adieu - another possible solution: a forum

[Gerald McGarvey]

see it or
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To: seqfan at ext.jussieu.fr
Date: Sun, 14 Jan 2007 21:21:12 -0500
Subject: A091352 - Formula Anyone?
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Seqfans, 
    I would like very much to know a closed formula or g.f. 
for the terms of the sequence A091352: 
[1, 2, 4, 9, 24, 77, 295, 1329, 6934, 41351, 278680, ...]
(A091352 is defined by powers of a triangular matrix.)
 
Below I show that a recurrence that generates the factorials,
and that a simple variant of that recurrence generates A091352.
 
This seems to imply that A091352 may possibly be expressed 
in terms of factorials ... 
 
Anyone like to try to find a formula for A091352?  
I have not been able to find one yet ... 
    Paul 
------------------------------------------------------------
TRIANGLE A.
Consider the following triangle related to factorials: 
1; 
1, 1; 
2, 2, 1, 1; 
6, 6, 4, 4, 2, 1, 1; 
24, 24, 18, 18, 12, 8, 8, 4, 2, 1, 1;
120, 120, 96, 96, 72, 54, 54, 36, 24, 16, 16, 8, 4, 2, 1, 1;
720, 720, 600, 600, 480, 384, 384, 288, 216, 162, 162, 108, 72, 48, 32,
32, 16, 8, 4, 2, 1, 1;
...
The recurrence illustrated by the following examples. 
Start with a single '1': 
1; 
To get row 1, insert 0 at position 1, 
and take partial sums in reverse order:  
0,_1;
1,_1;
To get row 2, insert 0 at positions [1,3], 
and take partial sums in reverse order:  
0,_1,_0,_1;
2,_2,_1,_1;
To get row 3, insert 0 at positions [1,3,6], 
and take partial sums in reverse order:  
0,_2,_0,_2,_1,_0,_1;
6,_6,_4,_4,_2,_1,_1;
To get row 4, insert 0 at positions [1,3,6,10], 
and take partial sums in reverse order:  
_0,__6,__0,__6,__4,_0,_4,_2,_1,_0,_1;
24,_24,_18,_18,_12,_8,_8,_4,_2,_1,_1;
Insert 0 at positions [1,3,6,10,15], 
and take reverse partial sums:
__0,__24,__0,_24,_18,__0,_18,_12,__8,__0,__8,_4,_2,_1,_0,_1;
120,_120,_96,_96,_72,_54,_54,_36,_24,_16,_16,_8,_4,_2,_1,_1;
etc.

Continuing in this way generates the factorials in the first column. 
------------------------------------------------------------
TRIANGLE B.
A triangle generated by a similar recurrece begins: 
1; 
1, 1; 
2, 1, 1; 
4, 2, 2, 1; 
9, 5, 5, 3, 1, 1; 
24, 15, 15, 10, 5, 5, 2, 1; 
77, 53, 53, 38, 23, 23, 13, 8, 3, 3, 1; 
295, 218, 218, 165, 112, 112, 74, 51, 28, 28, 15, 7, 4, 1, 1; 
1329, 1034, 1034, 816, 598, 598, 433, 321, 209, 209, 135, 84, 56, 28, 28,
13, 6, 2, 1;
... 
The recurrence is illustrated by the following examples. 
Start with 1's in the first 2 rows: 
1;
1, 1; 
To get row 2, insert 0 at position 2 in row 1, 
and take partial sums in reverse order: 
1,_0,_1;
2,_1,_1;
To get row 3, insert 0 at position 2, 
and take partial sums in reverse order: 
2,_0,_1,_1;
4,_2,_2,_1;
To get row 4, insert 0 at positions [2,5], 
and take partial sums in reverse order: 
4,_0,_2,_2,_0,_1;
9,_5,_5,_3,_1,_1;
To get row 5, insert 0 at positions [2,5], 
and take partial sums in reverse order:  
_9,__0,__5,__5,_0,_3,_1,_1;
24,_15,_15,_10,_5,_5,_2,_1; 
Continue by inserting zeros in current row at positions: 
  [2,5,9,14,..., (m+2)*(m+3)/2 - 1,...|m>=0],
then take the reverse partial sums to obtain the next row: 
24,__0,_15,_15,__0,_10,__5,_5,_0,_2,_1;
77,_53,_53,_38,_23,_23,_13,_8,_3,_3,_1; 
etc.
 
QUESTION:
Is there a nice formula for the first column of triangle B? 
[1, 1, 2, 4, 9, 24, 77, 295, 1329, 6934, 41351, 278680, ...].
------------------------------------------------------------
END.