[QUOTE=CRGreathouse]Let's see. (b) evaluates to b, v[b] evaluates to the b-th value in v. If v happens to be a vector, and #v is at least as large as the value of b, then it passes whatever happens to be in that location in the vector to forstep. Forstep ignores the values its given (unlike, say, sum() or prod()), so this code has the following behavior:

* If v is a vector with size at least as large as the final value of b (around 10^a), then do nothing.
* If lift(Mod(-1,x)/n) is greater than 10^a, then do nothing.
* Otherwise, if v is a vector, throw an "index out of bounds" error.
* If v is not a vector, and lift(Mod(-1,x)/n) is <= 10^a, then throw a "_[_]: not a vector" error.[/QUOTE]

Great. What if I pinned a return value on it instead? Oh. Right. Returns the first value

CRG I got my code to work to find primes and fill a vector of 0's with them I'm thinking of how to make a function VecApp(v,x,f) that takes one vector(v) into another vector(f) with 1 extra element which is appended with value x maybe that will make it easier on me. I think this would need to say something like f[i] = v[i] f[i]+1 = x but I'm without reading them all I have no idea how to get i not to go over the amount of indexes of v.

[QUOTE=3.14159;226190]Great. What if I pinned a return value on it instead? Oh. Right. Returns the first value[/QUOTE]

Just look at my code for how to use the vector!

scratch f we can just use v=f to redefine it back to v after.

[QUOTE=science_man_88;226201]CRG I got my code to work to find primes and fill a vector of 0's with them I'm thinking of how to make a function VecApp(v,x,f) that takes one vector(v) into another vector(f) with 1 extra element which is appended with value x maybe that will make it easier on me. I think this would need to say something like f[i] = v[i] f[i]+1 = x but I'm without reading them all I have no idea how to get i not to go over the amount of indexes of v.[/QUOTE]

[code]concat([1,2,3],10)[/code]

I'm an idiot thanks.

[QUOTE=CRGreathouse]Just look at my code for how to use the vector!
[/QUOTE]

my(v=vectorsmall(sz, i, 1))?

What is supposed to be the substitution here?

[QUOTE=science_man_88;226205]I'm an idiot thanks.[/QUOTE]

You were trying to write a function that already existed. That's pretty common. I wrote my own version of eint1, at the very least -- but probably others, too.

A 1 in 10[sup]65[/sup] chance:

I entered a(65), which is nextprime(random(10^65)). The result: 2.

Well, 3 in 10^65 anyway. Wow.

[QUOTE=CRGreathouse]Well, 3 in 10^65 anyway. Wow.
[/QUOTE]

It was in fact, too good to be true. It turned out PARI was messed for a while.

I entered a(7) and got 2, 6 times in a row. I tried a(8) through a(12), and got 2, about 20 times on end.

This would normally happen only about 1 in ≈10[sup]163[/sup] times. I'm going with PARI being a bit buggy. Or did I have a supreme stroke of luck and just defy odds of 1 in 10[sup]163[/sup]?

Also: The lottery should be changed to guessing 10 random numbers between 1 and 10[sup]15[/sup].

(To ensure that nobody wins, :devil:)

You may have inadvertently overwritten a. Try ?a to see the current definition.

[QUOTE=CRGreathouse]You may have inadvertently overwritten a. Try ?a to see the current definition.
[/QUOTE]

Nope. I changed nothing. It was either incredible luck or the computer fucking up.

[CODE](15:27) gp > v=vector(1,n,0);a=1;for(x=1,10000,if(isprime(x),v[a]=x;concat(v[],0);a=a+1))
[COLOR="Red"]*** unused characters: v=vector(1,n,0);a=1;for(x=1,10000,if(is
^--------------------[/COLOR][/CODE]

I'm leaning towards the latter being true.

@sm88: Too many arguments, failed to specify whether or not a=1 was part of the command, a=1 treated as its own command

[QUOTE=3.14159;226216]I'm leaning towards the latter being true.

@sm88: Too many arguments, failed to specify whether or not a=1 was part of the command, a=1 treated as its own command[/QUOTE]

it should be it's own command. it's only when i used concat() that this happened.

[QUOTE=3.14159;226214]Nope. I changed nothing. It was either incredible luck or the computer fucking up.[/QUOTE]

I'm leaning toward "you used a as a variable in a function without using local or my" which would explain it pretty well.

[QUOTE=3.14159;226216]@sm88: Too many arguments, failed to specify whether or not a=1 was part of the command, a=1 treated as its own command[/QUOTE]

[QUOTE=science_man_88;226217]it should be it's own command. it's only when i used concat() that this happened.[/QUOTE]

I don't know what either of these mean.

[QUOTE=CRGreathouse]I'm leaning toward "you used a as a variable in a function without using local or my" which would explain it pretty well.
[/QUOTE]

Nope. I defined f(x, n) as follows:

Print nextprime(x), followed by * , followed by nextprime(n), followed by printing the product of those two. In other words, a semiprime generator that gives the factors used to produce the semiprime as well.

a(n) = nextprime(random(10^n)).

a and f were well-defined. a was never a variable, it is a function.

a simply gives an n-digit prime most of the time.

Therefore, there was no error in my defined functions.

[CODE] v=vector(1250,n,0);a=1;for(x=1,10000,if(isprime(x),v[a]=x;a=a+1))[/CODE]

works but you already have to know the # of primes under the x upper bound to not get an error or a 0 at the end.

what I want to do is add a 0 into v every time a prime is found so as to have no 0's at the end of it all and to only have the number of indices necessary to list all the primes under a given number once the vector is made I can get a sieve like the sieve of Eratosthenes to try and work with it to create a bigger list instead of asking isprime hopefully I can make a way to check primality quicker and then make a function that uses the vector indices to check Prime2(x).

[QUOTE=CRGreathouse]I don't know what either of these mean.
[/QUOTE]

I was merely trying to explain sm88's error.

[code]v=vector(1,n,0);
a=1;
for(x=1,10000,
if(isprime(x),
v[a]=x;
concat(v[],0);
a=a+1
)
)[/code]

I'm not sure what "concat(v[],0);" is trying to do, but it's not right. v[] will cause an error. If you replace v[] with something that isn't an error, like v, then the result would take v and concatenate it with 0, then throw the result away. Maybe you mean

[code]v=concat(v,0);[/code]

which would take v, append 0 to the end, and replace the old v with the "old v followed by a 0" vector.

Here's how I would do what I think you're trying to do:
[code]v=[];
for(x=1,10000,
if(isprime(x),
v=concat(v,x)
)
)[/code]
or even
[code]v=[];
forprime(p=2,10000,
v=concat(v,p)
)[/code]
or even
[code]v=vector(primepi(10000));
i=0;
forprime(p=2,10000,
i++;
v[i]=p
)[/code]
which I would usually write in the confusing programmer-ese as
[code]v=vector(primepi(10000));
i=0;
forprime(p=2,10000,
v[i++]=p
)[/code]

[QUOTE=3.14159;226220]Nope. I defined f(x, n) as follows:

Print nextprime(x), followed by * , followed by nextprime(n), followed by printing the product of those two. In other words, a semiprime generator that gives the factors used to produce the semiprime as well.

a(n) = nextprime(random(10^n)).

a and f were well-defined. a was never a variable, it is a function.

a simply gives an n-digit prime most of the time.

Therefore, there was no error in my defined functions.[/QUOTE]

Did you put in ?a ? What did it say?

now if I only I could be smart like you so I could build you that helpful program lol.

[QUOTE=CRGreathouse]Did you put in ?a ? What did it say?
[/QUOTE]

?a = a(n) = nextprime(10^n))

[QUOTE=science_man_88;226226]now if I only I could be smart like you so I could build you that helpful program lol.[/QUOTE]

[i]I'm[/i] not smart enough to write that program, but maybe you are. You have a long way to go, though!

Here are some possibly-relevant papers in case you decide to think more about this:
[url=http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.132.8640&rep=rep1&type=pdf]mArachna – Ontology Engineering for Mathematical Natural Language Texts[/url]
[url=http://mathweb.org/projects/mws/pubs/mkm08.pdf]MathWebSearch 0.4: A Semantic Search Engine for Mathematics[/url]
etc.

[QUOTE=3.14159;226232]?a = a(n) = nextprime(10^n))[/QUOTE]

I'm sure it didn't say that -- it's not in the form of a valid closure. But if it said
[code]a(n) = nextprime(10^n)[/code]
then it would never return 2 for n >= 1 -- it's a constant, no randomness.

Now: Can we get back to the substitutions required to make the sieve for the k * n + 1?

[QUOTE=3.14159;226235]Now: Can we get back to the substitutions required to make the sieve for the k * n + 1?[/QUOTE]

Why don't you just download srsieve and use that? It'll be faster.

[QUOTE=CRGreathouse]Why don't you just download srsieve and use that? It'll be faster.[/QUOTE]

Because it can only sieve for Proth numbers.

[QUOTE=3.14159;226242]Because it can only sieve for Proth numbers.[/QUOTE]

It does more than Proth numbers. What, precisely, are you sieving for? Can you specify exactly for me?

[QUOTE=CRGreathouse]It does more than Proth numbers. What, precisely, are you sieving for? Can you specify exactly for me?
[/QUOTE]

k * n! + 1.

Maybe this?
[url]http://fatphil.org/maths/factorial/[/url]

Possibly even this?
[url]http://pgllr.mine.nu/software/fsieve/[/url]

[QUOTE=CRGreathouse]Maybe this?
[url]http://fatphil.org/maths/factorial/[/url]

Possibly even this?
[url]http://pgllr.mine.nu/software/fsieve/[/url][/QUOTE]

The first is source code, luckily I downloaded some freeware to deal with C source code.

[QUOTE=3.14159]The first is source code, luckily I downloaded some freeware to deal with C source code.[/QUOTE]

Nevermind that. I'm going to go to Generalized proths, using factorial and or primorial powers, just need a larger base range. Anything for that?

[QUOTE=3.14159;226258]Nevermind that. I'm going to go to Generalized proths, using factorial and or primorial powers, just need a larger base range. Anything for that?[/QUOTE]

I can never keep track of all the names (especially since some are contradictory... e.g. several definitions for generalized Fermats). What form are these?

can you make vectors of vectored strings ? if so we can do a find and replace like operation such that Pari can code it.

[QUOTE=3.14159]Nevermind that. I'm going to go to Generalized proths, using factorial and or primorial powers, just need a larger base range. Anything for that?
[/QUOTE]

To continue endlessly quoting myself...
PFGW's factor switch helps out for larger b.

[QUOTE=3.14159]I can never keep track of all the names (especially since some are contradictory... e.g. several definitions for generalized Fermats). What form are these?
[/QUOTE]

Generalized proths = k * b[sup]n[/sup] + 1 where b is any integer ≥ 2.

And: Please, show where I was self-contradictory.

got the vector within vector worked out now I just need to check them out lol.

got it working with one catch need to quotes around the phrase in the input lol. but then unless we insert them and check it will be hard.

oh and if you accidentally insert anything that gets typed over it wrecks it it seems.

[CODE]try = [Vec(x="string"),Vec(x="suck"),Vec(x="primes")][/CODE]

and

[CODE]x=3;y=Vec(input());if(y==try[x],print("really"))[/CODE]

try it out if you want remember quotes and hopefully no messing up to make it work.

[QUOTE=3.14159;226265]Generalized proths = k * b[sup]n[/sup] + 1 where b is any integer ≥ 2.[/QUOTE]

You can use srsieve for those.

[QUOTE=3.14159;226265]And: Please, show where I was self-contradictory.[/QUOTE]

I made no such accusation.

[QUOTE=science_man_88;226281][CODE]try = [Vec(x="string"),Vec(x="suck"),Vec(x="primes")][/CODE][/QUOTE]

You probably want
[CODE]try = [Vec("string"),Vec("suck"),Vec("primes")][/CODE]

[QUOTE=CRGreathouse;226294]You probably want
[CODE]try = [Vec("string"),Vec("suck"),Vec("primes")][/CODE][/QUOTE]


no it gave me a too few arguments error that way

got it working never mind still needed the quotes.

how do i check word/phrase by word/phrase through text if i need parentheses everywhere. get that under control then we can at least make a limited version.

basically we need a find and replace script to edit the text to something we can make Pari understand.

we need to build a gp2c like exe that is more of a nl2gp (natural language to pari) compiler.

[CODE]y=Vec(input());for(x=1,3,forstep(i=1,#y,#try[x],if(y[i]==try[x][1],print("really"))))[/CODE]

what I've got working.

had to make to:

[CODE]y=Vec(input());for(x=1,3,for(i=1,#y,if(y[i]==try[x][1],print("really"))))[/CODE]

[QUOTE=CRGreathouse]You can use srsieve for those.
[/QUOTE]

Or NewPGen.

[QUOTE=CRGreathouse]I made no such accusation.
[/QUOTE]

You said I made several contradictory names for various prime forms, and you were therefore accusing me of being internally inconsistent. Please show where I am internally inconsistent.

[QUOTE=3.14159;226307]You said I made several contradictory names for various prime forms, and you were therefore accusing me of being internally inconsistent. Please show where I am internally inconsistent.[/QUOTE]

So read again:

[QUOTE=CRGreathouse;226263]I can never keep track of all the names (especially since some are contradictory... e.g. several definitions for generalized Fermats). What form are these?[/QUOTE]

He never accused [b]you[/b]! It was meant generally!

Alright: My sieve only requires three arguments, as it is just Generalized proths. (Two for the kmin/kmax, one for the base and exponent)

Now it should be far easier to make all the appropriate substitutions and so on.

@CRG/sm88: Any hints on the appropriate substitutions, as the amount of arguments are the same?

Here is the [B]original[/B] code:


[code]sieve(lim, sz, sm=1)={
my(v=vectorsmall(sz,i,1));
forprime(p=3,p2,
b=znorder(Mod(2,p));
trap(,next,
a=znlog(2293,Mod(2,p))
);
if(Mod(2,p)^a!=2293, print("bad at "p);next);
a+=b*ceil((sm-a)/b);
forstep(n=a,sz,b,
v[n]=0
)
);
a=0;
for(i=sm,sz,if(v[i],a++;write("abc.txt", i)));
a
};[/code]

Make as many substitutions and modifications as you'd like. I'll try to make the appropriate substitutions/modifications to see if I can reach a finished product.

I thought of substituting the variables with the variables I designated, but this essentially does the same, and prints out the same factors, except it does so up to 10[sup]6[/sup].

Now it seems that substitutions didn't cut it. Any recommended modifications, anyone?

one thing i find is you have a variable p2 declared as a maximum but no definition of it before hand that I know of.

[QUOTE=science_man_88]one thing i find is you have a variable p2 declared as a maximum but no definition of it before hand that I know of.
[/QUOTE]

Damn. I forgot to change that to "lim".

since you use znlog() you can just check if a is a factorial because g^n=x n gets stored in a so if a=c! then the exponent to check is c! not a! so checking if a is a factorial means if true a!=c!! or a double factorial if such a thing exist anyway I'll shut up.

you write i but unless you have a seperator you won't be able to figure out what's what in abc.txt I know I couldn't so adding "," to that call might be warranted if you want it to be understood.

[CODE]a=0;y=Vec(input());for(x=1,3,for(i=1,#y,for(p=2,#try[x]-1,if(y[i+(p-1)]==try[x][p],a=a+1;print("really")));if(a==#try[x]-2)))[/CODE]

is my newest attempt at what I wanted to help CRG with. unfortunately this gives me errors.

can setsearch be used on strings if it could we can use that as find and possibly replace if we can find a way to make sure the replacement can fit into it.

got it without errors but not working.

I have other ideas about how to but I don't know enough on the routines.
I was also thinking vecextract.

[QUOTE=science_man_88;226401]can setsearch be used on strings[/QUOTE]

setsearch works on sets and also on ordered vectors, but not on arbitrary vectors or strings.

I suggest you start by writing a "stringsearch(string_to_search, string_to_find)" function.

[QUOTE=CRGreathouse]setsearch works on sets and also on ordered vectors, but not on arbitrary vectors or strings.

I suggest you start by writing a "stringsearch(string_to_search, string_to_find)" function[/QUOTE]

Reduced the amount of arguments to three. I think it will be easier to make the appropriate substitutions and modifications now.

[QUOTE=CRGreathouse;226404]setsearch works on sets and also on ordered vectors, but not on arbitrary vectors or strings.

I suggest you start by writing a "stringsearch(string_to_search, string_to_find)" function.[/QUOTE]

yeah I found lex so [CODE]y=Vec(input());if(lex(y,"yes"),print("no"))[/CODE]

now all we need is to check indexes against a vector of strings I think.

oh one thing I forgot is maybe using vecextract to another vector may help comparison. as if the input is longer than the text to check against then it can never be equal so maybe extraction of the first #try[x] characters to a new vector and if we want to reuse the input we can concat() /// to the end or something then place that other vector concat() on the end then take the next until we hit out end symbol. then check the next string to check for.

[CODE]y=Vec(input());t=vecextract(y,"1..10");return(t)[/CODE]

my best attempt with vecextract i can't figure out the variables use I get a incorrect range error.

well i got it working for 1 but I get a vector of numbers back not a string. got it working for a character but I needed quotes around the whole thing which is better than quotes around every phrase I guess. now I think i figured it out lol. well i can search for it now to replace it I'll put the find code together soon.

[CODE][COLOR="Yellow"](17:49) gp > [COLOR="White"]y=Vec(input());for(i=1,#try,c=vecextract(y,[1,2,3,4,5,6]);if(lex(c,try[i]),print(c)))
"string""suck""primes"
["s", "t", "r", "i", "n", "g"]
["s", "t", "r", "i", "n", "g"][/COLOR][/COLOR][/CODE]
best I have so far.

Okay: Returned from a camping trip.

Has there been any progress on the project I was working on?

A quick proposition:

Primes of the form k * b![sup]n[/sup] + 1, where k < b![sup]n[/sup], and where n > 1. (If n were allowed to be 1, every odd prime of the form 6n + 1 would be a prime of the form k * b![sup]n[/sup] + 1.)

Or: Simply that k * b![sup]n[/sup] must be divisible by a square number greater than 1.

I'll see if all numbers that fit those conditions are present in the OEIS. (Smallest one is 5.) Oh, wait.. Those would be all the 4n + 1 numbers. Nevermind.

[QUOTE=3.14159;226641]
Or: Simply that k * b![sup]n[/sup] must be divisible by a square number greater than 1.[/QUOTE]

this fits:

4x+1
9x+1
16x+1 (variant of 4x+1)
25x+1


and hence only primes^2 need to be used.

[QUOTE=science_man_88]this fits:

4x+1
9x+1
16x+1 (variant of 4x+1)
25x+1


and hence only primes^2 need to be used.[/QUOTE]

[I][B]k * b![sup]n[/sup] + 1, where k < b![sup]n[/sup].[/B][/I]

Which means, every 4n + 1 number could not be expressed as k * b![sup]n[/sup] + 1, where k ≤ b![sup]n[/sup]

Ex: 7 * 2![sup]2[/sup] + 1 is not a valid number of this form, although it is prime (7 > 4) (7 * 2![sup]2[/sup] + 1 = 29 = p.)

Now, I need to hunt numbers of that form and fit them all into a sequence and see if it is already in the OEIS.

Note: The n > 1 restriction was found to be unnecessary.

The sequence begins: 3, 5, 7, 9, 13, 17, 19, 25, 33, 37, 41, 49, 65..(This is where only b! is allowed, as opposed to any integer b.)

3: 1 * 2!^1 + 1
5: 1 * 2!^2 + 1 (Fermat number)
7: 1 * 3!^1 + 1
9: 1 * 2!^3 + 1
13: 2 * 3!^1 + 1
17: 1 * 2!^4 + 1 (Fermat number)
19: 3 * 3!^1 + 1
25: 4!^1 + 1
33: 2 * 2!^4 + 1
37: 3!^2 + 1
41: 5 * 2!^3 + 1
49: 2 * 4!^1 + 1
65: 4 * 2!^4 + 1

I never said [TEX]k*b!^n+1[/TEX] is of form 4x+1 you stated [TEX]k*b!^n/x^2[/TEX] for x>1= integer for some x.

this says that 4x+1 can be a [TEX]k*b!^n+1[/TEX]

and same for the rest so really I was just taking your last statement that the statements above and that statement were equal.

[url]http://www.research.att.com/~njas/sequences/?q=3,5,7,9,11,13,17,19&sort=0&fmt=0&language=english&go=Search[/url]

was the best i could find with all of some group in one sequence.

note also 65 = 2^6+1=2!^6+1 = 2!^3!+1

one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?

[QUOTE=science_man_88]one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?
[/QUOTE]

3, 5: Diff = 2
5, 7: Diff = 2
7, 9: Diff = 2
9, 13: Diff = 2^2
13, 17: Diff = 2^2
17, 19: Diff = 2^2.
[B]19, 25: Diff = 6 =/= 2^n.[/B]

Strong law of small numbers at work.

[QUOTE=3.14159;226658]3, 5: Diff = 2
5, 7: Diff = 2
7, 9: Diff = 2
9, 13: Diff = 2^2
13, 17: Diff = 2^2
17, 19: Diff = 2^2.
[B]19, 25: Diff = 6 =/= 2^n.[/B]

Strong law of small numbers at work.[/QUOTE]

okay sorry I forgot I took out 19 lol but the rest all have first difference of power of 2.

The sieve project is coming along.. Just that we have made zero progress.

Anyone have any spare ideas on it?

My best idea is to begin with the k-range loop, then a forprime loop which will be linked to a forstep loop, which, hopefully, will eliminate all the unwanted values. Then, it will save the remaining numbers in a file, and then I make the appropriate modifications to make the range testable, and then proceed to test.

[CODE]for(k=,,forprime(_=,,forstep(-=,,_,write(,))))[/CODE]

something on the lines of this ?

Or: I trial-divide each candidate to 1048576, and test those which have no small factors.

[QUOTE=science_man_88]something on the lines of this ?
[/QUOTE]

Yes. Although I doubt it will work as I want it to.

The forstep loop will have to contain the liftmod for the primes, to eliminate bad values of k (Those which are divisible by a certain prime p.)

I can print the bad values that are divisible by a certain prime p, so far. But I can't get a good enough text editor to remove those specific values.

Ex: 757.

[code]623
1380
2137
2894
3651
4408
5165
5922
6679
7436
8193
8950
9707
10464
11221
11978
12735
13492
14249
15006
15763
16520
17277
18034
18791
19548
20305
21062
21819
22576
23333
24090
24847
25604
26361
27118
27875
28632
29389
30146
30903
31660
32417
33174
33931
34688
35445
36202
36959
37716
38473
39230
39987
40744
41501
42258
43015
43772
44529
45286
46043
46800
47557
48314
49071
49828
50585
51342
52099
52856
53613
54370
55127
55884
56641
57398
58155
58912
59669
60426
61183
61940
62697
63454
64211
64968
65725
66482
67239
67996
68753
69510
70267
71024
71781
72538
73295
74052
74809
75566
76323
77080
77837
78594
79351
80108
80865
81622
82379
83136
83893
84650
85407
86164
86921
87678
88435
89192
89949
90706
91463
92220
92977
93734
94491
95248
96005
96762
97519
98276
99033
99790
100547
101304
102061
102818
103575
104332
105089
105846
106603
107360
108117
108874
109631
110388
111145
111902
112659
113416
114173
114930
115687
116444
117201
117958
118715
119472
120229
120986
121743
122500
123257
124014
124771
125528
126285
127042
127799
128556
129313
130070
130827
131584
132341
133098
133855
134612
135369
136126
136883
137640
138397
139154
139911
140668
141425
142182
142939
143696
144453
145210
145967
146724
147481
148238
148995
149752
150509
151266
152023
152780
153537
154294
155051
155808
156565
157322
158079
158836
159593
160350
161107
161864
162621
163378
164135
164892
165649
166406
167163
167920
168677
169434
170191
170948
171705
172462
173219
173976
174733
175490
176247
177004
177761
178518
179275
180032
180789
181546
182303
183060
183817
184574
185331
186088
186845
187602
188359
189116
189873
190630
191387
192144
192901
193658
194415
195172
195929
196686
197443
198200
198957
199714[/code]

Those would be the bad k-values for k * 11![sup]4[/sup] + 1, because they form candidates divisible by 757.

Proof: 15006 * 11![sup]4[/sup] + 1 = 757 * 50325944549556534016411764332893

if your printing inside a if to get these switch it to if(,,) form and switch which of the last 2 it's in.

[QUOTE=science_man_88]if your printing inside a if to get these switch it to if(,,) form and switch which of the last 2 it's in.
[/QUOTE]

Can you give some smaple code to elaborate what you posted? It will clear things up.

[CODE]if(x%6==1 || x%6==5 ,print(x))[/CODE]

is like the usual way for most I think.

[CODE]if(x%6==1 || x%6==5 ,print(x),)[/CODE]

is another way to put this in the longer notation

if i want the opposite i just change the position to the other and i get.

[CODE]if(x%6==1 || x%6==5 ,,print(x))[/CODE]

now it will print every number not in 6x+1 or 6x+5 so just that change in position is saying the same as:

[CODE]if(x%6!=1 && x%6!=5 ,print(x))[/CODE]

but it makes it easier as you don't have to add anything except a comma to an existing code rather than changing 4 characters in this case.

[CODE](12:10) gp > for(x=1,20,if(x%6==1 || x%6==5,print(x)))
1
5
7
11
13
17
19
(12:10) gp > for(x=1,20,if(x%6==1 || x%6==5,,print(x)))
2
3
4
6
8
9
10
12
14
15
16
18
20
(12:10) gp >[/CODE]

see the difference ?

Yes, but you shouldn't be printing at all. Instead of writing print(x) you should be writing something like v[x] = 0.

I was just giving example code for him to look over.

[QUOTE=CRGreathouse]Yes, but you shouldn't be printing at all. Instead of writing print(x) you should be writing something like v[x] = 0.
[/QUOTE]

It would only give the good old "Not a vector" error.

not if you make a vector and try to edit it on the fly like I sent in a pm to CRG the hard part for me was not filling it with 0's as he wanted me to.

can't remeber what i did to make it act dynamic but I can try to figure it out.

[CODE]v=vector(1,n,0);for(x=1,10,if(isprime(x),v=concat(v,x)));for(i=1,(#v)-1,v[i]=v[i+1])[/CODE]

there's one thing I still can't do and that's take off the end that repeats so no number repeats.

doh vecextract can work lol I just extract the last one.

didn't work.

made a code that works by replacing the 0 with 2 and starting at 3 and searching then I didn't need the index shift code.

[QUOTE=3.14159;226677]It would only give the good old "Not a vector" error.[/QUOTE]

Doesn't that message tell you what to do?

It says, "v isn't a vector, so I can't treat it like a vector". So... drumroll... you need to make a vector v. This is done with the vector command, unsurprisingly:
[code]v=vector(1000)[/code]

But I've already written this a goodly number of times on this thread. Actually, I think I'll make this one my last; in the future I'll just refer to this post number.

[QUOTE=science_man_88;226679][CODE]v=vector(1,n,0);for(x=1,10,if(isprime(x),v=concat(v,x)));for(i=1,(#v)-1,v[i]=v[i+1])[/CODE]

there's one thing I still can't do and that's take off the end that repeats so no number repeats.[/QUOTE]

Yes, vecextract could be used here. I'd just use vector, to be honest:
[code]v=vector(#v-1,i,v[i])[/code]
replaces v with v without the last element.

[QUOTE=science_man_88;226681]made a code that works by replacing the 0 with 2 and starting at 3 and searching then I didn't need the index shift code.[/QUOTE]

That's a better solution.

yeah the bad part for pi is unless there's a known lower bound for his idea then he can't use my second idea. though writing it both ways taught me a lot I think.

[QUOTE=science_man_88;226697]yeah the bad part for pi is unless there's a known lower bound for his idea then he can't use my second idea. though writing it both ways taught me a lot I think.[/QUOTE]

Can you explain this in more detail? What does he need a lower bound on?

well the reason I can set it to a number other than 0 is because i know the first answer already the reason i can change the lower in the loop to 3 is because i know it won't mess up my loop and ironically it's the next value. so unless you know the first time it comes up and possibly the second as the lower bound for a loop it's hard to use if not impossible.

[QUOTE=science_man_88;226700]well the reason I can set it to a number other than 0 is because i know the first answer already the reason i can change the lower in the loop to 3 is because i know it won't mess up my loop and ironically it's the next value. so unless you know the first time it comes up and possibly the second as the lower bound for a loop it's hard to use if not impossible.[/QUOTE]

Whoa, man, slow down. Define the problem for me, then what you're doing. Right now I don't know what "it", "the first answer", "the lower", or the "next value" are.

[QUOTE=CRGreathouse;226702]Whoa, man, slow down. Define the problem for me, then what you're doing. Right now I don't know what "it", "the first answer", "the lower", or the "next value" are.[/QUOTE]

it-the first number put in the vector to initialize it.
first answer- the first solution to what the problem is.
the lower- the smaller of the 2 values to denote a range given to the variable in the loop.
next value- the value after the current value.

my code is making a vector that contains only primes- no 0 values.