PARI's commands - Page 192

science_man_88 · 2010-12-22, 14:49

Code:

a=0;for(i=1,#mersenne,for(x=a+1,100,if(prime(x)==mersenne[i],a=x;break(),print1(x",");a=x;break())))

Basically checking for http://oeis.org/A135980 , but it's not working as intended. I'm using a so x doesn't repeat a value, this seems to work however it reports 6 as the second term not 9 so I've made a major error. doh I should have reversed the order of the loops I think. I see why it prints six because it checks 11==13 and that's false so it prints x then puts 5 in a so it then increments without finding the next match and hence it returns 6 on finding 13!=17.

science_man_88 · 2010-12-22, 14:58

Quote:

Originally Posted by science_man_88

Code:

a=0;for(i=1,#mersenne,for(x=a+1,100,if(prime(x)==mersenne[i],a=x;break(),print1(x",");a=x;break())))

Basically checking for http://oeis.org/A135980 , but it's not working as intended. I'm using a so x doesn't repeat a value, this seems to work however it reports 6 as the second term not 9 so I've made a major error. doh I should have reversed the order of the loops I think. I see why it prints six because it checks 11==13 and that's false so it prints x then puts 5 in a so it then increments without finding the next match and hence it returns 6 on finding 13!=17.

note putting a i=i-1 in the if false part worked.

science_man_88 · 2011-01-03, 16:29

Code:

aliquot(x,y)=b=0;v=vector(y);v[1]=x;for(i=2,#v,for(a=1,i-1,if(sigma(v[i-1])-v[i-1]==0 || sigma(v[i-1])-v[i-1]==v[a],break(2),b=b+1;if(b!=a,b=0;break(2),v[i]=sigma(v[i-1])-v[i-1];b=0;break()))))

this seem to work for aliquot checking repeat lol.Now I just have to eliminate the y variable and cleave it to the final length if not less than infinite I think. too bad i can't get it to start at any index already calculated.

science_man_88 · 2011-01-03, 20:15

Quote:

Originally Posted by science_man_88

Code:

aliquot(x,y)=b=0;v=vector(y);v[1]=x;for(i=2,#v,for(a=1,i-1,if(sigma(v[i-1])-v[i-1]==0 || sigma(v[i-1])-v[i-1]==v[a],break(2),b=b+1;if(b!=a,b=0;break(2),v[i]=sigma(v[i-1])-v[i-1];b=0;break()))))

this seem to work for aliquot checking repeat lol.Now I just have to eliminate the y variable and cleave it to the final length if not less than infinite I think. too bad i can't get it to start at any index already calculated.

crap a failure.

Code:

for(x=1,100, print(aliquot(x,5)))

gave 95,25,6,6,6, but it should have terminated after finding the second 6.

science_man_88 · 2011-01-03, 20:50

Code:

aliquot(x,y)=b=0;v=vector(y);v[1]=x;for(i=2,#v,for(a=1,i-1,if(sigma(v[i-1])-v[i-1]==0 || sigma(v[i-1])-v[i-1]==v[a],break(2),b=b+1;if(b==i-1,v[i]=sigma(v[i-1])-v[i-1];b=0;break()))))

this works better for now.

science_man_88 · 2011-01-03, 20:56

Code:

aliquot(x,y)=b=0;v=vector(y);v[1]=x;for(i=2,#v,for(a=1,i-1,if(sigma(v[i-1])-v[i-1]==0 || sigma(v[i-1])-v[i-1]==v[a],v=vector(i-1,n,v[n]);break(2),b=b+1;if(b==i-1,v[i]=sigma(v[i-1])-v[i-1];b=0;break()))))

this not only works but trims off excess.

science_man_88 · 2011-01-04, 15:14

Quote:

Originally Posted by CRGreathouse

kar_bon, I understand what an aliquot sequence is, but that doesn't tell me what sm88 wants.

I take it from his last post that he wants a vector with last term 1 and first term the input, where v[i+1] = sigma(v[i]) - v[i]. But of course it's not even proven that such a vector exists for all n...!

I see one possible exception but I don't know how to calculate when it would happen, a sequence with all abundant numbers as it would forever increase.

science_man_88 · 2011-01-04, 19:34

Quote:

Originally Posted by science_man_88

I see one possible exception but I don't know how to calculate when it would happen, a sequence with all abundant numbers as it would forever increase.

good news I've been testing x upto 3 million until the 4th spot in the sequence each and no x so far has made it past 3 terms that are abundant. the problem is if it decreases to another abundant number spiking it back up. Doh calculation error. redone and still accurate.

science_man_88 · 2011-01-04, 20:35

done up to 7 million now without a trace of a triply ? abundant number.

science_man_88 · 2011-01-29, 19:08

I couldn't get the addhelp to add help I've had it working before but I just tried the autosave that lavalamp got me doing and I can't get what I want to work.

science_man_88 · 2011-01-29, 19:48

Code:

(15:45)>for(x=1,400,if(((2^x)/(log(2^x)-1))-((2^x-1)/(log(2^x-1)-1))>=1,print(x)))
108
116

this is my attempt at something new if only I could get it to align with A000043.

2010-12-22, 14:49	#2102
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	Code: a=0;for(i=1,#mersenne,for(x=a+1,100,if(prime(x)==mersenne[i],a=x;break(),print1(x",");a=x;break()))) Basically checking for http://oeis.org/A135980 , but it's not working as intended. I'm using a so x doesn't repeat a value, this seems to work however it reports 6 as the second term not 9 so I've made a major error. doh I should have reversed the order of the loops I think. I see why it prints six because it checks 11==13 and that's false so it prints x then puts 5 in a so it then increments without finding the next match and hence it returns 6 on finding 13!=17. Last fiddled with by science_man_88 on 2010-12-22 at 14:57

2011-01-03, 16:29	#2104
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	Code: aliquot(x,y)=b=0;v=vector(y);v[1]=x;for(i=2,#v,for(a=1,i-1,if(sigma(v[i-1])-v[i-1]==0 \|\| sigma(v[i-1])-v[i-1]==v[a],break(2),b=b+1;if(b!=a,b=0;break(2),v[i]=sigma(v[i-1])-v[i-1];b=0;break())))) this seem to work for aliquot checking repeat lol.Now I just have to eliminate the y variable and cleave it to the final length if not less than infinite I think. too bad i can't get it to start at any index already calculated.

2011-01-03, 20:50	#2106
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	Code: aliquot(x,y)=b=0;v=vector(y);v[1]=x;for(i=2,#v,for(a=1,i-1,if(sigma(v[i-1])-v[i-1]==0 \|\| sigma(v[i-1])-v[i-1]==v[a],break(2),b=b+1;if(b==i-1,v[i]=sigma(v[i-1])-v[i-1];b=0;break())))) this works better for now.

2011-01-29, 19:48	#2112
science_man_88 "Forget I exist" Jul 2009 Dumbassville 20300₈ Posts	Code: (15:45)>for(x=1,400,if(((2^x)/(log(2^x)-1))-((2^x-1)/(log(2^x-1)-1))>=1,print(x))) 108 116 this is my attempt at something new if only I could get it to align with A000043.

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2011-01-04, 20:35	#2110
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	done up to 7 million now without a trace of a triply ? abundant number.

2011-01-29, 19:08	#2111
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	I couldn't get the addhelp to add help I've had it working before but I just tried the autosave that lavalamp got me doing and I can't get what I want to work.