May 2021 - Page 2

Kebbaj · 2021-05-04, 14:34

Quote:

Originally Posted by Kebbaj

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.

Sorry, in my answer, I meant if A0 and A1 are not coprime then the sequence would be trivial.

uau · 2021-05-04, 16:45

Quote:

Originally Posted by Dr Sardonicus

The condition that the initial terms be relatively prime seems to have been assumed, but should have been stated.

The sentence "Our goal is to see how to generate a Fibonacci-like sequence without any prime numbers." would make more sense if it ruled out the obvious and trivial common factors solution. But I don't think this affects the problem itself - the stated requirement is to find a set of tuples, not just A0 and A1, so the lack of the condition doesn't make much difference. Given the form of the required answer, it doesn't seem to allow any trivial solution that would otherwise be forbidden.

Dieter · 2021-05-05, 08:04

The text of the challenge is updated.

1) F_0 = 0
2) It's explicitly formulated that A_0 and A_1 have to be relatively prime.

Walter · 2021-05-05, 11:48

I believe I have solved the first part of the problem. I.e. I have a set of T triplets that satisfy the conditions 1-4.

However, I am a bit confused by the second part:

Quote:

Given this set, one can generate $A_0$ and $A_1$ that satisfy
$A_0$ is equivalent to $F_{m_k-a_k} modulo p_k$
$A_1$ is equivalent to $F_{m_k-a_k+1} modulo p_k$

Should there be a $A_0$ and $A_1$ that satisfies this for every $k$ ? Or is it sufficient if it is satisfied for one $k$ ? The problem specification isn't so clear about this, or am I missing something? I haven't checked my solution in detail yet, but currently, my intuition is that there isn't a pair of $A_0$ and $A_1$ that satisfies it for all k in my solution.

Dr Sardonicus · 2021-05-05, 12:06

Quote:

Originally Posted by Walter

Quote:

Given this set, one can generate $A_0$ and $A_1$ that satisfy
$A_0$ is equivalent to $F_{m_k-a_k} modulo p_k$
$A_1$ is equivalent to $F_{m_k-a_k+1} modulo p_k$

Should there be a $A_0$ and $A_1$ that satisfies this for every $k$ ?

Yes. Using the condition that the p_k are distinct, the existence of simultaneous solutions to all the congruences is guaranteed by the Chinese Remainder Theorem (CRT).

Walter · 2021-05-05, 17:13

Quote:

Originally Posted by Dr Sardonicus

Yes. Using the condition that the p_k are distinct, the existence of simultaneous solutions to all the congruences is guaranteed by the Chinese Remainder Theorem (CRT).

Ah, right. That makes perfect sense.

Kebbaj · 2021-05-15, 20:14

The error of ak condition (1.) is always in the text even after the update? ak can take the value 0. !!

ak =0 is equivalent to ak = 60 but the condition 1 limits ak to the maximum of mk.

On other hand i try to have a matrix of less than 17 elements triplets, but it may be impossible?

2021-05-15, 20:14	#18
Kebbaj "Kebbaj Reda" May 2018 Casablanca, Morocco 144₈ Posts	The error of ak condition (1.) is always in the text even after the update? ak can take the value 0. !! ak =0 is equivalent to ak = 60 but the condition 1 limits ak to the maximum of mk. On other hand i try to have a matrix of less than 17 elements triplets, but it may be impossible? Last fiddled with by Kebbaj on 2021-05-15 at 20:15

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2021-05-05, 08:04	#14
Dieter Oct 2017 139 Posts	The text of the challenge is updated. 1) F_0 = 0 2) It's explicitly formulated that A_0 and A_1 have to be relatively prime.