help with a new sequence

[franktaw at netscape.net]

How about if you look at the sequence (composition) of ascents, instead 
of the partition?

I think this starts (using the A066099 ordering):

1
1,1
1,2,1,1
1,3,2,3,1,2,1,1
1,4,3,6,2,5,3,1,1,3,2,2,1,3,4,1

Amazingly, this corresponds to Stern's diatomic array (version 3, 
A070879) for the first 18 terms.
Makes me wonder if A070879 can be interpreted as a function on 
compositions.

Franklin T. Adams-Watters


-----Original Message-----
From: deutsch at duke.poly.edu

Seqfans, 
 
I intend to submit to OEIS the following new sequence (triangle): 
 
1,1,1,1,3,1,1,4,2,6,1,1,5,5,10,10,10,1, 
 
Triangle read by rows: T(n,k) is the number of Dyck paths of 
semilength n whose ascent lengths form the k-th partition of 
the integer n; the partitions of n are ordered in the way 
exemplified by [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1], 
[1,1,1,1,1] (the "Mathematica" ordering). 
 
Equivalently, T(n,k) is the number of ordered trees with n 
edges whose node degrees form the k-th partition of the 
integer n. 
 
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and 
we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD, 
(UUU)D(UU)DDDD, and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and 
the ascents are shown between parentheses. 
 
Triangle starts: 
1; 
1,1; 
1,3,1; 
1,4,2,6,1; 
1,5,5,10,10,10,1; 
 
Row n has A000041(n) terms (=number of partitions of n). 
 
------------ 
 
So far this is all I have; terms have been found by 
straightforward counting. I am sure you can find more 
terms, more facts, etc. I'd appreciate any collaboration. 
 
Thanks, Emeric 


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