[seqfan] Re: Are the partials sums of A025147 equal to A038348?

[franktaw at netscape.net]

Sums of consecutive terms of A025147 gives the total number of 
partitions of n into distinct parts: A025147(n-1) + A025147(n) = 
A000009(n).  This is because any partition of n into distinct parts 
either includes a part of size 1 or it does not.  In the latter case, 
it is counted in A025147(n); otherwise, remove the part of size 1 and 
it is counted in A025147(n-1).  This includes the case n=0, if we take 
A025147(-1) = 0.

Now A038348, number of partitions of n with at most one even part, is 
readily seen to be the sum of partitions of n-2k into odd parts: 
A038348(n) = sum(k=0..n/2,A000009(n-2k)) - simply remove the even part 
of size 2k, removing nothing for k=0.

Put these two facts together, and you get the desired result.  
A038348(n) = sum(k=0,n/2,A025147(n-2k)+A025147(n-2k-1)) = 
sum(m=mod(n,2)..n step 2,A025147(m)) + sum(n=mod(n,2)+1..n-1 step 
2,A025147(m)) = sum(m=1..n,A025147(n)).

Franklin T. Adams-Watters

BTW, the parentheses in A038348 are unbalanced - there are two more 
open parentheses than close parentheses.  And I believe the style guide 
calls for "infinity" in preference to "inf".  So it should be 
"Expansion of (1/(1-x^2))*Product(1/(1-x^(2m+1)), m=0..infinity)."

-----Original Message-----
From: Jonathan Post <jvospost3 at gmail.com>

Are the partials sums of A025147 Number of partitions of n into
distinct parts >= 2.
equal to A038348  Expansion of (1/(1-x^2))*Product((1/(1-x^(2m+1)),
m=0..inf. (equivalently Number of partitions of n with at most one
even part)?

Both begin:

1, 1, 2, 3, 4, 6, 8, 11, 14, 19, 24, 31, 39, 49, 61, 76, 93, 114, 139,
168, 203, 244