20191017, 01:17  #1 
May 2013
1001_{2} Posts 
Need more items for sequence 17, 257, 641, 65537, …
Let n be an odd positive integer, Let o=ordn2 be the order of 2 modulo n and m the period of 1/n, k is number of distinct odd residues contained in set {2^1,2^2,...,2^{n−1}} modulo n.
If odd part of o,m and k is 1 and k divide n1, then n is item in the sequence 17, 257, 641, 65537, …. 167772161 also is item.It seems all known items in the sequence are Fermat factors and divide 2^(2^100)  1 and 10^(10^100)  1. Here's my PARI/GP code to check numbers(there's large space left for improve): Code:
oddres(n)=if(n<2,0,n>>valuation(n,2)) ck(n) = { my(l=List(),m=if(1<n/=5^valuation(n, 5)<<valuation(n, 2), znorder(Mod(10, n)), 0),o=znorder(Mod(2,n))); forstep(i=0,o,1,if((2^i % n) % 2 == 1, listput(l, 2^i % n))); [m,#(Set(l)),o] } forstep(n=1, 1e3, 2, [m,k,o] = ck(n); if( oddres(m) == 1 && (n  1) % m == 0 && oddres(k) == 1 && oddres(o) == 1, print1(n", "m", "k", "o",\n"))) \\out put: 17, 16, 4, 8, 257, 256, 8, 16, 641, 32, 32, 64, and Python code to get the o, k Code:
def ck(n): def f(n, t): k = 0 while (t << k) < n: k += 1 return (t << k)  n, k v, k, o = 1, 0, 0 while True: v, z = f(n, v) k += z o = o + 1 if v == 1 or v == 0: break return o, k ck(17) # out put (4, 8) Last fiddled with by miket on 20191017 at 01:35 Reason: add question 
20191017, 15:03  #2  
Feb 2017
Nowhere
5142_{10} Posts 
Quote:
Second, the "period" of 1/n apparently is to the base ten. If this period is a power of 2, then the multiplicative order of 10 (ten) (mod n) must also be a power of 2 (so 5 can not divide n), and the multiplicative order of 5 (mod n) must therefore also be a power of 2. The multiplicative orders may, of course, be different powers of 2. Third, if n is prime, and the multiplicative order o > 1 of 2 (mod n) is even, then exactly half the representatives of {1, 2, ... 2^(o1) (mod n)} in the interval [1,n1]  that is, o/2 of them  are odd. I leave the proof as an exercise for the reader. 

20191017, 19:41  #3 
Feb 2017
Nowhere
2×3×857 Posts 
Inspired by the examples p = 5*2^7 + 1 = 641 and p = 5*2^25 + 1 = 167772161 already given, I tried other known prime factors 5*2^k + 1 of Fermat numbers.
Numerical checking showed that the multiplicative orders of 5 (mod p) for p = 5*2^39 + 1, 5*2^75 + 1, 5*2^127 + 1, 5*2^1947 + 1, and 5*2^3313 + 1 are powers of 2. I did not attempt a numerical check for p = 5*2^23473 + 1, a factor of F_{23471} It is, alas, not always true that 5 is a quintic residue of p = 5*2^k + 1, which is what is needed to make the multiplicative order of 5 (mod p) a power of 2 when p = 5*2^k + 1. The primes p = 11 = 5*2^1 + 1, 41 = 5*2^3 + 1, and 40961 = 5*2^13 + 1 provide counterexamples. I am sure there are known criteria for when 5 is a quintic residue (mod p), but I am too lazy to look them up. It is possible (AFAIK) that the multiplicative order of 5 (mod p) is a power of 2 for other prime factors k*2^n + 1 (k odd), of composite Fermat numbers, but, apart from the known prime factors of Fermat numbers up to F_{23}, I haven't checked them. 
20191018, 06:09  #4 
May 2013
1001_{2} Posts 

20191019, 12:16  #5 
Feb 2017
Nowhere
2·3·857 Posts 
I checked p = 5*2^23473 + 1; it didn't take nearly as long as I had feared it would. It's on the list.

20191021, 20:20  #6 
Feb 2017
Nowhere
2×3×857 Posts 
This topic inspired me to formulate a hypothesis.
I checked some of the primes p = q*2^n + 1 which divide F_{m} = 2^(2^m) + 1, where q is an odd prime. In every case I checked, it was true that Mod(q,p)^2^n = Mod(1,p) [that is, that q is a qth power residue (mod p)]. I suspect this might always be true when q is an odd prime (my hypothesis), but don't see an obvious proof. Before trying to prove it myself (assuming it's not already known), I tried checking to make sure there weren't any obvious counterexamples. If somebody knows (of) a proof, please point me in the right direction. Meanwhile... I have counterexamples for composite odd q, where the prime p = q*2^n + 1 divides 2^(2^m) + 1, but Mod(q,p)^(2^n) is not Mod(1,p). I also have counterexamples for p = 5*2^n + 1 which do not divide Fermat numbers; the smallest is p = 11: Mod(5, 11)^2 = Mod(3,11), not Mod(1,11). The following list, culled from a web page, is of triples [m, q, n] where p = q*2^n + 1 divides 2^(2^m) + 1, and q is prime. Code:
v=[[5, 5, 7], [12, 7, 14], [12, 397, 16], [12, 17353230210429594579133099699123162989482444520899, 15], [18, 13, 20], [19, 33629, 21], [19, 8962167624028624126082526703, 22], [23, 5, 25], [25, 48413, 29], [36, 5, 39], [38, 3, 41], [55, 29, 57], [71, 683, 73], [73, 5, 75], [81, 271, 84], [117, 7, 120], [125, 5, 127], [144, 17, 147], [172, 20569603303, 174], [207, 3, 209], [228, 29, 231], [251, 85801657, 254], [284, 7, 290], [316, 7, 320], [416, 38039, 419], [547, 77377, 550], [556, 127, 558], [579, 63856313, 581], [637, 11969, 643], [642, 52943971, 644], [744, 17, 747], [1945, 5, 1947], [2023, 29, 2027], [2089, 431, 2099], [3056, 3370842847, 3058], [3310, 5, 3313], [3723, 13308899, 3725], [4724, 29, 4727], [5320, 21341, 5323], [6537, 17, 6539], [6835, 19, 6838], [9448, 19, 9450], [14276, 157, 14280], [18749, 11, 18759], [23288, 19, 23290], [23471, 5, 23473], [30256, 121531, 30260], [48624, 28949, 48627], [79221, 6089, 79223], [95328, 7, 95330], [114293, 13, 114296], [125410, 5, 125413], [138557, 7333, 138560], [157167, 3, 157169], [213319, 3, 213321], [287384, 211, 287388], [303088, 3, 303093], [382447, 3, 382449], [472097, 89, 472099], [585042, 151, 585044], [617813, 659, 617815], [960897, 11, 960901], [1494096, 131, 1494099], [2145351, 3, 2145353], [2167797, 7, 2167800], [2478782, 3, 2478785], [3329780, 193, 3329782]] I would appreciate it if someone could check whether (Mod(q,p)^(2^n) = Mod(1,p) for the ones with larger m. Last fiddled with by Dr Sardonicus on 20191021 at 21:07 Reason: infxig sytpo 
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