#its ok to post
#TaerimKim,10/4/2020,Assignment 8

#1.
#(i). equivalent to the CompsGFk statement that coefficient of x^201 for S={1,4,6} with k=10.
coeff(CompsGFk({1, 4, 6}, x, 10), x, 201);
                               0                #indicates that it does not exist.

#(ii). equivalent to the statement that coefficient of x^411 for S={2,3,7}
GFseq(CompsGF({2, 3, 7}, x), x, 411)[412];
   4741685087960650685822461715671211099459856575685906645393

#(iii). By using the CompsGFcon
GFseq(CompsGFcon(5, {1, 4}, x), x, 341)[342];
  234303065886413369536085349033389448858104244347193374824719

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#2. #Finally, I managed to do this
Pnk := proc(n, k) local a, A:
option remember

if n = 0 then
    return {[n]};
fi:

if k<0 or n<0 then
return fail
fi:

if k>n then
return {}
fi:

if k=n then
return k
fi:

A:={seq(seq([k, op(a)], a in Pnk(n - k, i)), i = 1 .. k)}:

end:

##examples

Pnk(5, 2);
                   {[2, 2, 1], [2, 1, 1, 1]}
Pnk(9, 3);
 {[3, 3, 3], [3, 2, 2, 2], [3, 3, 2, 1], [3, 2, 2, 1, 1], 

   [3, 3, 1, 1, 1], [3, 2, 1, 1, 1, 1], [3, 1, 1, 1, 1, 1, 1]}


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#3.
pnk := proc(n, k) local B:
option remember

if k<0 or n<0 then
return fail
fi:

B:=nops(Pnk(n,k)):
end:

#example
pnk(5, 2);
                               2
pnk(9, 3);
                               7

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#4.
pn:=proc(n) local C:
option remember

if n = 0 then
    return 1;
fi:

if n<0 then
return fail
fi:

C := add(pnk(n, i), i = 1 .. n)

end:

#for the given sequence
seq(pn(n), n = 0 .. 30);
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 

  297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 

  3718, 4565, 5604
#this is the sequence for the partition numbers, labeled as A000041

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#5. the original statement takes way too much of a time to process, so i will break it down

[seq(pn(5*n + 4), n = 1 .. 10)]; #n til 10 initially and above n=10, it will be impossible to calculate due to its big numbers
[30, 135, 490, 1575, 4565, 12310, 31185, 75175, 173525, 386155]

#from here we know that pn(5n+4) is all factored by 5 so we can deduce that pn(4n+4) mod 5 will generate 0's
#So
#[seq(pn(5*n+4) mod 5, n=1..50)]
#=[0,0,0,0,..,0]

#next, 
[seq(pn(7*n + 5), n = 0 .. 5)];
                [7, 77, 490, 2436, 10143, 37338]
% mod 7;
                       [0, 0, 0, 0, 0, 0]

#for n from 0 to 5, we see similar patterns so we can deduce that

#[seq(pn(7*n+5) mod 7, n=1..50)]
#=[0,0,0,0,..,0]

#last, 
[seq(pn(11*n + 6), n = 0 .. 3)];
                     [11, 297, 3718, 31185]
% mod 11;
                          [0, 0, 0, 0]
#Again,
#[seq(pn(11*n+6) mod 11, n=1..50)]
#=[0,0,0,0,..,0]

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#6. This is called "Ramanujan's congruences" and traced its origin through paper"An elementary proof of p(11m+6)¡Õ0 (mod 11)" by Lasse Winquist (https://www.sciencedirect.com/science/article/pii/S0021980069801055).
#mathematician Srinivasa Ramanujan discovered the congruences in his 1919 paper, "Congruence properties of partitions"(https://zenodo.org/record/1447425#.X3uGl8JKhEY).