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 2012-05-02, 21:51 #1 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 2·7·461 Posts Six-brilliant numbers That is, numbers with six prime factors all of the same number of digits. Ones marked with dots are by-definition otherwise by search; there are obviously no 7-digit without dupes. Code: .1771561 11,11,11,11,11,11 smallest 7-digit with dupes 9887999 11,11,11,17,19,23 largest 7-digit with dupes 10081357 11,11,13,13,17,29 smallest 8-digit with dupes .30808063 11,13,17,19,23,29 smallest 8-digit without dupes 99423181 11,13,19,23,37,43 largest 8-digit without dupes 99990407 11,11,19,23,31,61 largest 8-digit with dupes 100016059 11,11,13,13,67,73 smallest 9-digit with dupes 100145903 11,13,19,29,31,41 smallest 9-digit without dupes 999978551 11,13,29,51,61,67 largest 9-digit with or without dupes 1000052443 13,23,29,29,41,97 smallest 10-digit with dupes 1000202489 11,13,29,43,71,79 smallest 10-digit without dupes 9999884737 19,31,43,67,71,83 largest 10-digit with or without dupes 10000108889 23,29,59,59,59,73 smallest 11-digit with dupes 10000268641 11,31,43,79,89,97 smallest 11-digit without dupes 99919838357 53,61,67,71,73,89 largest 11-digit without dupes 99998113069 47,61,61,83,83,83 largest 11-digit with dupes 100000197487 41,53,67,73,97,97 smallest 12-digit with dupes 100042414319 53,59,61,71,83,89 smallest 12-digit without dupes .293391909323 71,73,79,83,89,97 largest 12-digit without dupes .832972004929 97,97,97,97,97,97 largest 12-digit with dupes For 13..18 I think this is a relatively easy sort-and-search, for 19..24 it's quite a large one, for larger numbers it starts to become interesting Last fiddled with by fivemack on 2012-05-02 at 21:56
 2012-05-02, 22:51 #2 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 2·7·461 Posts Code: 1061520150601 101 101 101 101 101 101 with (!) 1741209542339 101 103 107 109 113 127 without (!) 9999901067561 101 107 109 139 157 389 without 9999965520509 101 107 113 197 197 211 with 10000001342381 101 101 149 163 181 223 with 10000087934327 101 109 131 149 173 269 without 99999998106163 103 149 179 223 239 683 without 99999999253139 113 181 199 199 331 373 with 100000001274419 101 101 151 211 313 983 with 100000002228881 107 109 113 307 439 563 without 999999998566003 131 167 191 379 653 967 with/without 1000000000407943 151 311 317 379 421 421 with 1000000002307693 109 163 191 593 653 761 without 9999999998178521 131 353 467 647 751 953 with/without 10000000002328283 151 397 467 599 691 863 with/without 99999999886001279 359 631 673 821 877 911 with/without 100000000103742353 563 577 673 751 751 811 with 100000000374707123 401 449 761 773 947 997 without 890969009638765049 967 971 977 983 991 997 without(!) 982134461213542729 997 997 997 997 997 997 with (!) Code: #include #include #include #include #include using std::vector; using std::sort; using std::cout; using std::endl; bool isprime(int T) { if (T==2 || T==3) return true; if (T%2==0) return false; if (T%3==0) return false; int s=5,ds=2; while (s*s<=T) { if (T%s==0) return false; s+=ds; ds=6-ds; } return true; } int main(void) { vector logthreep; vector primes; for (int i=100; i<1000; i++) if(isprime(i)) primes.push_back(i); int pn = primes.size(); for (int pi1=0; pi1::iterator ilo,ihi; ihi = lower_bound(logthreep.begin(), logthreep.end(), t2); if (ihi != logthreep.begin()) { ilo = ihi-1; if (*ilo!=0) { if (logthreep[q]+*ilo > XXL) { XXL1=logthreep[q]; XXL2=*ilo; XXL=XXL1+XXL2; } if (logthreep[q]+*ihi < XXH) { XXH1=logthreep[q]; XXH2=*ihi; XXH=XXH1+XXH2; } long long U0 = (int)(0.1+exp(logthreep[q])); long long U1 = (int)(0.1+exp(*ilo)); long long U2 = (int)(0.1+exp(*ihi)); cout << q << " " << U0 << " " << U1 << " " << U2 << " " << U0*U1 << " " << U0*U2 << endl; cout << q << " " << (int)(0.1+exp(*ilo)) << " " << (int)(0.1+exp(*ihi)) << endl; } } } printf("%30.28f %30.28f\n%30.28f %30.28f\n",exp(XXL1),exp(XXL2),exp(XXH1),exp(XXH2)); } Behaviour isn't quite right for the larger values of target, so I had to wrap it in something like Code: ./meeble |awk 'NF==6 && $4!=1' | sort -gk6|awk '{print$6}' | uniq -c | head Last fiddled with by fivemack on 2012-05-02 at 22:52
 2012-05-02, 23:05 #3 science_man_88     "Forget I exist" Jul 2009 Dumbassville 24·3·52·7 Posts I'm guessing: Code: isbrilliant(x,y)=for(z=2^x,10^y,if(sum(a=1,if(#factor(z)[,2]<6,#factor(z)[,2],break()),factor(z)[a,2])==6,print(z))) is more of what you wanted ?
2012-05-02, 23:39   #4
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

11001001101102 Posts

Quote:
 Originally Posted by science_man_88 I'm guessing: Code: isbrilliant(x,y)=for(z=2^x,10^y,if(sum(a=1,if(#factor(z)[,2]<6,#factor(z)[,2],break()),factor(z)[a,2])==6,print(z))) is more of what you wanted ?
Yes; though I found that computing a GCD with the product of the usable primes (raised to the sixth power if you want to allow duplicates) was rather quicker than doing a full factorisation every time; this is obviously a sievable problem but I haven't implemented the sieve yet.

The C++ sort-and-search code I posted gives the wrong answer for the 22-digit case, because 53-bit doubles aren't wide enough, though obviously the answer from the C++ code bounds the size of the area you'd have to sieve.

Last fiddled with by fivemack on 2012-05-02 at 23:40

2012-05-03, 00:08   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

24·3·52·7 Posts

Quote:
 Originally Posted by fivemack Yes; though I found that computing a GCD with the product of the usable primes (raised to the sixth power if you want to allow duplicates) was rather quicker than doing a full factorisation every time; this is obviously a sievable problem but I haven't implemented the sieve yet. The C++ sort-and-search code I posted gives the wrong answer for the 22-digit case, because 53-bit doubles aren't wide enough, though obviously the answer from the C++ code bounds the size of the area you'd have to sieve.
I'm a little lost as to why that would be quicker unless maybe useable primes is not 2-sqrt(number to check) because GCD(sqrt(number to check)#,number to check) == number to check.

2012-05-03, 00:35   #6
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts

Quote:
 Originally Posted by science_man_88 I'm a little lost as to why that would be quicker unless maybe useable primes is not 2-sqrt(number to check) because GCD(sqrt(number to check)#,number to check) == number to check.
What he's saying is that if you're looking for (e.g.) 3-digit six-brilliant numbers, then you don't need to check a number for 2 digit, 4 digit, 5 digit, etc... factors; you only need to check for 3 digit factors which can be done (in this example) with $GCD\left(\prod_{prime\quad p>100}^{p<1000}p\quad,\quad n\right)$ and seeing what you get. (To check for duplicates you'd need to put a power of six/whatever on each prime.)

(Excuse the poor tex, it may or may not be better than "GCD(1000#/100#, n)". )

Last fiddled with by Dubslow on 2012-05-03 at 00:43 Reason: tex tex tex tex tex

2012-05-03, 01:15   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

24·3·52·7 Posts

Quote:
 Originally Posted by Dubslow What he's saying is that if you're looking for (e.g.) 3-digit six-brilliant numbers, then you don't need to check a number for 2 digit, 4 digit, 5 digit, etc... factors; you only need to check for 3 digit factors which can be done (in this example) with $GCD\left(\prod_{prime\quad p>100}^{p<1000}p\quad,\quad n\right)$ and seeing what you get. (To check for duplicates you'd need to put a power of six/whatever on each prime.) (Excuse the poor tex, it may or may not be better than "GCD(1000#/100#, n)". )
if you look for only 3 digit factors of a 3 digit number then you don't catch 3^6=729 you also miss any under 300 since 300/3=100 the lowest 3 digit divisor you can start with ( technically it starts at 303 that you could check).

okay never mind it's >202 that it tells you anything because it's the first multiple of a prime with 3 or more digits.

Last fiddled with by science_man_88 on 2012-05-03 at 01:27

2012-05-03, 01:22   #8
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

1C3516 Posts

Quote:
 Originally Posted by science_man_88 if you look for only 3 digit factors of a 3 digit number then you don't catch 3^6=729 you also miss any under 300 since 300/3=100 the lowest 3 digit divisor you can start with ( technically it starts at 303 that you could check).
???????
"numbers with six prime factors all of the same number of digits"

729 isn't prime; 3 is certainly not a 3 digit prime. And last time I checked, no 3 digit number will have 6 3-digit prime factors. Also last time I checked, if you do GCD(101, a number with 101 in its factorization), you'll get at least 101. I fail to see what 300 has to do with that.

2012-05-03, 01:33   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20D016 Posts

Quote:
 Originally Posted by Dubslow ??????? "numbers with six prime factors all of the same number of digits" 729 isn't prime; 3 is certainly not a 3 digit prime. And last time I checked, no 3 digit number will have 6 3-digit prime factors. Also last time I checked, if you do GCD(101, a number with 101 in its factorization), you'll get at least 101. I fail to see what 300 has to do with that.
3 has the same number of digits as 3 , 3 to the power of 6 is 729 is a number, therefore 729 is 6 brilliant

Code:
(22:31)>isbrilliant(6,3,4)
144
160
216
224
240
324
336
352
360
400
416
486
504
528
540
544
560
600
608
624
729
736
756
784
792
810
816
840
880
900
912
928
936
992
1000
and yes I messed up with the 300, but I did do a code that does:
Code:
for(n=100,999,print(gcd(prod(X=26,68,prime(X))^6,n)","n))
and the first time it doesn't give 1 or the number is 202. okay I'll admit it I got lost I messed 168 to 68 but I retested and 202 still stands at the minimum. the minimum 3 digit 6-brilliant is 3^2*2^4 = 144

Last fiddled with by science_man_88 on 2012-05-03 at 01:55

2012-05-03, 02:25   #10
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×29×83 Posts

Quote:
 Originally Posted by science_man_88
Whoops, I made a typo. Where I said "3-digit 6-brilliant numbers" I was thinking in my head "6-brilliant numbers whose factors have 3 digits". That's what I was writing about.

2012-05-03, 06:56   #11
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 Originally Posted by Dubslow Whoops, I made a typo. Where I said "3-digit 6-brilliant numbers" I was thinking in my head "6-brilliant numbers whose factors have 3 digits". That's what I was writing about.
Right, that's kind of the natural way of thinking about them.

The counts of such numbers are not in the OEIS, but the counts for the analogous semiprimes are A087434. The formula is straightforward and gives

Code:
84, 230230, 13172431272, 2009505083440218, 476013062485109252824, 148697710093074782891561317, 56289121765869792295257792982777, 24349676137284030200879331533337194376, 11666012947107695154385353794415395985137344, 6057277355183085204144847469602671151899349499197, 3354985557158369214407984988027190131369315871722691152, 1959509071991741722270413421131623309713272464704021335869310, 1196301271416935612331559019142578123179924493160149725594041285390, 758297477798207294085616092160894284729546088426880198004979229145638184, 496405409134346365011354234191682214886538111380694256570445787621168130062528, 334180505840507943479437772268647724355589483819448761562142733090916595143509225212, 230551283386125977830557515195210002315457614489571967352960948012129577632233414552542796, 162537577146540321845227986334505185151917953317422163374965016524615696383252603524891046763532, 116816284673708680421835123424581599081284210871219523674771358297205544299884690188130918699786791500, 85416640764723824021108446456250433057219850414437417656346074325343176420847706796002701169332531133021551, 63434778901223585384265131927059356802305353342435316304189778022372093879998625701188622017546319272210868240784, 47777139494090360016305651100722437896951187322404498059597560979956679547504643361209098188623110927359491777939720841, 36447608389806648140714977201273896196346361099368631002618281535486043280009949162450062261360663543889016612413795952580291, 28131574280009154082710448682314774375712261908758572574980667541365874655616083694412009836355371649813984129660663498677543353152, ...
six-brilliant numbers composed of 1, 2, ... digit prime factors.

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