"mountain man" <hobbit@southern_seaweed.com.op> wrote in message
news:GIV6c.116370$Wa.77309@news-server.bigpond.net.au...
> Greetings,
>
> I have encountered a problem which is starting to stretch
> the limits of my current abilities in an area of TSQL and have
> decided to post it here with a request for assistance.
>
> Essentially over time if it is possible I need to translate a
> fair number of maple algorithms (see below to TSQL).
>
> These algorithms all involve combinatorial mathematics.
> If I can get the hang of one the rest should follow.
> So below I have presented an example.
> Thanks for any assistance provided.
>
>
> PROBLEM 1: No of Ordered factorizations of n
> ==================================
> Background URL:
> http://www.research.att.com/projects/OEIS?Anum=A074206
>
> *Sequence: 0,1,1,1,2,1,3,1,4,2,3,1,8,1,3,3,8,1,8,1,8,3,3,1,20,2,3,4,8,
> 1,13,1,16,3,3,3,26,1,3,3,20,1,13,1,8,8,3,1,48,2,8,3,8,1,20,
> 3,20,3,3,1,44,1,3,8,32,3,13,1,8,3,13,1,76,1,3,8,8,3,13,1,48,
> 8,3,1,44,3,3,3,20,1,44,3,8,3,3,3,112
> Name: Number of ordered factorizations of n.
> References L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 126, see
#27.
> R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 141.
> J. Riordan, An Introduction to Combinatorial Analysis, Wiley,
> 1958,
> Links: E. W. Weisstein, Link to a section of The World of Mathematics.
> Index entries for "core" sequences
> Eric Weisstein's World of Mathematics, Ordered Factorization
>
> Formula: With different offset: a(n) = sum of all a(i)
> such that i divides n and i < n (Clark Kimberling).
> a(p^k)=2^(k-1).
> Dirichlet g.f.: 1/(2-zeta(s)). - Herb Wilf, Apr 29, 2003
>
> The Riemann zeta function looks a little foreboding for TSQL
> (an aside .... is it?) and there are always sample maple routines
> provided, such as the following:
>
> Maple: a:=array(1..150): for k from 1 to 150 do a[k]:=0 od: a[1]:=1:
for
> j from 2
> to 150 do for m from 1 to j-1 do if j mod m = 0 then
> a[j]:=a[j]+a[m] fi: od: od:
> for k from 1 to 150 do printf(`%d,`,a[k])
>
> What the above routine does is generate, for the numbers 2,3,4 ...150
> the corresponding count of ordered factor sequences as defined above.*
>
> Example: Number of ordered factorizations of 8 is 4:
> 8 = 2*4 = 4*2 = 2*2*2.
>
> I recall writing a similar bit of code in basic 20 years ago, and it used
> arrays. Not sure how this might be implemented with TSQL.
>
> Can anyone assist here?
> Many thanks!!
>
>
>
>
> --
> Pete Brown
> Winluck P/L
> IT Managers & Engineers
> Falls Creek
> Australia
> www.mountainman.com.au/software
>
>
>
>
>
>
>
>
>
>

"Dan Guzman" <danguzman@nospam-earthlink.net> wrote in message
news:45i7c.31783$%06.8596@newsread2.news.pas.earthlink.net...
> The script below works with SQL Server 2000. You can substitute a temp
> table for the table variable if you are using an earlier Microsoft SQL
> Server version.
>
> SET NOCOUNT ON
> DECLARE @k int, @j int, @m int
> DECLARE @a TABLE(Ordinal int, Value int)
> SET @k = 0
> WHILE @k < 151
> BEGIN
> INSERT INTO @a VALUES(@k, 0)
> SET @k = @k + 1
> END
> SET @k = 1
> WHILE @k < 2
> BEGIN
> UPDATE @a SET Value = 1 WHERE Ordinal = 1
> SET @j = 2
> WHILE @j < 151
> BEGIN
> SET @m = 1
> WHILE @m < @j
> BEGIN
> IF @j % @m = 0
> UPDATE @a
> SET Value = Value + (SELECT Value FROM @a WHERE Ordinal = @m)
> WHERE Ordinal = @j
> SET @m = @m + 1
> END
> SET @j = @j + 1
> END
> SET @k = @k + 1
> END
>
> SELECT
> Ordinal AS n,
> Value as "a(n)"
> FROM @a
>
>
> --
> Hope this helps.
>
> Dan Guzman
> SQL Server MVP
>
> "mountain man" <hobbit@southern_seaweed.com.op> wrote in message
> news:GIV6c.116370$Wa.77309@news-server.bigpond.net.au...
> > Greetings,
> >
> > I have encountered a problem which is starting to stretch
> > the limits of my current abilities in an area of TSQL and have
> > decided to post it here with a request for assistance.
> >
> > Essentially over time if it is possible I need to translate a
> > fair number of maple algorithms (see below to TSQL).
> >
> > These algorithms all involve combinatorial mathematics.
> > If I can get the hang of one the rest should follow.
> > So below I have presented an example.
> > Thanks for any assistance provided.
> >
> >
> > PROBLEM 1: No of Ordered factorizations of n
> > ==================================
> > Background URL:
> > http://www.research.att.com/projects/OEIS?Anum=A074206
> >
> > *Sequence: 0,1,1,1,2,1,3,1,4,2,3,1,8,1,3,3,8,1,8,1,8,3,3,1,20,2,3,4,8,
> > 1,13,1,16,3,3,3,26,1,3,3,20,1,13,1,8,8,3,1,48,2,8,3,8,1,20,
> > 3,20,3,3,1,44,1,3,8,32,3,13,1,8,3,13,1,76,1,3,8,8,3,13,1,48,
> > 8,3,1,44,3,3,3,20,1,44,3,8,3,3,3,112
> > Name: Number of ordered factorizations of n.
> > References L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 126, see
> #27.
> > R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 141.
> > J. Riordan, An Introduction to Combinatorial Analysis, Wiley,
> > 1958,
> > Links: E. W. Weisstein, Link to a section of The World of
Mathematics.
> > Index entries for "core" sequences
> > Eric Weisstein's World of Mathematics, Ordered Factorization
> >
> > Formula: With different offset: a(n) = sum of all a(i)
> > such that i divides n and i < n (Clark Kimberling).
> > a(p^k)=2^(k-1).
> > Dirichlet g.f.: 1/(2-zeta(s)). - Herb Wilf, Apr 29, 2003
> >
> > The Riemann zeta function looks a little foreboding for TSQL
> > (an aside .... is it?) and there are always sample maple routines
> > provided, such as the following:
> >
> > Maple: a:=array(1..150): for k from 1 to 150 do a[k]:=0 od: a[1]:=1:
> for
> > j from 2
> > to 150 do for m from 1 to j-1 do if j mod m = 0 then
> > a[j]:=a[j]+a[m] fi: od: od:
> > for k from 1 to 150 do printf(`%d,`,a[k])
> >
> > What the above routine does is generate, for the numbers 2,3,4 ...150
> > the corresponding count of ordered factor sequences as defined above.*
> >
> > Example: Number of ordered factorizations of 8 is 4:
> > 8 = 2*4 = 4*2 = 2*2*2.
> >
> > I recall writing a similar bit of code in basic 20 years ago, and it
used
> > arrays. Not sure how this might be implemented with TSQL.
> >
> > Can anyone assist here?
> > Many thanks!!
> >
> >
> >
> >
> > --
> > Pete Brown
> > Winluck P/L
> > IT Managers & Engineers
> > Falls Creek
> > Australia
> > www.mountainman.com.au/software
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
>
>