Internat. J. Math. & Math. Scl.
VOL. 16, NO. 4 (1993) 763-774
763
ON A CONJECTURE OF ANDREWS
PADIIAVATHAIIIIAandT.G. SUOHA
Department
of Studies in MathematicsUniversity of
Mysore,
Manasagangotri Mysore-570 006,India(Received
July 15,1991)
ABSTRACT. In
this paper,weproveaparticularcaseofaconjectureof Andrewsontwo partition functionsAA, k,a(n
andBA, k,a(n
).KEY WORDS AND PHRASES.
Partitionfunctions.1991
AMS SUBJECT CLASSIFICATION CODE.
11P82.1.
INTRODUCTION.
For
aneveninteger,
letA,,k,a(n
denote thenumberof partitions ofnintoparts such thatnopart
0
(rood ,+ I) may berepeatedandnopartis _=0,=k(a-)(,+
I)rood[(2k-+I)(,/I)].For
anodd integer,
letA,k,a(n
denotethe number of partitions ofnintoparts suchthatnopart0 (rood
--)
may be repeated,nopart is+
(rood 2,+
2) andnopartis --0, +(2a-,)’ +
[rood(2t-
, +
1)(+
1)].Let B)hk, a(n
denotethenumber ofpartitions ofnof the form b +...+bswithb >_b+
1, no part 0(rood)+
1)isrepeated,bi-b +
k->-
+ withstrictinequality if+ lib
and-j+
/i -<
a-jfor _<j_< andYl + + 1, + -<
a- where’i
isthenumberof appearances i=ofjin the partition.
Andrews
[1]
conjectured thefollowingidentities forA,k,a(n)
andB,k,a(n
).CONJURE. For
<a<t<,
,,(.) A,,(.)
for0<n<(k+’-a+l)
2+
(k- A+
I)(A+
1), whileB,,k,a(n A,t,a(n) +
(k
+,-a+1)
whenn 2
+
(t-, +
1)(,+
1).This conjecture has been verified
[1]
for3<,
_<7, <t<min(-1,5), <a<_k.In
this paperweprove thecaset aof the above conjecture.2. PROOF.
We
provetheconjecturefort aby establishingthefollowingidentities.CASE
1.Let ,
beeven. Then(1) B,k,a(n)= AA, k,a(n)
forn<(a-)( +
1)(2) B,k,a(n AA, k,a(n
whenn (a-;)(, +
1)764 PADMAVATHAMMA AND T.G. SUDHA
(3) (4)
CASE
2. Let,
beodd.(s) () (;)
BA, k,a(n) AA, k,a(n)
forn_<.
BA, k,a(A +
I)=AA, k,a(A +
I)a, t,a( + +
e)=A,k,a( + +
e),f. [
(s) ,, [f
+ 1)]’’ +
A+I(9) hA, ,(.) AA, t,o(.), .
(2-+ ])() + e, O<T
(i0) For
n (2a-A+ 2)()
]AA, k,a(n)
whenk>aBA, k,a(n)
+
whenk aCASE
1. Let Abeeven.A+I
a
>-
and foranya
A-A----l-
andk>awhent=a=A+I
2
PROOF OF
(1).
LetPBA
t,a(n)
andPAA,
k,a(n)
denote the set of partitions enumeratedbyBA, k,a(n)
andAA, k,a(n
respectively. Toprove(1)
weprove thefollowing strongerresult.(11) PBA,
k,a(n)PA
A, (n) forn<(a-)(A +
1)k,a
In
factweshowthat bothareequalto(12) PD(n)U PE(n)
where
PD(n)
is the set of partitions ofninto distinct partsandPE(B)
isthe set of partitions ofnin whichonly(A+
I)canberepeated.From the definitionof
AA, t,a(n)
it is clear thatPA(")
isequal to(12).
AlsoPB(n)
impliesthat
e PD(n)
ifA+
isnotrepeatedand ePE(")
otherwise.Hence PB(B)C PD(n)U PE(B).
On
the otherhand,letPD(n).
I[. b+ +
bk+ +
bshasmorethantparts, thenn>l+2+.--+t=l+2+...+(+a), wheret= +a,
-(-a+l+-I-a)+(-+2++-l)+...+(++1)+1+2+...
=(A+I)+.-- +(A+I)+I+2+.-.
+(-a)
=a(A+l)+l+2+.--+(-a)>(a-)(A+l).
Thus for n<
(a-)(A +
1) and for efD(n),
no partition ofn contains more than tparts and hencetheconditiononb’sissatisfied.Letus nowverify the conditionA+I on
l’s
forPD(n).
Leta+O,
O<.
Iffi
>a- orE fi >
a-i=1 i=1
CONJECTURE OF ANDREWS 765
thenthenumber beingpartitionedis
_>2+3+
=e(+1)+2+3+...
+(-e)
>
(a-)(A +
I)if-e>_2.
-I
2for 2and
Let
Hence - -e .
<(-2)( +
).-e:
I.Thena=-lforPD(n). ]iIfri=1,2,’’"-I
dhence-I
liA-2=a-I
If
. Ji
2,thenthenumber being partitionedis>_2+3+.-. +(A-l)
(A-1+2)+(A-2+3)+...
+(+ +) (-
)(+
): e(+
): (-)( +
).Thus forn<(a
)( +
1),- fi<_A-3=a-2.
Proceeding on the same lines we can show that the other conditions on
f’s
are satisfied for partitions inPD(n).
This proves thatPD(n)CPB(n).
Similarly,PE(n)CPB(n). Hence
PROOF OF (2). Let p4(n) [reap. P’/(n)]
denote the set of partitionsenumerated by[reap. BA, t,a(n)]
but notbyB,t,a(n [resp. A,t,a(n)].
Thenweclaim’A{> Io +
{o->+
+{-o+>+{-o+ >]aza {> Io->{ +
>] for{o->{ +
Clearly ==a+(a-1)+-..+(-a+1)
PA(n)
butrPB(n
asit violatesthe condition onfs
whenj=,-a+l.
In
f&ctfA_a+l+...+fa=a-(-a):2a-a-(-a+l)=2a--l. On
theotherhand,
(a-)(+ 1), PB(n)
but it does not belong toPA(n)
since for partitions enumerated byA,k,a(n
nopartis _--(a-)(A +
1)mod[(2b-, +
1)(,+
1)].As
in the proofof(1),
wecan show that partitions r#a+(a-1)+...+(A-a+I)$PA(n
are thesame asthepartitions=
#(a-)(, +
1)PB(n).
This proves(2).
PROOF OF (3). To
prove(3)
we establish a bijection oflA(n
onto/B(n)
wheren
(a-)(A+
1)+O, O<A+
1.Now F4(n
impliesthat it violatesoneofthe conditionson]’sorb’s.
Let Sj(j
1,2,.,)
denote the condition -j-I-E fi
<-a-ji=j
andlet$denote thecondition
A+I
E fi
<-a-1i=j
766 PADMAVATHAMMA AND T.G. SUDHA
andlet $* be the conditionon b’s.
In
the following steps to+
2 weenumeratethe partitions inPA
violatingSA,...,SI,S
andS*and also give thenecessarybijection ofP4(t)
onto2
STEP
1. ConsiderSA: ].A + fA +1
_<2<a-. For a-
>2thereare nopartitionsin FAviolating
S,V
Ifa--I
then thesetof partitions violatingSA
is{(+ 1)++=: = PD(O)
withparts
<}u{(+e’)+(+1)+/,:,ePD(e--e’)
withparts<, 2_<’_<}. For
anelementin the first setweassociate(/1)/ in
P
whileforanelement in the second setwe associate(A+
1)+ ( + O’) +
inP.
STEP
2. ConsiderSA ].A + IA + IA + IA
_<4<a+
1. Fora >3thereareno-1 -1 +1
partitionsin
PA
violatingSA Leta-
1. Then the set of partitions violatingSA
iswith parts <
-1}
withparts <
-1)
withparts <
-1)
withparts <
-1}
Wenote that the partitions inthe first two sets
violate
SA" Forapartitioninthe third set we associate (A+
1)++
inPb
while weassociate (,+
1)+(
+1)+ inP
for apartitionin the lastset.
Leta
-
2. The set of partitions of 2(A+
1)+
OinP4
violatingSA isg-1
{(+
2)+ (+
1)++ (-
1)+
,:,e PD(e
withpartsFor
anelement inthe first setweassociate2(A+
1)+
inP/while
foranelement in the secondset weassociate 2(A+ 1)+(+O’)+
inP.
Proceedinglike thiswearriveat thefollowingstep.STEP .
ConsiderSl:fl+...+fA<_a-1.
Sincefi<_l
for all. i--I,2,---,A we havefl+12+... +].A_<A. Let ].1+].2 + +].A=A.
Then 1+2++A=A+I)>n.
Thus there areno partitions of n in PA in which all parts 1,2,...,A appear.
Let ]’1+"" +IA
A-1.Let
thedeletedpart
among
1,2,...,Abez. Consider(13) I+2+...+(z-1)+(z+I)+...+(A-I)+A=(-I)(A+I)+(A+I-z)withl_<A+I-_<A.
If
a- -
I, then theonlypartition ofnViolatingS is+(-)+...+(+)+(-)+...+2+
withA+l-z-eforwhichweassociate
(-I)(A+
1)+einvb.
When
a-< -I,
thereare no partitions of ,tviolatings
since(13)>
n.More generally,
if].I+...+].A
=A-y,2<y<_A-a, and if Xl,...,zy are the parts which are left out with _<:I
<x2 < <y -<
A, then(14) A+(A-l)+--.+(zy+l)+(zy-1)+-..+(z1+I)+(z 1-I)+...
+2+1(-
y)(A+
I)+ (A+1-zi)+
+(A+1-y)
CONJECTURE OF ANDREWS 767
Ifa <
;- ,
then thereare nopartitions ofnviolatingS since(14)
>n. Ifa-
y, thenn=(a-)(A+l)+(A+l-Zl)+ +(A+l-zy).
Therearenopartitions ofnviolatingS if (A+
l-l)+
+(A+l-y)
>O. The ptition(14)
violates
S1
when (A+ 1) + +
A+ zy)
Oand for this ptitionwe siate(--y)(A +
I)+(A+l--z1)+
+(A+l--Zy)
inP.
If (A
+
1-zl)+... +
(A+
1-zy)<
O, then there are no ptitions ofn violating S since pts havetoreated.
Let a-
>-y.
Then-y+l
Sa-
S-1
d theree noptitions ofnviolatingS sincefl+...+fA=A-ya-1.
STEP
+1.
ConsiderS:/I+-..+/A+ISa-1.
Clely/il
for i=1,2,..., dA
+2+-.-+(A++
a-. 1)=(+
Letl
1)(A+l)>n,+"+ A +
it follows+ ’
that therewhere/A
e+
no ptitions ofwith SaSna-violating.
SinceS if1 + + A +
A+
1. Thus letusconsiderthecwhen1 + +/A + A +
Ath+
a.Then the humoringptitionedis
>+2+...+(A-)+(A+)
=l+2+...+a+(-a)(A+l)+(A+1)
=(A+
1)+ +2+ +a>n.Thus thereare nopartitions ofnviolating$in thiscasealso.
More
generally, letfl+...+fA+l=A-y, fA+l=a
with l__A-a.Let
Xl,...,zy+a be theparts deletedamong 1,2,,,
with <Zl<2 < <+
a <A. Consider(15) ,(A+
)+ +(+)+
+(- )++ (u+ +
)+ (u
+,-)+times
+(z
l+l)+(zI-l)+-.-
+2+1-a(A+l)+(-c-)(A+l)+(A+1-Zl)+...+(A+l-z+) (-
u)(A+
I)+(A+I-i)+
+(A+I-+ ).
As
inthecofS we cshowthat thereenoptitions ofnviolatingSwhena-
islessor
eater
th-
d even whena-=-
d(A+1-Zl)+... +(A+l-z+a)is
ls oreater
thenO. If(A+1-zi)+
+(A+1-z+a)=O
then theptiton
on the extreleR hindsideof
(15)
violates Sfor whichwesiatethelt ptition of(15)
whichongs
toB"
+2.
We nowprovethat a ptition dolat the conditionS*on 6’s then itdolatoneof the nditions on
l’s.
Before prodngts
we first note that when >afor aptition of.
(a)(A +
I)+
O, O<A+
hating pts+2+
+
=I+2+...+(+a) wheret=
+a,la<.
(-)( +
)+++ + (-.)
>
(.-)(A +
)+A+ >.,Andhenthereenoptitions ofnviolating S*inthisce.
Thus it
suces
toconsider thece whent a. Ifa ptition violatS*
then thereexists a pition768 PADMAVATHAMMA AND T G SUDHA
(16)
n b+ +b;+ +b; +
k-+ +bk + +bs
andaninteger with -b +/_ < A
+
1. If+
k- >A+1, then the number being partitionedis>_(+)+ +(+l)+
>/(A
+
l)_>(a-+
1)(A+
1) >n.Thus let
+
k- <A+
1. If b <A+
then(16)
contains at least kparts <A and hence: fi ->
kwhich implies that suchapartitionviolates SI.
Let bi+k_l<A+l
andbi_>A+l.
Sincen=(a-)(A+l)+O,O<A+l,
the number of parts>
+1 amongbi,...,b +
k-1 is <a-.
Ifa-
parts areequal to+
1, thenI
A +1a-
and theremainingk-a
+.
parts are <A andhenceand suchapartitionviolates S.
If a partition ofa number violates $* and if there are parts >A
+
then the number being partitionedis(17)
(A+Xc)+
(A+ Zc 1)+
+(A+Zl)+y + +Yk-c
where a<
a-,l
< 1<2< < and Yl’"’Yt-
c are among 1,2, .,A. Sincebi bi +
k <A+
we have A+ za
Yk c<A+
whichimplieszc
Yk a< and hencezc
yt.
If Yk-
zc
> then(17)
is>_c(A
+
l)+
(k-c+ 1)+ +3+2+=(A+I)+( +B-+I)+...+2+I wherein= +,1_<<2.
c(A
+
1)+ (/-c+
1)(A+
l)++
2+..-+(-+c-
l)(/+I)(A+I)+I+2+-..
+(-+a-
1)--(k-+l)(A+l)+l+2+.-. +(-/+o-l):>n.
From
this it is clear that if a partition of(a-)(A+l)+O,
O<A+1, violatesS*
then it does notcontain apart >
+
andhence all thepartswill be among1,2,.. ,A+
1. Thisimplies thatfl+...+fA+l >k=aa-I
andhencesuchapartition violatesS. Thiscompletes the proofof
(3).
PROOF
OF(4).
Firstpartof(4)
canbeprovedon thesamelinesof(3).
The secondpartof(4)
isthecase k aofthe Conjecture.As
intheproofof(3)
we canshow that every partition inP
hasanassociateinP
except(a-+
I)(A+
1)andthisproves
(4).
CASE
2. Let Abeodd.PROOF
OF(5). We
prove(5)
by establishingthefollowingstrongerresult(18) PBA
(n)PD(n)= PA
(n) forn<A.From the definitions of
aA, k,a(n)
andBA, k,a(n
it is clear thatPAx ,t,a(n)= PD(n)
and thatPB(n)
CPD(n).
On the otherhand, ifxePD(n)
then$i < for 1,2,. ,A andfA +
0 as n<A.Also
CONJECTURE OF ANDREWS 769
I+I+ +IA<I
and
II+"’+I,=ll+."+I,_I+I,++’"+IA<_ .
+1=T
A+IBut
I1
+""+ !,
"
implies that thenumber beingpartitionedis_>+
2+...+ +
>,
Thus
I + + I,
<’ _1
<a- since’ __1
<a. ConsiderAs beforeif
12 + + l,_
thenthe numberbeingpartitioned >_2+
3+ + A >
Aand2
hence
12 + + !,- -< - -<
a-2+isince <a. Proceedinglikethiswearriveat!,
--f-+
< asn_<Afromwhichweobtain
IA + -< a---T-"
For =
6.PD(n)
and,<e
conditionon b’sis satisfied since nopartition of. hasmorethan A+I--2--parts. This proves that
PD(n)C PB(n)
and hence(5)
isestablished.PROOF
OF(6). From
the definitions ofAA, t,a(r,)
andl,,t,a(r,
it isclearthat_+Alwhen.
A+I4( +
)- +---
whena>--2--
A+I andP(A
+1)={(A +1)}
PROOF
OF (7). For
n (+ +
O),O<A+
2e’(.) -
A+32++ AI
r:r 6.PD(O)
withparts <-AI,
O<.,
a "-2"--A+I-3 A+3
._._!1 ._!, ,a
+1+ +, + +
:PD(O)
withpts < OA+I
1
A+I++
:PD(O)
da>
PB(n) {(A +
1)+:6.PD(O)}
Pb(-)
A
+
1), (A+
1)+
:r6.PD(
withparts < andaT
A +
l),(A+
1)+a’:a"6.PD(-)and ,
r+Ix
+3(A+l)
+
a-:a- 6..DtT#anda>T
770 PADMAVATHAMMA AN
’
G. SLIDHAWhen a :/r, theninthe conjecture becomes
4
+1) and4
+1)EP
hasno associateinPA
andthis establishes theconjecturewhen/ a 4+1
T"
PROOF
OF(9). Let
(2-+ )(-) +
O,O<--. Now. P()
implies violatesoneofthe conditions
81,. ",$4 + 1,8’8*’8**
where$**is thecondition"no
parts 0 (rood4+
1)arerepeated".
A
proof similartothat ofStep +
2ofeven4willshow that partitions violating5,* will also violate81.
Since nopartis _= 4+
(rood24+
2) for partitions enumerated byA4,k,a(n
wehave
!4 +
0 and hence $reduces to5’1.In
thefollowing steps to-,
weenumerate the partitions inPA
violating5"4 +
1"""’5’1’5"**
andalsogivethe bijectionofPA(n)
ontoPB(n).
2
STEP
1. Consider5"4 + 1:I4 + -<
_<(a--).
Clearly therearenopartitionsinPA
violating2
5’4 +
fora-4 .__1
_>1. Since4._1
isnotapartofpartitionsenumeratedbybothA4,1:,a(n)
and2
B4,1,a(n)
whena itfollows that therearenopartitionsviolating5"4 +
whena also.STEP
2. Consider5"4
1:!4-1 + ’4 + + ]’4 +
3<3<a 4-12 --f- --f-
T
F,
ora>_4____55
thereare nopartitionsinPA
violating5"4-1"
Ifa---,
then n--(4+ 1)+O,O<-
and the set of partitionsviolatingS4_
is{--+-- +
: EPD(O)}
For
each partition in the above set we associate (4+1)+t inP. Let a=-.
Thenn 2(4
+
1)+
O, O<4{__.__1
andthe set ofpartitions violating5’4
isWe
associate4+
1)+rP
for every partition inthefirst set while fora partition in the second setweassociate4 +
1)+ (- + 09 +
inP.
Proceedinglike thiswearriveatthe following step.
STEP __-"
Consider5"1:I1+’"+I4
<a-1.By
the definition of-A4,t,a(a)’ li-<
for all1,
.,
4exceptfor4__. But
_<!4 + -<
2a-4+
1. Thecase!4 +
> willbeconsidered in step--.
Henceletusnowassume4 + -<
1.In
this caseI1+...+I4_<4. Iffl+...+14=4,
then1+2+...+4=(4+1)=-(4+1)+-
>_
(a--1)(4 +
1)+
4-1
>n. Thus there are no partitions violating$1
in/"A" Let I1 + + !4
4-1 and let the deletedpartbez. Consider(19) +
2+...+(z- l)+(z+
1)+ +(4-1)+4(4
2)(-) +
(4+
z) where <(4+
t)_<4.If2a-4
+
<4-2 then(19)
is>nand hence there will be nopartitions ofn violating$1"
Clearly2a-4
+
#4-2. When2a-4+
>4- 2theonlypartition ofnviolating5,1
is4+(4-1)+... +(z+1)+(z-1)+...+2+ with
--z
0for whichweassociatethefollowingpartitionin
P
CONJECTURE OF ANDREWS
_
A+l 771 (a+l)+ +(a+l)+(+0)+.--.2---
A 3 times 2
More generally, let
I1 + +
<z <z2< <
z
<a
betheparts deletedamong1,2,..,a.
Thenand let Zl’
"’ zlt
with(20)
a+(a- 1)++(Zy+ 1)+(zF-
1)++(Zl + l)+(Zl-
l)+ +2+(a-
2y)(----1-) +
(a+ l-Zl)+
+(a+-zy).
If 2a-a
+
<a-2U then(20)
is>n d hence there e no ptitions of n violating S1. Aim 2a-A+l#A-2y.t
2a-A+l>A-2y. Then A-2y+l2a-A+lA-l. If 2a-A+l>A-2y+l then11
+.."+Ia=a-Ya-I
and hence there will no ptitions of n violating S I. If2a
a + a
2y+
and if (a+ Zl) + +
(a+ z)
>a_ +
O then(20)
is>n. On the otherhd, if (a+
l-Zl)+... +(A+l-u)<+O
thenMso
there e noptitions of n violatingI
since in this ce pts have to
reated.
Since+O<a+l
we note that(A
+ I) + +
(A+ )
A+
O isssible
onlyif(a)
1<A, z2:d zi>
fori=3,...(b)
1<dzi >
fori=2,...,(=)
=d >(d)
ri>
fori=1,...,In
eh of the ces(a)-(d)
the ptition on the left hd side of(20)
violatesS
for which werctivelyiatethefollowingptitionsin
P.
,(a+l)+...+(a+l),+(a+1-Zl)+(a+1-z3)+... +(a+1-zu) (a 221t + 1)
times,(A+l)+--.+(A+l),,+(A+l-Zl)+A--l+(A+l-z2)+... +(A+l-zy)
(a 22# 1)
times,,(a+l)+... +(A+l),+(a+l-z2)+.../(A+
,(a+
I)+-.-+(a+I),+ a--l+
(a+ I-,1)+...
+(a+1-1
(a- 22it- 1)
timesSTEP -.
Consider$**:’no
part 0 (da +
1)rerepeated’. Thi implie thatla + >- "
When
. a __1
there are no partition violating $** ince-
i not pt for p’titio enumeratedbybothAa, t,.(
andBa, k,.(
).Let.
a__3.
Thenn 2(a+
1)+ ,
<a__l.
The set of partition inPt
violating$**i{A + A++.:.
epD(+O
withpts <}
7 72 PADMAVATHAMMA AND T G SUDHA
For each of the above sets of partitions in
P4
we respectively associate the following sets of partitionsinU{(A+
1)+.:.PD(A+I+O)parts <-}
For
anygiven ’a’we cansimilarly enumerate the partitionsinP4
violating S**and alsocanobtain the bijection ofP4
ontop.
Theproofof(9)
nowfollowsfromSteps
to---.
PROOF
OF(10).
The first part of(10)
follows on a line similar to the proof of(9).
Thesecondpartof
(10)
isthecase t aof the conjecture.As
intheproofof(9)
we canshowthat everypartitionin
p
hasanassociateinP4
except(2a-+ 2)(--)
and this proves(10).
We
nowconsidersomenumericalexamples.EXAMPLE
1. Let 4, k 3 a,n(k +
X-2a+
)+(k-X+1)(X+ 1)--10.TABLE
1PA4,3,3
(n)PB4,3,3
(n)1 {1} {1}
2 {2} {2}
3 {3,2+1} {3,2+1}
4 {4,3+1} {4,3+I}
5 {4+1}O{3+2} {4+ 1}O{5}
6 {6,4+2}13{3+2-t-1}
7 {7,6+1,4+3}13{4+2+I} {7,6+ 1,4+3}13{5+2}
8 {8,7+1,6+2}13{4+3+ I} {8,7+1,6+2}13{5+3}
9 {9,8
+
1,7+
2,6+
3,6+2+
1}13{4+
3+
2} {9,8+
1,7+
2,6+
3,6+2+
1}0{5 +4}10 {9
+
1,8 +2,7 +3,7+2+
1,6+4,6+3+
I} {9+
1,8+2,7+3,7+
2+
1,6+4,6+3+
I}13{4+3+2+ I} O{10,5+5}
Accordingtotheproofsof
(1)-(4),
wehave(a) PA
4,3,3(n)PB4,3,3
(n) forn<4(b) P’A4,
3,3(5)
{3+
2},P’B4,
3, 3(5) {5}CONJECTURE OF ANDRENS 773
(c)
The partitions enumeratedbyA4,3,3(n)
for n 6,7,8,9violatingS2accordingtoStep
1in theproofof(3)
are(3+2+1}u{4+3+2}
forwhichtheirassociatesin
P
are(d)
The partitions enumerated byA4,3,3(n)
forn 6,7,8,9violatingS asprovedinStep
2 are(4+2+I)U(4+3+I) for whichthe correspondingpartitionsin
F
are(e)
ThepartitionsenumeratedbyA4,3,3(n)
forn 6,7,8,9 violatingSalsoviolate$I
or S2.(f)
The partition 10 2 (4+
I)EF’B4,
3,3(10)
hnosociate inP
whileall other partitionshave.
From
Table 1 it isclear that(a)-(f)
areindeedtrue.EXAMPLE
2. Let,X=5,t=a=3,n=(t+’-a+l) +
(-+
)(A+ )=9.TABLE
2PA5,3,3
(n)PB5,3,3
(n)1 {} {}
2 {2) {2)
3 (2+1)
4 {4)
5 {5,4+ 1) {5,4+1)
6 {5+ 1}U{4+2} {5+1)U(6)
7 {7,5+2}U{4+2+ I} {7,5+2}U{6+ 1}
8 {8,z+}u{s+2+} {s,z+}u{+}
9 {8+1,7+2,5+4} {8+ 1,7+2,5+4}U{9}
From
the proofs of(5)-(8)
wehavethe
following:(g) PA5,3,3
(n)PB5,3,3
(n) forn<5(h) P’ A5,3 3(6)
{4/2}PBs,
3,3(6)
{6}(i) P’A5,3 3(7)
{4+
2+
1}/B5,
3,3(7)
{6+
1}(j) PA5,
3,3(8)
{5+
2+
1}P’BS,3 3(8)
{6+
2}(k)
Thepartition(2x3-5+ 2)(-)=
9inPB5,
3,3(9)
hasnoassociate inPAs,
3,3(9)
whileallothers have.
From
Table 2it isevidentthatthe results(g)-(k)
aretrue.774 PADMAVATHAMMA AND T.G. SUDHA
ACKNOWLEDGEMENT.
The authors areindebtedandgratefulto ProfessorGeorge E.
Andrews for suggestingthe
problem and forhisconstanthelp duringthecourseofthe work.REFERENCES
ANDREWS, G.E.,
On the general Rogers-Ramanujantheorem,
No. 152