some loose ideas

Wouter Meeussen eu000949 at pophost.eunet.be
Sun Jun 25 20:40:52 CEST 2000

```hi sequers,

Series of  E^(1/k^z - 1) for k around k=1
equals    Sum[ Poly[j,z] / j!  (k-1) ^j , {j, 0, \[Infinity]}]

Poly[j,z] stands for a polynome in z of degree j :
{1,
-z,
z +   2 z^2,
-2 z -   6 z^2 -   5 z^3,
6 z +  22 z^2 +  30 z^3 +  15 z^4,
-24 z - 100 z^2 - 175 z^3 - 150 z^4 - 52 z^5}

The Coefficient List table, after transforming z->-z, is:

{1}
{0, 1}
{0, -1, 2}
{0, 2, -6, 5}
{0, -6, 22, -30, 15}
{0, 24, -100, 175, -150, 52}
{0, -120, 548, -1125, 1275, -780, 203}
{0, 720, -3528, 8120, -11025, 9100, -4263, 877}
{0, -5040, 26136, -65660, 101535, -101920, 65366, -24556, 4140}
{0, 40320, -219168, 590620, -1009260, 1167348, -920808, 478842, -149040, 21147}
{0, -362880, 2053152, -5863500, 10855200, -14004900, 12844419, -8287650,
3601800, -951615, 115975}
{0, 3628800, -21257280, 63767880, -126142500, 177680360, -183117165,
138366921, -75141000, 27914040, -6378625, 678570}

each row sums to 1,
rowsums of unsigned table = {1,1,3,13,73,501, .. = A000262 = "Sets of
lists:a(n)=(2n-1)a(n-1)-(n-1)(n-2)a(n-2)"
and shares this property with A008297 = Triangle of Lah numbers (see
below)
second column= factorials,
Last column = {1,1,2,5,15,52,203,...} = Bell Numbers = A000110
--------------------------------------------------------------------------------
pro memori:

Triangle of Lah numbers :
In[26]:=
Table[Sum[Abs[StirlingS1[n, k] ]* StirlingS2[k, m], {k, m, n}], {n, 0, 7}, {m,
0, n}]

Out[26]=
{{1}, {0, 1}, {0, 2, 1}, {0, 6, 6, 1}, {0, 24, 36, 12, 1}, {0, 120, 240, 120,
20, 1}, {0, 720, 1800, 1200, 300, 30, 1}, {0, 5040, 15120, 12600, 4200,
630, 42, 1}}
--------------------------------------------------------------------------------
Row sum of Triangle of Lah numbers, extra terms (more..) for A000262 :
In[25]:=
Plus @@@ Table[
Sum[Abs[StirlingS1[n, k] ]* StirlingS2[k, m], {k, m, n}], {n, 24}, {m, n}]

Out[25]=
{1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, 58941091, 824073141,
12470162233, 202976401213, 3535017524403, 65573803186921, 1290434218669921,
26846616451246353, 588633468315403843, 13564373693588558173,
327697927886085654441, 8281153039765859726341, 218456450997775993367443,
6004647590528092507965393, 171679472549945695230447313}

Wouter.

ps,
I hold these truths about E^(1/k^z ) to be self-evident:

Sum[Sum[(1/k^z), {z, 2, \[Infinity]}] , {k, 2, \[Infinity]}] == 1 ;
Sum[(1/k^z), {k, 1, \[Infinity]}] == Zeta[z] ;
Sum[-1 + Zeta[z], {z, 2, \[Infinity]}] == 1 ;
Product[Product[E^(1/k^z ), {k, 2, \[Infinity]}], {z, 2, \[Infinity]}] == E ;
Series in z of E^(1/k^z )==
E  Sum[ Bell[j] /j! (-z Log[k])^j , {j,0,\[Infinity]}]

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

```