a different triangle of StirlingS2

[Wouter Meeussen]

hi,

A008277 contains the StirlingS2[n,m] triangle :

%I A008277
%S A008277 1,1,1,1,3,1,1,7,6,1,1,15,25,10,1,1,31,90,65,15,1,1,63,301,
%T A008277 350,140,21,1,1,127,966,1701,1050,266,28,1,1,255,3025,7770,
%U A008277 6951,2646,462,36,1,1,511,9330,34105,42525,22827,5880,750
%N A008277 Triangle of Stirling numbers of 2nd kind.
%R A008277 AS1 835. DKB 223.
%O A008277 1,5
%K A008277 nonn,tabl,nice
%A A008277 njas

Their row sums add up to the Bell numbers : 

%I A000110 M1484 N0585
%S A000110 1,1,2,5,15,52,203,877,4140,21147,115975,678570,4213597,27644437,
%T A000110 190899322,1382958545,10480142147,82864869804,682076806159,
%U A000110 5832742205057,51724158235372,474869816156751,4506715738447323
%N A000110 Bell or exponential numbers: ways of placing n labeled balls into n
 indistinguishable boxes.
%R A000110 MOC 16 418 1962. AMM 71 498 1964. PSPM 19 172 1971. GO71.
%K A000110 core,nonn
%F A000110 E.g.f.: exp (exp x - 1 ). Recurrence: a(n+1) = Sum a(k)C(n,k). Also
 a(n) = Sum Stirling2(n,k), k=1..n.
%O A000110 0,3
%t A000110 NestList[ Factor[D[#1,x]]&, Exp[Exp[x-1]-1], n] /. (x->1)
%p A000110 series(exp(exp(x)-1),x,40).
%A A000110 njas


These do indeed contain a reference to the row-sum of the StirlingS2, but
also contain a recursion formula (ref: ??) that yields a *different*
triangular table, not present (as such) in EIS:

" Recurrence: a(n+1) = Sum a(k)C(n,k)"

this produces:

b[n_]:=b[n]=Plus@@Table[b[k] Binomial[n-1,k],{k,0,n-1}];b[0]:=1;b[1]:=1

Table[b[k] Binomial[n-1,k],{n,10},{k,0,n-1}]//ColumnForm

{1}
{1, 1}
{1, 2, 2}
{1, 3, 6, 5}
{1, 4, 12, 20, 15}
{1, 5, 20, 50, 75, 52}
{1, 6, 30, 100, 225, 312, 203}
{1, 7, 42, 175, 525, 1092, 1421, 877}
{1, 8, 56, 280, 1050, 2912, 5684, 7016, 4140}
{1, 9, 72, 420, 1890, 6552, 17052, 31572, 37260, 21147}


I do not see an easy transformation from one to another.

** I feel this deserves an entry of its own **

wouter.


Dr. Wouter L. J. MEEUSSEN
w.meeussen.vdmcc at vandemoortele.be
eu000949 at pophost.eunet.be