Re: List manipulation question - 2013

*To*: mathgroup at smc.vnet.net*Subject*: [mg129632] Re: List manipulation question - 2013*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Thu, 31 Jan 2013 20:47:21 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <20130130092653.09BA46877@smc.vnet.net>

Clear[f] f[a_, b_, outputLen_: (-1), c_: 2000] := Module[{list1, list2}, list1 = Table[24 + (a*n1), {n1, c}]; list2 = Select[list1, IntegerQ[#/24] &]; list3 = Last /@ (Select[ Solve[{12 + (b*n2) == #, Element[n2, Integers]}, n2] & /@ list2, Length[#] > 0 &] // Flatten); If[outputLen == -1, list3, list3[[1 ;; Min[outputLen, Length[list3]]]]]]; f[15, 3, 6] {44, 84, 124, 164, 204, 244} f[13, 7, 6] {180, 492, 804, 1116, 1428, 1740} Bob Hanlon On Thu, Jan 31, 2013 at 2:56 AM, Lea Rebanks <lrebanks at gmail.com> wrote: > Hi Bob, > > Many thanks for your excellent reply. > Your coding for lists 1,2,3 is exactly what I want. > And your statement - "Apparently you want list2 to consist of the members of > list1 that are > integer multiples of 24 rather than the integers in list1/24." is correct. > > But I tried changing the input values from the example given > From 24 + (15*n1) to 24 + (13*n1) .....15 to 13 > From 12 + (3*n2) to 12 + (7*n2) .....3 to 7 > This immediately caused me error messages, some of which I was able to > correct by expanding the list range of the Take function. > But I was unable to calculate the last part for list3 :- > > list1 = Table[24 + (13*n1), {n1, 2000}]; > list2 = Take[Select[list1, IntegerQ[#/24] &], 40]; > list3 = n2 /. > Solve[{12 + (7*n2) == #, Element[n2, Integers]}, n2][[1]] & /@ > list2 > > Hopefully the result should look something like :- > {1272, 3456, 5640, 7824, 10008, 12192} > These values above being the first & then following repetitions thereafter > of equal points of repetitions of n1 & n2 in list2 & list3 respectively. > > Please could you rewrite the enclosed code so that :- > (i) The above new values work & provide a similar result to that which I > have shown. > (ii) Also, if possible, allow in the rewrite the flexibility within the > function to accept much larger & challenging input values for list1 & 2. > This would include a domain allowance relative to the size of input values > as in this case its values are 13 & 7, but they could be integer values of > 127 & 56 for list1 & list3 respectively. > (Note higher value first.) Actually I will probably want to use much much > higher values later. > Ideally I am only looking for about 6 elements in the final answer list like > - {1272, 3456, 5640, 7824, 10008, 12192} > > Many thanks for your help & attention. > Really appreciated. > Best regards, > Lea... > > > > > On Thu, Jan 31, 2013 at 11:05 AM, Bob Hanlon <hanlonr357 at gmail.com> wrote: >> >> list1 = Table[24 + (15*n1), {n1, 2000}]; >> >> Take[Select[list1/24, IntegerQ], 9] >> >> {6, 11, 16, 21, 26, 31, 36, 41, 46} >> >> Take[Select[list1, IntegerQ[#/24] &], 9] >> >> {144, 264, 384, 504, 624, 744, 864, 984, 1104} >> >> Apparently you want list2 to consist of the members of list1 that are >> integer multiples of 24 rather than the integers in list1/24. >> >> list2 = Select[list1, IntegerQ[#/24] &]; >> >> list3 = n2 /. Solve[ >> {12 + (3*n2) == #, Element[n2, Integers]}, >> n2][[1]] & /@ list2; >> >> Length[list2] == Length[list3] >> >> True >> >> >> Bob Hanlon >> >> Consequently, there is an integer n2 for every element of list2. >> >> On Wed, Jan 30, 2013 at 4:26 AM, Lea Rebanks <lrebanks at gmail.com> wrote: >> > >> > Dear All, >> > >> > I have the follow problem with the combination of a few lists. >> > I shall outline the problem or what I am trying to do. >> > I know that to solve this problem requires a number of processes, but I >> > am >> > not sure how to setup the coding to achieve my desired result. >> > Please could someone show me the coding for this problem. >> > >> > The problem:- >> > 1 - I have a list1 created from 24+(15*n1) All n1 are integers >> > 1,2,3,4.....to a large number of integers. >> > 2 - I want to divide the list1 by 24 and create another list2 with only >> > the >> > integer results in list2. >> > Table[24 + 15*i, {i, 100}]/24 ... so that integer results in >> > list2 >> > = { 144, 264, 384, 504, 624, 744, 864, 984, 1104 } >> > 3 - I have another list3 created from 12+(3*n2) All n2 are integers >> > 1,2,3,4.....to a large number of integers. >> > 4 - With list3 I want to find :- >> > (i) The integer number of n2 that either equates 12+(3*n2) = 144 or >> > the next FIRST available number in list2 that meets this equality. >> > then also (ii)The integer number(s) of n2 that equates 12+(3*n2) = >> > 24+(15*n1) after the first equal value for say 6 repetitions. >> > then also (iii)The integer number(s) of n1 that equates 12+(3*n2) = >> > 24+(15*n1) after the first equal value for say 6 repetitions. >> > >> > The above example is quite simple, but I am hoping to setup the coding >> > to >> > work with other integer values instead of 3 & 15 which will present more >> > of >> > a challenge. >> > >> > Any help or advice greatly appreciated. >> > Best regards, >> > Lea... >> > >> > >> >

**References**:**List manipulation question - 2013***From:*Lea Rebanks <lrebanks@gmail.com>

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