- #1
phoenixy
The Q is: show that the number of partitions of n within Z+ where no summand is divisible by 4 equals the number of partitions of n where no even summand is repeated
Here is what I got so far
Let the partition where no summand is divisible by 4 be P1(x)
Let the partition where no even summand is repeated be P2(x)
My goal is to show that P1(x) = P2(x)
P1(x) = (1+x+x^2 ... )(1+x^2+x^4 ... )(1+x^3+x^6 ...)(1+x^5+x^10 ...)...
(skip x^4, x^8, ... etc.)
P1(x) = [ let i go from 1 to infinity, the products of ( 1 / (1-x^i) ) ] /
[ let i go from 1 to infinity, the prodcuts of ( 1 / (1-x^(4i) )
P2(x) = [ (1+x^2)(1+x^4) ... ][ (1+x+x^2 ...)(1+x^3+x^6) ... ]
P2(x) = [ let i go from 1 to infinity,the products of ( 1 + x^(2i) ) ] *
[ let i go from 1 to infinity, the products of ( 1 / (1-x^(2i-1)) )
So are my equations correct? If so, how do I solve them?
Here is what I got so far
Let the partition where no summand is divisible by 4 be P1(x)
Let the partition where no even summand is repeated be P2(x)
My goal is to show that P1(x) = P2(x)
P1(x) = (1+x+x^2 ... )(1+x^2+x^4 ... )(1+x^3+x^6 ...)(1+x^5+x^10 ...)...
(skip x^4, x^8, ... etc.)
P1(x) = [ let i go from 1 to infinity, the products of ( 1 / (1-x^i) ) ] /
[ let i go from 1 to infinity, the prodcuts of ( 1 / (1-x^(4i) )
P2(x) = [ (1+x^2)(1+x^4) ... ][ (1+x+x^2 ...)(1+x^3+x^6) ... ]
P2(x) = [ let i go from 1 to infinity,the products of ( 1 + x^(2i) ) ] *
[ let i go from 1 to infinity, the products of ( 1 / (1-x^(2i-1)) )
So are my equations correct? If so, how do I solve them?
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