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 2021-05-02, 10:03 #1 tgan   Jul 2015 3210 Posts May 2021
 2021-05-02, 12:30 #2 Walter   "Walter S. Gisler" Sep 2020 Switzerland 11112 Posts I believe there are some mistakes in the problem statement, can anyone confirm this? 1. For every natural number $n$, we have that for some $k$, $n$ is equivalent to $a_k$ modulo $m_k$ (i.e. $m_k$ divides $n-a_k$). Shouldn't this be: For every natural number $n$, we have that for some $k$, $a_k$ is equivalent to $n$ modulo $m_k$ 2. one can prove that this sequence does not contain any primes by using the following easy-to-prove identity, which holds for any Fibonacci-like sequence: $A_{m+n} = A_mF_{n-1} + A_{m+1}F_n$ . And this one: $A_{m+n+1} = A_mF_{n-1} + A_{m+1}F_n$ ?
 2021-05-02, 13:51 #3 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 9,973 Posts Isn't (1) the same either way? My "argument" with them is F0=1, which is wrong. Everybody knows F0=0 (otherwise the "only prime indexes can be prime" won't hold). In fact, how I always remember it is that the 5th fibo number is 5 In fact, they say as much, further when affirming that F3=2 and F11=89. If F0=1, those would be false. Last fiddled with by LaurV on 2021-05-02 at 14:06
 2021-05-02, 14:44 #4 Walter   "Walter S. Gisler" Sep 2020 Switzerland 1510 Posts Ah, you are right. I had missed that F0 = 1. With that, 2 is indeed correct. Regarding (1), I am pretty sure it isn't the same either way. m_k >= a_k, hence a_k mod m_k is either 0 or a_k , so with a set of triplets that is limited in size, a_k mod m_k couldn't be any natural number.
 2021-05-02, 18:03 #5 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 9,973 Posts "a (mod m)" is defined as the reminder (an integer number between 0 and m-1, inclusive) that is left when you divide a to m, therefore, a$$\equiv$$b (mod m) is the same as b$$\equiv$$a (mod m), those are "equivalence classes", not numbers, it is just that we are lazy and don't write the "hats" (i.e. $$\hat{a} \equiv \hat{b}$$ (mod m))
 2021-05-02, 18:43 #6 Walter   "Walter S. Gisler" Sep 2020 Switzerland 3·5 Posts Thanks, LaurV. 0scar also kindly explained it to me. This really confused me, but it is clear now.
 2021-05-02, 20:48 #7 SmartMersenne   Sep 2017 131 Posts Also note that a_k can't be zero but can be m_k due to 1 <= a_k <= m_k
 2021-05-03, 05:52 #8 Dieter   Oct 2017 2×32×7 Posts Have A_0 and A_1 to be relatively prime? He didn't write it. Otherwise A_0 = 4 and A_1 = 6 yielded a Fibonacci-like sequence without primes. But that cannot be the challenge.
2021-05-03, 14:08   #9
slandrum

Jan 2021
California

1A516 Posts

Quote:
 Originally Posted by Dieter Have A_0 and A_1 to be relatively prime? He didn't write it. Otherwise A_0 = 4 and A_1 = 6 yielded a Fibonacci-like sequence without primes. But that cannot be the challenge.
That was just the start of it, you need to read the rest of the requirements of the series to see if the series you made meets the other requirements.

2021-05-03, 18:12   #10
Kebbaj

"Kebbaj Reda"
May 2018
Casablanca, Morocco

32·11 Posts

Quote:
 Originally Posted by Dieter Have A_0 and A_1 to be relatively prime? He didn't write it. Otherwise A_0 = 4 and A_1 = 6 yielded a Fibonacci-like sequence without primes. But that cannot be the challenge.
If A0 and A1 are coprime, the sequance would be trivial: “easy common factor of the sequence with no prime “. So for the sequence to be interesting it is essential that A0 and A1 would be coprime.
GCD[A0,A1] must be = 1.

I think that the webmaster did not write the relation of divisiblity between A0 and A1 because the question was to find a sequence of triplets with the conditions quoted 1 to 4 ”and the A0 A1 are simply to explain the goal of the question.

Last fiddled with by Kebbaj on 2021-05-03 at 18:18

 2021-05-04, 12:50 #11 Dr Sardonicus     Feb 2017 Nowhere 26·7·13 Posts The May 2021 Challenge seems quite similar to finding Riesel or Sierpinski numbers by means of "covering sets" of congruences. The misstatement of the subscripts is the kind of oopsadaisy a number theorist might make The condition that the initial terms be relatively prime seems to have been assumed, but should have been stated.

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