n−k1−1)! + k2!(2n−k2−1)! = k1!(2n−k2−1)! (2n−k1−1)!
(2n−k2−1)! +k2!
k1!.
Se ing k2=k1+ 1, his becomes k1!(2n−k1−2)!(2n). Clea ly, 2n2|k1!(2n−
k1−2)!(2n). Because nis an odd p ime, he sum o Z+(2n) will ha e an odd
numbe o e ms, allowing us o pai e e y o he e m wi h i s succeeding e m as
desc ibed, lea ing only he las one, (n−1)!n!, unpai ed. The e o e, nZ+(2n)≡ −n!
(mod 2n2) by Wilson’s heo em, and since Wilson quo ien s o odd p imes a e
always odd, 2n2|n(1 + (n−1)! + n). A simple algeb aic manipula ion allows us o
see ha −n!≡n2+n(mod 2n2), which p o es Equa ion (9).
Theo em 6. We ha e he cong uence Z+(p)≡B2(p−1) −Bp−1−1
2Hp−1
2(mod p),
whe e pis an odd p ime.
P oo . I pis p ime, we in oke Equa ion (7) o obse e ha
Z+(p)≡
p−1
X
k=0
(−1)kB2(p−1) −Bp−1−Hk
≡B2(p−1) −Bp−1−
p−1
X
k=1
(−1)kHk(mod p).
Obse e ha
p−1
X
k=1
(−1)kHk=−1 + 1 + 1
2−1 + 1
2+1
3+. . . +1 + . . . +1
p−1
=1
2
p−1
2
X
n=1
1
n=1
2Hp−1
2
when pis an odd p ime, so we ge ou desi ed cong uence.
I p= 2, hen Z+(2) = 1 ≡B2−B1+ 1 (mod 2). I is wo h no ing ha his
cong uence bea s a close esemblance o he one o Mp−1
2(p). Thus, by combining
Equa ion (6) wi h he abo e cong uence, we ob ain he in e es ing ela ion
Z+(p)≡1
2M0(p) + Mp−1
2(p)(mod p).
4. Gene a ing Func ions
In his sec ion we de i e he exponen ial gene a ing unc ions o ou M-numbe s
and Z-numbe s.
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4.1. M-numbe s
Conside ing he sequences o quo ien s Mk(n) o n∈N, i s exponen ial gene a ing
unc ion1 o |x| ≤ R, whe e R > 0 is he adius o con e gence, is de i ed as ollows:
EG(Mk(n); x) =
∞
X
n=0
Mk(n+k+ 1)xn
n!=
∞
X
n=0 1+(−1)kk!n!
n+k+ 1 xn
n!.
No e ha he i s e m appea ing in he summand is Mk(k+ 1), as Mk(n) is
unde ined o n≤k. Also obse e ha
∞
X
n=0 1+(−1)kk!n!
n+k+ 1 xn
n!<
∞
X
n=0 1+(−1)kk!xn,
which has a adius o con e gence R= 1. The e o e, EG(Mk(n); x) has a posi i e
adius o con e gence o a leas 1. This allows he summa ion o be spli as
EG(Mk(n); x) =
∞
X
n=0
xn
n!(n+k+ 1) + (−1)kk!
∞
X
n=0
xn
n+k+ 1 ,
whe e, om he second summa ion e m abo e, he adius o con e gence, R, equals
1. As dic a ed by he de ini ions o he lowe incomple e gamma unc ion γ(a, x) =
xaP∞
n=0
(−x)n
n!(n+a)and he Le ch anscenden Φ(x, s, a) = P∞
n=0
xn
(n+a)s[2], we de i e
he exponen ial gene a ing unc ion by se ing a=k+ 1 and s= 1,
EG(Mk(n); x)=(−1)kk!Φ(x, 1, k + 1) −γ(k+ 1,−x)
xk+1 .
Fo he exponen ial gene a ing unc ion o he signless M+
k(p), i is no di icul
o see om Equa ion (2) ha
EG(M+
k(n); x)=(−1)kEG(Mk(n); x) = k!Φ(x, 1, k + 1) −γ(k+ 1,−x)
xk+1 ,
since he summa ions a e independen o k.
4.2. Z-numbe s
We can also de i e he exponen ial gene a ing unc ion o Z(n) o n∈Nand
|x| ≤ R, whe e R > 0 is he adius o con e gence,
EG(Z(n); x) =
∞
X
n=0
Z(n+ 1)xn
n!=
∞
X
n=0 1 + (1 + (−1)n)n!
n+ 2 xn
n!.
1No e ha o dina y gene a ing unc ions a e o li le u ili y in his ins ance, as hei adius o
con e gence is ze o.