[seqfan] Re: A000041 partition sequence

[William Keith]

Bob:

     Okay.  It seems to me that what you have is simply an obfuscated
version of the pentagonal recurrence.  For example, for 6 <= n <= 11, the
PNT says

a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-7)

and your recurrence says

a(n) = 1 + a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) +
a(n-10) - a(n-13) - a(n-14) + a(n-15).

But of course for 6 <= n <= 11, a(n-13) and lower terms are zero.  So drop
them.  Then if we subtract, we get

0 = 1 - a(n-6) + a(n-8) + a(n-9) + a(n-10)

or

a(n-6) - a(n-8) - a(n-9) + a(n-10) = 1

 for this segment of the partition numbers, which is true.

There might be a nugget of combinatorial interest here involving sets of
partition numbers which sum to 1 under suitable signs, but I don't really
feel inclined to hunt for it.  Sorry.  One can come up with all kinds of
increasingly baroque constructions involving known sequences.  Since the
expression for each F(G) implies that some moving frame of coefficients
must sum to 1, perhaps the triangle of terms which sum to 1 would make a
suitable sequence.

Cordially,
William Keith