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EdH

"Ed Hall"
Dec 2009

64248 Posts Quote:
 Originally Posted by bsquared Code: >> x=2^7* 5 * 23 * 127 * 659 * 53323 * 1876187 * 97544836889 * 665320793909 7998766649128898059663516612687535453720960 >> totient(x) 3031634148236289733373855928919180891127808 >> sigma(x,1)-x 12142697391577851168337274092012830083559040 >>
What program gave you totient() and sigma() functions? I expected YAFU from the ">>" prompts, but my YAFU doesn't like them.   2020-06-25, 19:21 #13 bsquared   "Ben" Feb 2007 1100110110102 Posts It is yafu starting from a r381 (wip-branch).   2020-06-25, 19:31   #14
EdH

"Ed Hall"
Dec 2009

22·33·31 Posts Quote:
 Originally Posted by bsquared It is yafu starting from a r381 (wip-branch).
Of course! The two machines I checked were only at 380! Thanks!   2020-06-25, 19:38 #15 yae9911   "Hugo" Jul 2019 Germany 378 Posts PARI/GP confirms Code: x=2^7* 5 * 23 * 127 * 659 * 53323 * 1876187 * 97544836889 * 665320793909 %111 = 7998766649128898059663516612687535453720960 (21:22) gp > eulerphi(x) %112 = 3031634148236289733373855928919180891127808 (21:22) gp > sigma(x)-x %113 = 12142697391577851168337274092012830083559040 @Dieter: The solution doesn't have prime factors > 2^64.   2020-06-26, 12:28 #16 Dieter   Oct 2017 89 Posts Thank you for the confirmations! @yae9911: Thank you. Otherwise I would have given up. It is a critical point, if a number fits into one register or not.   2020-07-05, 22:21 #17 uau   Jan 2017 7410 Posts Here's the program I used: Code: #!/usr/bin/python3 from gmpy2 import is_prime def rec(n, target, used=set(), res=1, divsum=1, last1=None, last2=None): if n == 1: if divsum - res == target: print(res) if 2 not in used and divsum * 3 - 2*res == target: print(2*res) return p, i = divs[n] e = p-1 if n % e == 0: used.add(p) rec(n//e//p**i, target, used, res*p**(i+1), divsum*((p**(i+2)-1)//(p-1))) used.remove(p) r = n // p for k in divisors(r): if p == last1 and k >= last2: continue t = k*p + 1 if t in primes and t not in used: rec(r//k, target, used, res*t, divsum*(t+1), p, k) factors = [(2, 18), (3, 3), (7, 2), (11, 1), (13, 1), (47, 1), (3169, 1), (8887, 1), (66643, 1), (72161, 1), (2495839, 1), (3847619, 1)] target = 12142680281284711468101282998309016699980172 divs = {1:None} for p, i in factors: prev = list(divs.keys()) for j in range(1, i+1): pp = p ** j for d in prev: divs[d * pp] = (p, j) def divisors(n, divs=divs): if n == 1: yield 1 return p, i = divs[n] pows =  for j in range(i): pows.append(pows[-1]*p) for d in divisors(n // pows[-1]): for m in pows: yield d * m primes = set(d+1 for d in divs if is_prime(d+1)) rec(max(divs), target) Given the number of smaller relative primes, RP=3031634148236289733373855928919180891127808, the program finds all possible numbers for which euler_phi(n) equals this given value, and checks whether the divisor sum has the desired property. Since each prime power factor p^i in n produces a factor of (p-1)p^(i-1) in RP, it follows that (p-1) divides RP, and p is of the form d+1 for some divisor d or RP. This limits the set of primes possibly appearing in n to that set. Thus the problem essentially becomes a packing problem - find a subset of powers of those primes such that the factors of (p-1)p^(i-1) add up to match exactly those in RP. The program iterates through possible solutions by trying different ways to select a prime power for n that reduces or eliminates the largest remaining prime factor in RP, then recursively solving the remaining values. There are two separate cases - eliminating a factor p with p^i with i > 1 (under "if n% e == 0:") and eliminating it a single larger prime of the form k*p+1. The "used" set is needed to avoid a case like 3^2 being used to eliminate a factor of 3 from RP, then trying to use 3 again to eliminate a factor of 2. The program is careful to avoid repeating work by trying the same solution again. It won't try both selecting 2 and then 3, and later 3 then then 2. Eliminating a factor p with a power p^i with i > 1 always eliminates all the remaining powers of p, so it'll always be the last operation for that prime. The "last1" and "last2" arguments are used to ensure that factors removing same prime are in decreasing order - to eliminate powers of 2, adding 5 then 3 is allowed, but not vice versa. The divisor sum is also calculated recursively, as it can also be expressed as a product of independent terms from each p^i factor in n.   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Xyzzy Puzzles 5 2019-06-03 05:54 Batalov Puzzles 8 2018-07-04 06:45 R. Gerbicz Puzzles 14 2017-07-03 20:01 Xyzzy Puzzles 16 2016-07-07 02:51 Batalov Puzzles 10 2015-07-07 14:59

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