A091352 - Formula Anyone? - Number of Terms in Rows

[Paul D. Hanna]

Seqfans, 
     The number of terms in row n (offset n=0) in triangle B 
is a sequence in itself (submitted as A127419) that begins: 
1,2,3,4,6,8,11,15,19,24,30,37,45,53,62,72,83,95,108,122,137,153,169,
  
Recurrence: 
a(n) = a(n-1) + floor( (sqrt(8*a(n-1) - 7) - 1)/2 ) for n>=2 with a(0)=1,
a(1)=2.
 
G.f.: A(x) = (1-x+x^3)/(1-x)^3 - x^3/(1-x)^2*Sum_{k>=0}x^(2^k+k-1)
 
where Sum_{k>=0} x^(2^k+k-1) = 
1 + x^2 + x^5 + x^10 + x^19 + x^36 + x^69 +...+ x^(2^k+k-1) +...
 
The number of terms in sister triangle A (with factorials in first
column)
is simply n*(n+1)/2 + 1.
 
The apparent complexity of A091352 is due in part to the complicated 
g.f. of the number of terms in each row of triangle A. 
 
Still, it is quite possible that A091352 has a nice formula involving
binomials.
    Paul 
  
> ------------------------------------------------------------
> 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.



Seqfans,

I apologize for my last email, I did not see Olivier's closure of this thread.

Sincerely,
Gerald