Determine all possible pairs (m,n) of positive integers such that C(m,n) equals the product of all prime numbers ≤ m.
Note:
C is the choose function, that is:
C(m,n) = m!/(n! * (m-n)!)
10 prod=1
20 for m=2 to 1000
30 if prmdiv(m)=m then prod=prod*m
40 for n=1 to int(m/2)
50 c=combi(m,n)
60 if c=prod then ?m,n,prod
70 if c>prod then cancel for: goto 90
80 next
90 next m
finds
m n C(m,n)
2 1 2
4 2 6
10 4 210
of course this last implies
10 6 210
The product of all the primes below the last m checked, 1000, is 1959034064499908343126250819820638104612397239058936822388260532896866631637987
06618519516487894823215962295591154360191491895297252152667282922829908526490233
62731392404017939142010958261393634959471483757196721672243410067118516227661133
13519248884898991489215718830867989687513743951933890396809490554975038640710603
38365866606835392010116359179000399044950652032997495429859931346698148053184740
80581207891125910, while the maximum C(1000,n) is C(1000,500) = 5400369844039560999703199766858308398635691825448998708471063233168304473737080
16152716759298257230609836264852508119063686650618364731731471944339160115756285
60708532538020975951821665432659726888011723608905609216689826495417997683655807
4278734888461642260398682633662848128817178049440446349376320. Thus is looks like the product of the primes is increasing faster than the possible combination numbers.
But does that mean the combinations will never make a comeback and possibly equal the product of primes? I don't know and therefore have no proof that the list is exhaustive:
2 = C(2,1)
6 = 2*3 = C(4,2)
210 = 2*3*5*7 = C(10,4) = C(10,6)
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Posted by Charlie
on 2014-07-23 15:32:03 |
computer exploration (possible spoiler)
