# A036040 revisited

wouter meeussen wouter.meeussen at pandora.be
Sun Mar 23 14:41:00 CET 2003

```rule1: Abramowitz & Stegun is always right
rule 2:  even in case of FUBAR, rule 1 applies.

1,
1,1,
1,3,1,
1,4,3,6,1,
1,5,10,10,15,10,1,
1,6,15,10,15,60,15,20,45,15,1
** **      ** **
1,6,15,15,10,60,20,15,45,15,1 is what Mathematica gives.

Mathematica orders the partitions of 6 as
{6},{5,1},{4,2},{4,1,1},{3,3},{3,2,1},{3,1,1,1},{2,2,2},{2,2,1,1},{2,1,1,1,1
},{1,1,1,1,1,1}
****     **                  ******    ****
The A036040 as it now stands is given by :
runs[li:{__Integer}]:=((Length/@ Split[#]))&[Sort@ li]
Table[temp=Map[Reverse,Sort@ (Sort/@ Partitions[w]),{1}];
Apply[Multinomial,temp,{1}]/Apply[Times,(runs/@ temp)!,{1}],{w,6}]

while the 'natural' (??) Mathematica ordering is produced by
runs[li:{__Integer}]:=((Length/@ Split[#]))&[Sort@ li]
Table[Apply[Multinomial,Partitions[w],{1}]/Apply[Times,(runs/@
Partitions[w])!,{1}],{w,6}]

Maybe both should be in the OEIS?

An alternate description would be:
T[n,m]=count of set partitions of n with block lengths given by the m-th
partition of n.

The current description "Triangle of multinomial coefficients" gives the
impression that a calculation
is possible without generating the partitions of n. Did I look over it ?

fumblingly yours,

Wouter.

```