Triangle

[Gottfried Helms]

Hi Richard -

 thanks for that idea. I considered the maple-implementation
 and found a good fixed matrix-representation for this.
 In a matrix-multiplication scheme we have

 St2 = // Stirling numbers 2'nd kind
 | 1   .   .   .   .  . |
 | 1   1   .   .   .  . |
 | 1   3   1   .   .  . |
 | 1   7   6   1   .  . |
 | 1  15  25  10   1  . |
 | 1  31  90  65  15  1 |

 X1 =  // Stirling numbers 2'nd kind, row shifted
 | 1   .    .    .    .    . |
 | .   1    1    .    .    . |
 | .   .    1    3    1    . |
 | .   .    .    1    7    6 |
 | .   .    .    .    1   15 |
 | .   .    .    .    .    1 |

Then simply

     H3 = St2 * X1

In a matrix-multiplication-scheme
                             X1=
                            | 1   .    .    .    .    . |
                            | .   1    1    .    .    . |
                            | .   .    1    3    1    . |
                            | .   .    .    1    7    6 |
                            | .   .    .    .    1   15 |
                            | .   .    .    .    .    1 |
  St2=                       H3 =
 | 1   .   .   .   .  . |   | 1   .    .    .    .    . |
 | 1   1   .   .   .  . |   | 1   1    1    .    .    . |
 | 1   3   1   .   .  . |   | 1   3    4    3    1    . |
 | 1   7   6   1   .  . |   | 1   7   13   19   13    6 |
 | 1  15  25  10   1  . |   | 1  15   40   85   96   75 |
 | 1  31  90  65  15  1 |   | 1  31  121  335  560  616 |

It is nice, that X1 is a simple rowshift of St2.
Now it would be good, if also H4 could be computed by a
similar simple scheme:

H4 =
| 1   .   .    .    .    .    .    .    .   .   .   .  . |
| 1   1   1    1    .    .    .    .    .   .   .   .  . |
| 1   3   4    6    4    3    1    .    .   .   .   .  . |
| 1   7  13   26   31   31   25   13    6   1   .   .  . |
| 1  15  40  100  171  220  255  215  156  85  35  10  1 |

I fiddled a bit but didn't see it yet ...

Gottfried


Am 15.04.2008 21:43 schrieb Richard Mathar:
> After noticing the sum formula in A000258 and further
> decomposition of the Bell numbers in there in terms of S2
> (Stirling numbers of the second kind) one can summarize
> the "Helms" array  H(n,j),  rows n=1,2,3,4,...,
> terms j=2,3,4,..,2n enumerated from the left to the right as:
> 
> H(n,j)= sum_{k+l=j) X(k,l,n)
> 
> where the sum is over the diagonal of an auxiliary upper right
> triangle defined as
> X(k,l,n) := S2(n,k)*S2(k,l) for 1<=k<=n and 1<=l<=k.
> 
> where H(.,.) becomes
> 
> 1 
> 1 1 1 
> 1 3 4 3 1 
> 1 7 13 19 13 6 1 
> 1 15 40 85 96 75 35 10 1 
> 1 31 121 335 560 616 471 240 80 15 1 
> 1 63 364 1253 2891 4221 4502 3353 1806 665 161 21 1 
> 1 127 1093 4599 13923 26222 36225 36205 26895 14756 5887 1638 294 28 1 
> 1 255 3280 16845 64366 153531 264033 336792 322576 236421 131587 55272 16989 3654 498 36 1 
> 1 511 9841 62095 290590 865332 1810747 2850870 3391455 3136381 2258413 1269960 552280 182595 44367 7500 795 45 1 
> ...
> 
> Maple Implementation:
> 
> X := proc(k,l,n)
>         if k >=1 and k <=n and l >=1 and l <= n then
>                 combinat[stirling2](n,k)*combinat[stirling2](k,l)  ;
>         else
>                 0 ;
>         fi ;
> end:
> 
> H := proc(n,j)
>         add( X(j-l,l,n),l=1..floor(j/2)) ;
> end:
> 
> for n from 1 to 10 do
>         for j from 2 to 2*n do
>                 printf("%d ",H(n,j)) ;
>         od:
>         printf("\n") ;
> od:
> 
> 
>