Finite differences for the win

George M · 2018-01-01, 13:00

I don’t think this goes by the rules because I am not making a guess. I just think a good way to find a probable value for p such that M51 = 2^p - 1 is as follows:

Arrange $p_1, p_2,p_3,\ldots p_{50}$ in a sequence just like so where $p_n$ would be a prime p for the n-th mersenne prime. Then, find the difference between two adjacent pairs of those prime numbers $(p_1, p_2), (p_2, p_3), (p_3, p_4)\ldots (p_{49}, p_{50})$

Now apply the same rule to the difference of these prime numbers. Let $a_1 = p_2 - p_1$ and then $a_2 = p_3 - p_2$ and so on. Now, find the difference between $a_1$ and $a_2$ and find the difference of each adjacent pair of $(a_n, a_{n + 1})$ just like what you did with the primes.

Keep doing this and eventually you will reach a constant C where the difference between $C_n$ and $C_{n + 1}$ is 0 because $C_n = C_{n + 1}$ . Now work your way backwards by assuming that if there existed $p_{51}$ then this process would still reach 0, implying that it reaches C, implying etc...

After that, you will reach a value V that is likely to be equal to $p_{51}$ .

axn · 2018-01-01, 13:41

Quote:

Originally Posted by George M

I just think a good way to find a probable value for p such that M51 = 2^p - 1 is as follows:

Ok... So you want to do polynomial interpolation?

Code:

v=[2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281];
polinterpolate(v,,#v+1)

Code:

-4668069701844879790

Now what? Maybe we had too many primes in there. Let's try with > 10m digits

Code:

v=[37156667, 42643801, 43112609, 57885161, 74207281];
polinterpolate(v,,#v+1)

Code:

47248547

Hmmm... Not quite what you'd expect?

Batalov · 2018-01-01, 21:37

Oh no, this curve is definitely coming down. Deja vu all over again.

2018-01-01, 13:00	#1
George M Dec 2017 110010₂ Posts	Finite differences for the win I don’t think this goes by the rules because I am not making a guess. I just think a good way to find a probable value for p such that M51 = 2^p - 1 is as follows: Arrange $p_1, p_2,p_3,\ldots p_{50}$ in a sequence just like so where $p_n$ would be a prime p for the n-th mersenne prime. Then, find the difference between two adjacent pairs of those prime numbers $(p_1, p_2), (p_2, p_3), (p_3, p_4)\ldots (p_{49}, p_{50})$ Now apply the same rule to the difference of these prime numbers. Let $a_1 = p_2 - p_1$ and then $a_2 = p_3 - p_2$ and so on. Now, find the difference between $a_1$ and $a_2$ and find the difference of each adjacent pair of $(a_n, a_{n + 1})$ just like what you did with the primes. Keep doing this and eventually you will reach a constant C where the difference between $C_n$ and $C_{n + 1}$ is 0 because $C_n = C_{n + 1}$ . Now work your way backwards by assuming that if there existed $p_{51}$ then this process would still reach 0, implying that it reaches C, implying etc... After that, you will reach a value V that is likely to be equal to $p_{51}$ . Last fiddled with by George M on 2018-01-01 at 13:01

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2018-01-01, 21:37	#3
Batalov "Serge" Mar 2008 San Diego, Calif. 101×103 Posts	Oh no, this curve is definitely coming down. Deja vu all over again.