[seqfan] Re: Are the partials sums of A025147 equal to A038348?
franktaw at netscape.net
franktaw at netscape.net
Fri Apr 2 04:44:19 CEST 2010
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
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