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MathGroup Archive 2007

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Re: Re: Re: Working with factors of triangular numbers.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80083] Re: [mg80043] Re: [mg79864] Re: Working with factors of triangular numbers.
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Sun, 12 Aug 2007 07:14:56 -0400 (EDT)
  • References: <200707030923.FAA17995@smc.vnet.net> <15087134.1186826869931.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

I'd already posted a different correction:

Clear[factorCounts, nine]
nine[{a___, {3, j_}, b___}] := {a, {3, j + 2}, b}
nine[x_List] := Join[{{3, 2}}, x]
factorCounts[1] = factorCounts[2] = factorCounts[3] = {1};
factorCounts[4] = {2};
factorCounts[7] = {3};
factorCounts[k_Integer] :=
  If[EvenQ@k, factorCounts[(k + 2)/2, k - 1],
   factorCounts[(k - 1)/2, k + 2]]
factorCounts[j_Integer, k_Integer] :=
  Switch[GCD[j, k],
     1,
     Join[FactorInteger@j, FactorInteger@k],
     3, nine@Join[FactorInteger[j/3], FactorInteger[k/3]]][[All, -1]] //
    Sort

That's a little faster than my previous (broken) solution.

Bobby

On Sat, 11 Aug 2007 01:10:07 -0500, Oleksandr Pavlyk <pavlyk at gmail.com> 
wrote:

> The reason for discrepancy is that FactorInteger[j/3] and  
> FactorInteger[k/3]
> might both not have any factors of 3, thus Replace would fails to add 
> the two
> instances of 3.
>
> factorCounts[1] = factorCounts[2] = factorCounts[3] = {1};
> factorCounts[4] = {2};
> factorCounts[7] = {3};
> factorCounts[k_Integer] :=
>  If[EvenQ@k, factorCounts[(k + 2)/2, k - 1],
>   factorCounts[(k - 1)/2, k + 2]]
> factorCounts[j_Integer, k_Integer] :=
>  Switch[GCD[j, k], 1, Join[FactorInteger@j, FactorInteger@k], 3,
>     With[{r =
>        Replace[Join[FactorInteger[j/3],
>          FactorInteger[k/3]], {3, s_} :> {3, s + 2}, {1}] },
>      If[MemberQ[r[[All, 1]], 3], r, Join[r, {{3, 2}}]]]
>     ][[All, -1]] // Sort
>
>
>
>
> On 8/10/07, Oleksandr Pavlyk <pavlyk at gmail.com> wrote:
>> By design factorCounts[n]  should give the same as
>>
>> Sort[FactorInteger[ (n(n+1)/2) - 1][[All,2]]]
>>
>> Here are few cases where they disagree:
>>
>> In[134]:= Cases[
>>  Table[{chi, FactorInteger[chi*(chi + 1)/2 - 1][[All, 2]] // Sort,
>>    factorCounts[chi]}, {chi, 100}], {chi_, x_, y_} /; x =!= y]
>>
>> Out[134]= \
>> {{13,{1,1,2},{1,1}},{22,{1,2,2},{1,2}},{31,{1,1,2},{1,1}},{40,{1,1,2},\
>> {1,1}},{49,{1,2,3},{1,3}},{58,{1,1,1,2},{1,1,1}},{67,{1,1,2},{1,1}},{\
>> 76,{1,2,2},{1,2}},{85,{1,1,1,2},{1,1,1}},{94,{1,2,4},{1,4}}}
>>
>> Oleksandr
>>
>>
>> On 8/10/07, Oleksandr Pavlyk <pavlyk at gmail.com> wrote:
>> > Hi,
>> >
>> > First of all, very nice work.
>> >
>> > It threw me off initially that lld5 did not work for my example:
>> >
>> > In[70]:= lld5[chi = 2^84*5^18*7^6]
>> >
>> > Out[70]= 14
>> >
>> > In[71]:= lldaux5[FactorInteger[chi][[All, 2]] ]
>> >
>> > Out[71]= 26
>> >
>> > But then I realize that factorCounts is supposed to work only for
>> > triangular numbers,
>> > as is implied by the link given in the mail, and I would like to
>> > explicitly state for the record:
>> >
>> > In[72]:= factorCounts[chi]
>> >
>> > Out[72]= {1,1,1,1,1,1,1,1,1,1,1,1,5}
>> >
>> > In[73]:= FactorInteger[chi]
>> >
>> > Out[73]= {{2,84},{5,18},{7,6}}
>> >
>> > Oleksandr Pavlyk
>> >
>> > On 8/10/07, DrMajorBob <drmajorbob at bigfoot.com> wrote:
>> > > Here's a significantly faster code (for n=12,13 anyway).
>> > >
>> > > Clear[lldaux5, lld5, t5, factorCounts]
>> > > factorCounts[1] = factorCounts[2] = factorCounts[3] = {1};
>> > > factorCounts[4] = {2};
>> > > factorCounts[7] = {3};
>> > > factorCounts[k_Integer] :=
>> > >   If[EvenQ@k, factorCounts[(k + 2)/2, k - 1],
>> > >    factorCounts[(k - 1)/2, k + 2]]
>> > > factorCounts[j_Integer, k_Integer] :=
>> > >   Switch[GCD[j, k],
>> > >      1,
>> > >      Join[FactorInteger@j, FactorInteger@k],
>> > >      3, Replace[
>> > >       Join[FactorInteger[j/3],
>> > >        FactorInteger[k/3]], {3, s_} :> {3, s + 2}, {1}]][[All, -1]]  
>> //
>> > >    Sort
>> > > lldaux5[{}] := 0
>> > > lldaux5[x_List] /; MemberQ[x, 1] :=
>> > >   Count[x, 1] + lldaux5@DeleteCases[x, 1]
>> > > lldaux5[Lpow_] :=
>> > >   lldaux5[Lpow] =
>> > >    With[{Mtup = Tuples[Range[0, #] & /@ Lpow]},
>> > >     Total[Quiet@
>> > >        LinearProgramming[ConstantArray[-1, Length[Mtup]],
>> > >         Transpose@Mtup, Thread@{Lpow, 0},
>> > >         Array[{0, 1} &, Length[Mtup]], Integers]] - 1]
>> > > lld5[n_Integer] := lldaux5@factorCounts@n
>> > > t5[n_] := # (# + 1)/2 &@NestWhile[# + 1 &, 2, lld5[#] < n &]
>> > >
>> > > T2c /@ Range[10] // Timing
>> > > t5 /@ Range[10] // Timing
>> > >
>> > > {0.422, {3, 15, 55, 253, 1081, 13861, 115921, 665281, 18280081,
>> > >    75479041}}
>> > >
>> > > {1.046, {3, 15, 55, 325, 1081, 18145, 226801, 665281, 18280081,
>> > >    75479041}}
>> > >
>> > > T2c[11] // Timing
>> > > t5[11] // Timing
>> > >
>> > > {1.625, 2080995841}
>> > >
>> > > {3.063, 2080995841}
>> > >
>> > > Timing differences aren't reliable either way for n<=11, but beyond  
>> that
>> > > it's a different story:
>> > >
>> > > T2c[12] // Timing
>> > > t5[12] // Timing
>> > >
>> > > {27., 68302634401}
>> > >
>> > > {17.187, 68302634401}
>> > >
>> > > T2c[13] // Timing
>> > > t5[13] // Timing
>> > >
>> > > {150.172, 924972048001}
>> > >
>> > > {64.032, 924972048001}
>> > >
>> > > t5[14] // Timing
>> > >
>> > > {516.546, 48318396825601}
>> > >
>> > > The trick (already used by someone else, I think)
>> > >
>> > > lldaux5[x_List] /; MemberQ[x, 1] :=
>> > >   Count[x, 1] + lldaux5@DeleteCases[x, 1]
>> > >
>> > > means that if the factor counts includes n ones, the LP problem  
>> used by
>> > > T2c has 2^n times as many variables as the one used by t5. Yet this
>> > > yielded only a 10-15% improvement, by itself.
>> > >
>> > > A bigger gain was achieved in factorCounts, based on one of Carl 
>> Woll's
>> > > ideas in
>> > >
>> > > http://forums.wolfram.com/mathgroup/archive/2007/Jul/msg00411.html
>> > >
>> > > factorCounts[k_Integer] :=
>> > >   If[EvenQ@k, factorCounts[(k + 2)/2, k - 1],
>> > >    factorCounts[(k - 1)/2, k + 2]]
>> > >
>> > > Several fine adjustments were required to get any improvement at 
>> all.
>> > > Defining factorCounts for 1,2,3,4, and 7 removed steps from the  
>> code for
>> > > other cases, and it was crucial to take full advantage of the fact  
>> that
>> > > the GCD of the two factors could only be 1 or 3.
>> > >
>> > > Bobby
>> > >
>> > > On Tue, 07 Aug 2007 00:31:22 -0500, sashap <pavlyk at gmail.com> wrote:
>> > >
>> > > > On Aug 5, 3:58 am, Carl Woll <ca... at wolfram.com> wrote:
>> > > >> sashap wrote:
>> > > >> >Carl's code is very good on numbers not involving
>> > > >> >products of large powers of primes. Otherwise it
>> > > >> >it is memory intensive and not as efficient:
>> > > >>
>> > > >> >In[109]:= Table[Timing[LargestPartition[{k, k}]], {k, 7, 15}]
>> > > >>
>> > > >> >Out[109]=
>> > > >> >{{0.031,7},{0.141,7},{0.172,8},{0.187,9},{1.531,9},{1.875,10},
>> > > >> >{12.016,10},{16.234,11},{141.391,11}}
>> > > >>
>> > > >> >One can use LinearProgramming to improve on the
>> > > >> >situation.
>> > > >>
>> > > >> >Additional improvement comes from the following
>> > > >> >observation. The longest factorization can be built
>> > > >> >by taking the longest sequence of divisors of degree 1,
>> > > >> >then of degree 2 and so on. The degree of the divisor
>> > > >> >is defined as
>> > > >>
>> > > >> >  deg[n_] := Total[ FactorInteger[n][[All,2]] ]
>> > > >>
>> > > >> >In other words, let n == p1^e1 *...* pn^en.
>> > > >> >First take {p1, p2,.., pn}, then the longest
>> > > >> >sequence of pairs with repetitions and so on.
>> > > >>
>> > > >> Nice work! However, I think there is a problem with your  
>> algorithm.
>> > > >> There are many possible ways to a longest sequence of (distinct)  
>> pairs
>> > > >> with repetitions. It is possible that some of these sequences 

>> will not
>> > > >> produce the longest factorization. For instance, consider 2^10 *  
>> 3^3 *
>> > > >> 5^2. After removing the degree 1 factors we are left with 2^9 *  
>> 3^2 * 5.
>> > > >> Two possible length 3 sequences of pairs are:
>> > > >>
>> > > >> a) 2^2, 2*3, 2*5 leaving 2^5*3
>> > > >>
>> > > >> and
>> > > >>
>> > > >> b) 2^2, 2*3, 3*5 leaving 2^6
>> > > >>
>> > > >> Now, for case a) we can create two more unique factors, 2^3 and  
>> 2^2*3,
>> > > >> while for case b) we can't create two more unique factors.
>> > > >>
>> > > >> So, I think one more thing is needed in your algorithm, a way of  
>> knowing
>> > > >> which longest sequence of pairs to pick.
>> > > >>
>> > > >> Now, it turns out that the above example is handled correctly by  
>> your
>> > > >> algorithm, that is, your algorithm by design or chance picks a 
>> longest
>> > > >> sequence that leads to a longest factorization. I played around  
>> with a
>> > > >> bunch of different examples, and came up with one example where  
>> your
>> > > >> algorithm picks a longest sequence that doesn't lead to a longest
>> > > >> factorization. This wasn't easy because my algorithm is soo slow  
>> (but I
>> > > >> think, correct).
>> > > >>
>> > > >> In[594]:= counterexample   {2,3,4,5,6,7,8,10,12,14,15,35};
>> > > >>
>> > > >> In[595]:= Times @@ counterexample
>> > > >>
>> > > >> Out[595]= 35562240000
>> > > >>
>> > > >> In[596]:= Length[counterexample]
>> > > >>
>> > > >> Out[596]= 12
>> > > >>
>> > > >> In[597]:= LengthOfLongestDecomposition[Times @@ counterexample]
>> > > >>
>> > > >> Out[597]= 11
>> > > >>
>> > > >> My guess is that a simple heuristic probably exists which will

>> allow you
>> > > >> to pick a good longest sequence instead of a bad longest  
>> sequence.
>> > > >> Perhaps something like choosing the longest sequence that keeps  
>> as many
>> > > >> different types of prime factors available for the next stage? 
>> If so,
>> > > >> then the biggest bottleneck will be just factoring the triangular
>> > > >> numbers.
>> > > >>
>> > > >> Carl
>> > > >>
>> > > >
>> > > > Carl,
>> > > >
>> > > > good points ! Indeed:
>> > > >
>> > > > In[98]:= ffaux /@ Permutations[FactorInteger[chi][[All, 2]] 
>> > > >
>> > > > Out[98]= {12,12,11,12,12,12,12,12,12,11,11,11}
>> > > >
>> > > > The heuristics you mention might be different sorting of  
>> exponents,
>> > > > even though we can not guarantee it, that is
>> > > > changing LengthOfLongestDecomposition to the following:
>> > > >
>> > > > LengthOfLongestDecomposition[nn_Integer] :=
>> > > >   ffaux[Reverse@Sort@FactorInteger[nn][[All, 2]]]
>> > > >
>> > > > Another approach is to write n == d1^s1 * ... * dk^sk and just use
>> > > > LinearProgramming once. Here
>> > > > s1, ..., sk are either 0 or 1, and d1, ..., dk are all distinct
>> > > > divisors of the given integer n.
>> > > >
>> > > > Clear[lldaux, lld];
>> > > > lldaux[Lpow_] := (lldaux[Lpow] =
>> > > >    With[{Mtup = Tuples[ Range[0, #] & /@ Lpow]},
>> > > >     Total[
>> > > >       Quiet@LinearProgramming[ConstantArray[-1, Length[Mtup]],
>> > > >         Transpose@Mtup, Thread@{Lpow, 0},
>> > > >         Array[{0, 1} &, Length[Mtup]], Integers]] - 1])
>> > > >
>> > > > lld[n_Integer] := lldaux[FactorInteger[n][[All, 2]] // Sort]
>> > > >
>> > > > T2c[n_] := # (# + 1)/2 &@
>> > > >   NestWhile[# + 1 &, 2, lld[# (# + 1)/2 - 1] < n &]
>> > > >
>> > > > The performance of T2c is competitive to Carl's code for small 

>> values
>> > > > of the argument
>> > > > and has an edge over it for large ones:
>> > > >
>> > > > In[168]:= {Timing@T2[11], Timing@T2c[11]}
>> > > >
>> > > > Out[168]= {{3.829,2080995841},{3.921,2080995841}}
>> > > >
>> > > > In[169]:= {Timing@T2[12], Timing@T2c[12]}
>> > > >
>> > > > Out[169]= {{26.532,68302634401},{22.797,68302634401}}
>> > > >
>> > > > In[170]:= {Timing@T2[13], Timing@T2c[13]}
>> > > >
>> > > > Out[170]= {{144.406,924972048001},{123.875,924972048001}}
>> > > >
>> > > > Oleksandr Pavlyk and Maxim Rytin
>> > > >
>> > > >> >In order to choose the longest sequence of factors of a given
>> degree,
>> > > >> >we set this up as an integer linear programming problem. We  
>> illustrate
>> > > >> >this for degree 2. Let p1*p1, p1*p2, and so on be candidate  
>> factors of
>> > > >> >degree 2. Then
>> > > >>
>> > > >> >   (p1*p1)^s1 (p1*p2)^s2 ...
>> > > >>
>> > > >> >must divide the original number. Additionally, each sk is  
>> either 0 or
>> > > >> >1.
>> > > >> >This gives us the constraints for variables s1, s2, ...   and
>> > > >> >the objective function being maximized is s1+s2+....
>> > > >>
>> > > >> >The code implementing those ideas:
>> > > >>
>> > > >> >CombinationsWithRepetitions >> > > >> >  Compile[{{n, _Integer}, {k, _Integer}},
>> > > >> >   Module[{ans, i  1, j = 1, l = 1}, >> > > >> >    ans = Array[0 &, {Binomial[n + k - 1, k], k}];
>> > > >> >    While[True, While[j <= k, ans[[i, j++]] = l];
>> > > >> >     If[i == Length@ans, Break[]];
>> > > >> >     While[ans[[i, --j]] == n,];
>> > > >> >     l = ans[[i++, j]] + 1;
>> > > >> >     ans[[i]] = ans[[i - 1]];];
>> > > >> >    ans]];
>> > > >>
>> > > >> >Clear[LengthOfLongestDecomposition, ffaux]
>> > > >> >LengthOfLongestDecomposition[nn_] :=
>> > > >> > ffaux[Sort@FactorInteger[nn][[All, 2]]]
>> > > >> >ffaux[$Lpow_] :=
>> > > >> > ffaux[$Lpow] =
>> > > >> >  Module[{Lpow = $Lpow, Ldiv, n, m, k = 1, Lind, ans = 0},
>> > > >> >   n = Length@Lpow;
>> > > >> >   While[Total@Lpow >= k,
>> > > >> >    Ldiv = CombinationsWithRepetitions[n, k++];
>> > > >> >    m = Length@Ldiv;
>> > > >> >    Lind =
>> > > >> >     Quiet@LinearProgramming[ConstantArray[-1, m],
>> > > >> >       Total[1 - Unitize[Ldiv - #], {2}] & /@ Range@n,
>> > > >> >       Thread@{Lpow, -1}, Array[{0, 1} &, m], Integers];
>> > > >> >    ans += Total@Lind;
>> > > >> >    Lpow -= BinCounts[Flatten@Pick[Ldiv, Lind, 1], {Range[n 
>> + 1]}]];
>> > > >> >   ans]
>> > > >>
>> > > >> >Compare the timing given earlier with the following
>> > > >>
>> > > >> >In[132]:= Table[Timing[ffaux[{k, k}]], {k, 7, 15}]
>> > > >>
>> > > >> >Out[132]= \
>> > > >>  =

>> >{{0.016,7},{1.78746*10^-14,7},{1.78746*10^-14,8},{0.015,9},{3.15303* \
>> > > >> >10^-14,9},{0.016,10},{0.,10},{0.015,11},{0.,11}}
>> > > >>
>> > > >> >Clearly, the use of heavy machinery of LinearProgramming comes  
>> at the
>> > > >> >cost of
>> > > >> >noticeable overhead which is why the performance of this  
>> algorithm is
>> > > >> >comparable
>> > > >> >to that of Carl's code when measured over a sequence of  
>> triangular
>> > > >> >numbers because
>> > > >> >problematic numbers are rare.
>> > > >>
>> > > >> >T2b[n_] := # (# + 1)/2 &@
>> > > >> >  NestWhile[# + 1 &, 1,
>> > > >> >   LengthOfLongestDecomposition[# (# + 1)/2 - 1] < n &]
>> > > >>
>> > > >> >In[66]:= {Timing[T2[11]], Timing[T2b[11]]}
>> > > >>
>> > > >> >Out[66]= {{3.172,2080995841},{2.328,2080995841}}
>> > > >>
>> > > >> >In[67]:= {Timing[T2[12]], Timing[T2b[12]]}
>> > > >>
>> > > >> >Out[67]= {{26.312,68302634401},{18.235,68302634401}}
>> > > >>
>> > > >> >In[68]:= {Timing[T2[13]], Timing[T2b[13]]}
>> > > >>
>> > > >> >Out[68]= {{142.765,924972048001},{111.422,924972048001}}
>> > > >>
>> > > >> >Oleksandr Pavlyk and Maxim Rytin
>> > > >>
>> > > >> >On Jul 10, 5:37 am, Carl Woll <ca... at wolfram.com> wrote:
>> > > >>
>> > > >> >>Andrzej and Diana,
>> > > >>
>> > > >> >>Here is a faster algorithm. First, an outline:
>> > > >>
>> > > >> >>For a given number, the easy first step in writing  it as a
>> > > >> >>factorization of the most distinct factors is to use each  =

>> prime as a
>> > > >> >>factor. Then, the hard part is to take the remaining factors=
  =

>> and
>> > > >> >>partition them in a way to get the most distinct factors.  =

>> However,
>> > > >> note
>> > > >> >>that this step only depends on the counts of each prime  =

>> factor, and
>> > > >> not
>> > > >> >>on the values of the primes themselves. Hence, we can memoiz=
e  =

>> the
>> > > >> number
>> > > >> >>of distinct factors possible for a set of counts. I do this =
as
>> > > >> follows:
>> > > >>
>> > > >> >>Let f be the list of remaining factors, of length l.
>> > > >>
>> > > >> >>1. Determine possible integer partitions of l, with the  =

>> smallest
>> > > >> >>partition allowed being 2. We also need to order these  =

>> partitions by
>> > > >> >>length, from largest to smallest:
>> > > >>
>> > > >> >>partitions = IntegerPartitions[l, All, Range[2, l]];
>> > > >> >>partitions = Reverse@partitions[[ Ordering[Length/@partiti=
ons]  =

>> ]];
>> > > >>
>> > > >> >>2. Now, we need to test each partition to see if it's possib=
le  =

>> to fill
>> > > >> >>in the partitions with the factors such that each partition =
is  =

>> unique.
>> > > >> >>Once we find such a partition, we are done, and we know how =
 =

>> many
>> > > >> >>distinct factors that number can be written as. This step I =
do  =

>> by
>> > > >> >>recursion, i.e., take the first member of a partition and fi=
ll  =

>> it in
>> > > >> >>with one of the possible subsets of factors, and then repeat=
  =

>> with the
>> > > >> >>remaining members of a partition. I do this with the functio=
n
>> > > >> Partition=
>> > > > able.
>> > > >>
>> > > >> >>Here is the code:
>> > > >>
>> > > >> >>LargestPartition[list : {1 ..}] := Length[list]
>> > > >> >>LargestPartition[list_List] :=  Length[list] +
>> > > >> >>LargestTwoPartition[Reverse@Sort@list - 1 /. {a__, 0 ..} -> =
 =

>> {a}]
>> > > >>
>> > > >> >>Clear[LargestTwoPartition]
>> > > >> >>LargestTwoPartition[list_] :=  LargestTwoPartition[list] ==
  =

>> Module[{=
>> > > > set,
>> > > >> >>partitions, res},
>> > > >> >>   set = CountToSet[list];
>> > > >> >>   partitions = IntegerPartitions[Total[list], All, Range[=
2,  =

>> Total[li=
>> > > > st]]];
>> > > >> >>   partitions = Reverse@partitions[[Ordering[Length /@  =

>> partitions]]];
>> > > >> >>   res = Cases[partitions, x_ /; Partitionable[{}, set, x]=
, 1,  =

>> 1];
>> > > >> >>   If[res === {},
>> > > >> >>      0,
>> > > >> >>      Length@First@res
>> > > >> >>   ]
>> > > >> >>]
>> > > >>
>> > > >> >>Partitionable[used_, unused_, part_] := Module[{first, res=
t},
>> > > >> >>  first = Complement[DistinctSubsets[unused, First@part], =
 =

>> used];
>> > > >> >>  If[first === {}, Return[False]];
>> > > >> >>  rest = CountToSet /@ Transpose[
>> > > >> >>    SetToCount[unused, Max[unused]] -
>> > > >> >>    Transpose[SetToCount[#, Max[unused]] & /@ first]
>> > > >> >>  ];
>> > > >> >>  Block[{Partitionable},
>> > > >> >>    Or @@ MapThread[
>> > > >> >>      Partitionable[Join[used, {#1}], #2, Rest[part]] &,
>> > > >> >>      {first, rest}
>> > > >> >>    ]
>> > > >> >>  ]
>> > > >> >>]
>> > > >>
>> > > >> >>Partitionable[used_, rest_, {last_}] := ! MemberQ[used, re=
st]
>> > > >>
>> > > >> >>CountToSet[list_] := Flatten@MapIndexed[Table[#2, {#1}] &,=
  =

>> list]
>> > > >> >>SetToCount[list_, max_] := BinCounts[list, {1, max + 1}]
>> > > >>
>> > > >> >>DistinctSubsets[list_, len_] := Union[Subsets[list, {len}]=
]
>> > > >>
>> > > >> >>T2[n_] := Module[{k=1},
>> > > >> >>  While[k++; LargestPartition[FactorInteger[k*((k + 1)/2) - =
 =

>> 1][[All,
>> > > >> >>2]]] < n];
>> > > >> >>  k*((k + 1)/2)
>> > > >> >>]
>> > > >>
>> > > >> >>Then:
>> > > >>
>> > > >> >>In[11]:= T2[6] // Timing
>> > > >> >>Out[11]= {0.047,13861}
>> > > >>
>> > > >> >>In[12]:= T2[7] // Timing
>> > > >> >>Out[12]= {0.031,115921}
>> > > >>
>> > > >> >>In[13]:= T2[8] // Timing
>> > > >> >>Out[13]= {0.11,665281}
>> > > >>
>> > > >> >>In[14]:= T2[9] // Timing
>> > > >> >>Out[14]= {0.625,18280081}
>> > > >>
>> > > >> >>In[15]:= T2[10] // Timing
>> > > >> >>Out[15]= {1.157,75479041}
>> > > >>
>> > > >> >>In[16]:= T2[11] // Timing
>> > > >> >>Out[16]= {5.875,2080995841}
>> > > >>
>> > > >> >>In[17]:= T2[12] // Timing
>> > > >> >>Out[17]= {48.703,68302634401}
>> > > >>
>> > > >> >>For Diana, note that
>> > > >>
>> > > >> >>In[20]:= 481 482/2 - 1 == 115921 - 1 == 2 3 4 5 6 =
7 23
>> > > >> >>Out[20]= True
>> > > >>
>> > > >> >>so 115921-1 can indeed be written as a product of 7 distinct=
  =

>> factors.
>> > > >>
>> > > >> >>Partitionable can be improved, but another bottleneck for  =

>> larger cases
>> > > >> >>is evaluating FactorInteger. One possibility for improving
>> > > >> FactorInteger
>> > > >> >>speed is to note that
>> > > >>
>> > > >> >>In[21]:= n (n + 1)/2 - 1 == (n + 2) (n - 1)/2 // Expan=
d
>> > > >>
>> > > >> >>Out[21]= True
>> > > >>
>> > > >> >>So, rather than applying FactorInteger to n(n+1)/2-1 you can=
  =

>> instead
>> > > >> >>somehow combine FactorInteger[n+2] and FactorInteger[n-1].
>> > > >>
>> > > >> >>At any rate, it takes a bit longer, but with enough patience=
  =

>> one can
>> > > >> >>also find:
>> > > >>
>> > > >> >>In[53]:= 1360126 1360127/2 - 1 == 924972048001 - 1 ==
=  2 3 4 =
>> > > > 5 6 7 10 11
>> > > >> >>12 13 15 23 31
>> > > >> >>Out[53]= True
>> > > >>
>> > > >> >>and
>> > > >>
>> > > >> >>In[56]:= 9830401 9830402/2 - 1 == 48318396825601 - 1 ==
=  2 3 =
>> > > > 4 5 6 8 9
>> > > >> >>10 11 12 17 22 32 59
>> > > >> >>Out[56]= True
>> > > >>
>> > > >> >>Carl Woll
>> > > >> >>Wolfram Research
>> > > >>
>> > > >> >>Andrzej Kozlowski wrote:
>> > > >>
>> > > >> >>>Well, I stayed up longer than I wanted and I think I have n=
ow  =

>> fixed
>> > > >> >>>it, I hope for the final time. Here is the new FF:
>> > > >>
>> > > >> >>>FFF[n_] := Module[{u = FactorInteger[n], s, k, partialQ=
,  =

>> finalQ, s=
>> > > > pace},
>> > > >> >>>   s = u[[All,2]]; k = Length[u]; partialQ[l_List] :==

>> > > >> >>>     And @@ Flatten[{Last[l] == Array[0 & , k] ||
>> > > >> >>>          !MemberQ[Most[l], Last[l]], Thread[Total[l] <= =
s -  =

>> 1]}];
>> > > >> >>>    finalQ[l_List] := And @@ Flatten[{Last[l] == Arra=
y[0 & ,  =

>> k] =
>> > > > ||
>> > > >> >>>          !MemberQ[Most[l], Last[l]], Total[l] == s - 1=
}];
>> > > >> >>>    space =
>> > > >>
>> > > >> ...
>> > > >>
>> > > >> read more =BB
>> > > >
>> > > >
>> > > >
>> > > >
>> > >
>> > >
>> > >
>> > > --
>> > > DrMajorBob at bigfoot.com
>> > >
>> >
>>
>
>



-- =

DrMajorBob at bigfoot.com


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