> #Weiji Zheng, Homework 24, 12/11/2020
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> #OK TO POST HOMEWORK
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> 
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> #Q1
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> #(i) How many (integer) partitions are there of 97 where the each part, upon division by 7, gives remainder in the set {1,4,5}?
> #(ii) How many (integer) partitions are there of 97 where the the difference between consecutive parts is at least 3?
> 

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> 
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> #pnC(97, {1, 4, 5}, 7)
> #108236

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> #pnD(97, 3)
> #19990

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> #Q2
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> #Write a joint generalization of procedures pnD(n,d) and pnC(n,C,m) in M24.txt, call it
> #pnDC(n,d,C,m)
> 
> #that inputs a positive integer n, a non-negative integer d, a set C that is a subset of {0,..., m-1} and a positive integer m and outputs the #NUMBER of partitions of n such that the difference of two consecutive entries is at least d, and each entry mod m belongs to C.
> 
> #(i) Check that pnDC(n,0,C,m) gives the same output as pnC(n,C,m) and pnDC(n,d,{0},1) give the same output as pnD(n,d)
> 
> #(ii) Use it to find the first 40 terms of the sequence enumerating partitions that are both odd and distinct. Is it in the OEIS? If it is, What #is the A-number?
> 
> #(iii) Use it to find the first 40 terms of the sequence enumerating partitions such that the difference between consecutive entries is at least #2, and each entry gives remainder 1 or 4 when divided by 5. Is it in the OEIS? If it is, What is the A-number?
> 

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> 
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> pnkCJD:=proc(n,d,k,C,m) local L,S:
> option remember:
> if k > n then
>    RETURN(0):
> fi:
> if k = n then
>    if member(k mod m, C) then
>       RETURN(1):
>    else
>       RETURN(0):
>    fi:
> fi:
> if n = 1 then
>    if member(1, C) then
>       RETURN(1):
>    else 
>       RETURN(0):
>    fi:
> fi:
> if not member(k mod m, C) then
>    RETURN(0):
> fi:
> 
> S := 0:
> for L to k-d do
>     if member(L mod m, C) then
>        S := S + pnkCJD(n-k, d, L, C, m):
>     fi:
> od:
> 
> S:
> 
> end:

> pnDC := proc(n,d,C,m) local k:
> add(pnkCJD(n,d,k,c,M),k=1..n):
> end:

> #(1) #check
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> #pnDC(20, 0, {1, 2}, 2) = pnC(20, {1, 2}, 2)
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> 
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> #pnDC(20, 2, {0}, 1) = pnD(20, 2)
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> 
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> 
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> #(2)
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> #seq(pnDC(i, 1, {1}, 2), i = 1 .. 40)
> #1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 8, 8, 9, 11, 12, 12, 14, 16, 17, 18, 20, 23, 25, 26, 29, 33, 35, 37, 41, 46

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> #A000700
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> #Expansion of Product_{k>=0} (1 + x^(2k+1)); number of partitions of n into distinct odd parts; number of self-conjugate partitions; number of symmetric Ferrers graphs with n nodes.

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> #(3)
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> #seq(pnDC(i, 2, {1, 4}, 5), i = 1 .. 40)
> #1, 0, 0, 1, 1, 1, 1, 0, 1, 2, 2, 1, 1, 2, 3, 3, 2, 2, 3, 5, 5, 3, 3, 5, 7, 7, 6, 5, 7, 11, 11, 8, 8, 12, 15, 15, 13, 12, 16, 22

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> #A203776		Number of partitions of n into distinct parts 5k+1 or 5k+4.
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> 
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> #Q3
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> #Use the argument in the lecture that proved that the generating function of p(n) is
> #1/((1-q)*(1-q^2)*(1-q^3)*....)
> 
> #To prove that the generating function of the sequence enumerating distinct partitions is
> 
> #(1+q)*(1+q^2)*(1+q^3)*... =Product(1+q^i,i=1..infinity)
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> 
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> 
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> #Q4
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> #Use this time of argument to prove that the generating function of the sequence enumerating odd partitions (i.e. partitions whose components are all odd) is
> #1/((1-q)*(1-q^3)*(1-q^5)*....)=Product(1/(1-q^(2*i+1),i=0..infinity)
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> 
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> 
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> #Q5
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> #By combining the above two problems, prove Euler's theorem that for every n, the number of partitions of n into odd parts, equals the number of partitions of n into distinct parts.
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