[QUOTE=science_man_88;442448]okay well based on this going back to the huge number in post 2 has a smallest factor near: ...[/QUOTE]

[CODE] %22 = 40445364023527381951378636564391212010397122822120720357
<999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999)/17.5
%23 = 4.2326886774241279797411942169097598446 E308
<289912154831438167899885040445364023527381951378636564391212010397122822120720357/9 E308)/17.5
%24 = 1.4110247948887228971825699932010487318 E154 [/CODE]%24 = 1.4110247948887228971825699932010487318 E154[U][B]

EDIT

<999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999))/17.5
%1 = 4.2330743846661686915477099796031461955 E231 near it
(00:30) gp >
[/B][/U]

Do you know the smallest factor ? Is It near ?

[QUOTE=Godzilla;442449][CODE] %22 = 40445364023527381951378636564391212010397122822120720357
<999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999)/17.5
%23 = 4.2326886774241279797411942169097598446 E308
<289912154831438167899885040445364023527381951378636564391212010397122822120720357/9 E308)/17.5
%24 = 1.4110247948887228971825699932010487318 E154 [/CODE]%24 = 1.4110247948887228971825699932010487318 E154[U][B]

EDIT

<999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999))/17.5
%1 = 4.2330743846661686915477099796031461955 E231 near it
(00:30) gp >
[/B][/U]

Do you know the smallest factor ? Is It near ?[/QUOTE]

my result from your method gave the same digits but with the equivalent of about 80 digits fewer. and no I don't know the factors I'm one of the typically stupid ones around here.

[QUOTE=science_man_88;442451]my result from your method gave the same digits but with the equivalent of about 80 digits fewer. and no I don't know the factors I'm one of the typically stupid ones around here.[/QUOTE]

But i must test It with \frac22

[QUOTE=Godzilla;442452]But i must test It with \frac22[/QUOTE]

no I used 17.5 maybe I didn't copy it correctly I'll try it again. edit: tried it again got the same result length with what could fool me for the same digits.

[QUOTE=science_man_88;442455]no I used 17.5 maybe I didn't copy it correctly I'll try it again. edit: tried it again got the same result length with what could fool me for the same digits.[/QUOTE]

Perhaps It work only with the numbers One very big and One very small ,i think . Because i have see a thing , the First formula in the previews post (17,5) start at 13 *3 the second step is 19*5 and so on 101*23 ; 1001*205 ; 10001*2055 ;100001*20375 and step by step Always plus. So the numbers have a proportion. But i don t understand the cicle

I have found out one thing. So to find the prime factor smaller than a product .

is very simple and similar to the others . Three steps :

1) [TEX] 1997 * 16257 = 32465229[/TEX]

[TEX] \frac{\sqrt{2178*(32465229-9999999)}}{22} = 10054,…[/TEX] I take only four digits (1005) and not the last (never the last)


2)

[TEX] \frac{\sqrt{2178*(324652290-99999999)}}{22} = 31795,…[/TEX] I take only four digits (3179) and not the last (never the last) , and i added a zero and a 9


3)

subtract step 2 with step 1

[TEX]3179-1005 = 2174 [/TEX] result is near [TEX]1997[/TEX]

[QUOTE=Godzilla;442951][TEX]3179-1005 = 2174 [/TEX] result is near [TEX]1997[/TEX][/QUOTE]But still wrong. There set of numbers that your formula produces wrong answers for is infinite.

I have to mention them with French mathematicians, it said that, be able to improve the formula so :





[TEX] 2\sqrt{product}[/TEX] if the numbers are similar


Thread : [URL]http://www.les-mathematiques.net/phorum/read.php?5,1334720[/URL]

[QUOTE=Godzilla;445200]I have to mention them with French mathematicians, it said that, be able to improve the formula so :





[TEX] 2\sqrt{product}[/TEX] if the numbers are similar


Thread : [URL]http://www.les-mathematiques.net/phorum/read.php?5,1334720[/URL][/QUOTE]

well yeah for example if they are the same we have p*p =p^2 for the product and 2*sqrt(p^2) =p+p for the sum. so in a prime square situation this is completely accurate. and if they are both one away from a number q we get that (q-1)(q+1)=q^2+q-q-1 = q^2-1 as the product and the sum is 2q we get just under that for using the sqrt method. in fact this can be extended to any prime power p^n by [TEX]n\sqrt[n]{product} = n(\text{geometric mean of the values}).[/TEX]

[QUOTE=science_man_88;445212]well yeah for example if they are the same we have p*p =p^2 for the product and 2*sqrt(p^2) =p+p for the sum. so in a prime square situation this is completely accurate. and if they are both one away from a number q we get that (q-1)(q+1)=q^2+q-q-1 = q^2-1 as the product and the sum is 2q we get just under that for using the sqrt method. in fact this can be extended to any prime power p^n by [TEX]n\sqrt[n]{product} = n(\text{geometric mean of the values}).[/TEX][/QUOTE]

Test it please :

[TEX] p1 * p2 = n [/TEX]

[TEX]\frac{n}{22*(n2+1)} = n3 (if n3 = 22,.. stop)[/TEX]

[TEX]\frac{n2}{2} = n4 [/TEX]near p1 or p2

[B]example :[/B]

[TEX]997 * 991 = 988027[/TEX]

[TEX]\frac{988027}{22*2000} = 22,45..[/TEX]

[TEX]\frac{2000}{2} = 1000[/TEX] is near 997 or 991

[B]other example :[/B]

[TEX] 997 * 13 = 12961[/TEX]

[TEX]\frac{12961}{22*26}= 22,65..[/TEX]

[TEX]\frac{26}{2} = 13[/TEX] is equal p2

work , is a possibility ?

[QUOTE=Godzilla;446436]Test it please :

[TEX] p1 * p2 = n [/TEX]

[TEX]\frac{n}{22*(n2+1)} = n3 (if n3 = 22,.. stop)[/TEX]

[TEX]\frac{n2}{2} = n4 [/TEX]near p1 or p2

[B]example :[/B]

[TEX]997 * 991 = 988027[/TEX]

[TEX]\frac{988027}{22*2000} = 22,45..[/TEX]

[TEX]\frac{2000}{2} = 1000[/TEX] is near 997 or 991

[B]other example :[/B]

[TEX] 997 * 13 = 12961[/TEX]

[TEX]\frac{12961}{22*26}= 22,65..[/TEX]

[TEX]\frac{26}{2} = 13[/TEX] is equal p2

work , is a possibility ?[/QUOTE]

I don't even know how to interpret this you don't show how to get n2 etc.