#HomeWork#12
#Please do not post homework
#Abrar Almahmeed, March 7

#Question 1:

A41c1:=proc(n) option remember: local j,s:
if n<0 then
RETURN(0):
fi:

if n=0 then
RETURN(1):
fi:

s:=0:

for j from 1 while (3*j^2-j)/2 <= n do 
s:=s+(-1)^(j-1)*A41c1(n-(3*j^2-j)/2):

if (3*j^2+j)/2 <= n then
s:=s+(-1)^(j-1)*A41c1(n-(3*j^2+j)/2):
fi:
od:
s:
end:

A41c:=proc(N)
local n:
[seq(A41c1(n),n=1..N)]:
end:

time(A41c(1000));
                             0.078

time(A41ok(1000));
                             4.046

#Question 2:

# phi: implements the bijection described in the paper
# L = partition [lambda1, lambda2, ..., lambdat]
# j = the index in a(j)=(3*j^2+j)/2
phi:=proc(L, j)
local t,i:
t := nops(L):

if t+3*j >= L[1] then
return [t + 3*j-1, seq(L[i]-1,i=1..t)]:
else
return [seq(L[i]+1,i=2..t),1$(L[1]-3*j-1)];
fi:
end:

##check##

phi([5,5,4,3,2],1);
                       [7, 4, 4, 3, 2, 1]

#Question 3:

OddToDis:=proc(p)
local i,a,T:
T:=[]:
for i from 1 to nops(p) do
a:=(p[i]-1)/2:
T:=[op(T), a+1]:
od:
return sort(T,`>`):
end:

DisToOdd:=proc(n)
local i,a,T:
T:=[]:
for i from 1 to nops(n) do
a:=n[i]-1:
T:=[op(T),2*a+1]:
od:
return sort(T,`>`):
end:

##check##

OddToDis([5,3,1]);
                           [3, 2, 1]

DisToOdd(%);
                           [5, 3, 1]