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

**Increasing scattered subsequence**

**Re: Foucault pendulum**

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

**rebuild the help index**