Re: Working with factors of triangular numbers.
- To: mathgroup at smc.vnet.net
- Subject: [mg78593] Re: Working with factors of triangular numbers.
- From: "Diana Mecum" <diana.mecum at gmail.com>
- Date: Thu, 5 Jul 2007 03:58:17 -0400 (EDT)
- References: <f6d4ll$hka$1@smc.vnet.net> <468AC019.1040906@gmail.com>
Jean-Marc,
It is great that you can calculate more values.
You show the fourth term as 561. It is actually 253 ?
Thanks for your time.
Diana
On 7/4/07, Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote:
>
> On 7/4/07, Diana Mecum <diana.mecum at gmail.com> wrote:
> > Jean-Marc,
> >
> > It seems to stop working at 5. I can't get seek[6,t] to display
> anything.
> >
> > Thanks,
> >
> > Diana
>
> The following version of the function seek is memory efficient and
> allows range of triangular numbers.
>
> In[1]:= T[n_Integer /; n > 0] := Binomial[n + 1, 2]
> seek[m_Integer /; m > 0, min_Integer /; min > 1, max_Integer] :=
> Module[{numb, divs, prod, comb}, For[i = min, i <= max, i++,
> numb = T[i] - 1; divs = Most[Rest[Divisors[numb]]];
> comb = Binomial[Length[divs], m]; If[comb > 0,
> For[j = 1, j <= comb, j++,
>
> prod = First[Apply[Times, Subsets[divs, {m}, {j}], {1}]];
> If[prod > numb, Break[]]; If[prod == numb,
> Return[numb + 1]]; ]; ]; ]]
>
> In[3]:= seek[4, 2, 50]
>
> Out[3]= 561
>
> In[4]:= seek[5, 46, 46]
>
> Out[4]= 1081
>
> In[5]:= seek[5, 22, 50]
>
> Out[5]= 1081
>
> In[6]:= seek[6, 100, 200]
>
> Out[6]= 18145
>
> In[7]:= seek[7, 100, 1000]
>
> Out[7]= 115921
>
> Regards,
> Jean-Marc
>
> > On 7/3/07, Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote:
> > > Diana wrote:
> > > > Math folks,
> > > >
> > > > I first generate a list of triangular numbers:
> > > >
> > > > 1, 3, 6, 10, 15, 21, ...
> > > >
> > > > and then subtract one from each as:
> > > >
> > > > 0, 2, 5, 9, 14, 20, ...
> > > >
> > > > I am trying to find the smallest triangular number (minus one) which
> > > > can be written as a product of "n" distinct factors, each factor >
> 1.
> > > >
> > > > For example:
> > > >
> > > > a(2) = 15, because 2*7 + 1 = 15.
> > > > a(3) = 55, because 2*3*9 + 1 = 55.
> > > >
> > > > I have worked with Divisors and FactorInteger, but am getting bogged
> > > > down with repeated terms. Can someone think of a neat trick to work
> > > > this problem?
> > > >
> > > > Diana M.
> > >
> > > Hi Diana,
> > >
> > > To me, your requirements are crystal clear, so I may not have
> correctly
> > > understood what you are trying to achieve; nevertheless, the following
> > > function 'seek' should return the expected results. (Note that this
> code
> > > is not memory efficient.)
> > >
> > > In[1]:= T[n_Integer /; n > 0] := Binomial[n + 1, 2]
> > >
> > > In[2]:= t = T /@ Range[30]
> > >
> > > Out[2]= {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120,
> 136,
> > > 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465}
> > >
> > > In[3]:= seek[m_Integer /; m > 0, p_] :=
> > > Module[{l = p - 1 /. 0 -> Sequence[], prod, target},
> > > prod = (Union[Apply[Times,
> > Subsets[Most[Rest[Divisors[#1]]], {m}],
> > > {1}]] & ) /@ l;
> > > target = Transpose[{l, prod}];
> > > Select[Transpose[{l, prod}], MemberQ[#1[[2]], #1[[1]]] & ,
> > > 1] /. q_ /; Length[q] > 0 :> q[[1, 1]] + 1]
> > >
> > > In[4]:= seek[2, t]
> > >
> > > Out[4]= 15
> > >
> > > In[5]:= seek[3, t]
> > >
> > > Out[5]= 55
> > >
> > > In[6]:= seek[4, t]
> > >
> > > Out[6]= 253
> > >
> > > Regards,
> > > Jean-Marc
> > >
> > >
> >
> >
> >
> > --
> > "God made the integers, all else is the work of man."
> > L. Kronecker, Jahresber. DMV 2, S. 19.
>
--
"God made the integers, all else is the work of man."
L. Kronecker, Jahresber. DMV 2, S. 19.