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May 20

f(Xj)=2Xi+1

Let f(Xj)=2Xi+1 (i.e., the previous Xj becomes the succeeding Xi, where the first Xi=0), which yields the following:

1

3

7

15=3*5

31

63=3*3*7

127

255=3*5*17

511=7*73

1023=3*11*31

...

It was claimed (on a "Numberphile" episode) that each and every value of f(Xj) has a new prime among its factor(s), except for 63.

What is the proof that each and every value of f(Xj) has a new prime among its factor(s)?

If I understood the episode correctly, there is a reason why 63 does not have a new prime factor. What is it?

TIABh12 (talk) 14:34, 20 May 2014 (UTC)[reply]

See Mersenne prime and the OEIS entry here [1], for starters. SemanticMantis (talk) 14:47, 20 May 2014 (UTC)[reply]

So the formal statement of the conjecture is that any Mersenne number 2ⁿ-1 (and not equal to 63) has at least one prime factor that does not divide any previous Mersenne number, right? AlexTiefling (talk) 16:35, 20 May 2014 (UTC)[reply]

It's asserted, but not proved or explained, in this PDF: [2]AlexTiefling (talk) 16:39, 20 May 2014 (UTC)[reply]

It turns out that this is Bang's theorem, and the case of 63 is one of only two exceptions to Zsigmondy's theorem. AlexTiefling (talk) 16:46, 20 May 2014 (UTC)[reply]

Looks like you've got it, thanks! SemanticMantis (talk) 21:30, 20 May 2014 (UTC)[reply]

1. So far as I can tell, neither the article on Mersenne primes nor that on Zsigmondy's theorem say anything about each new term (3, 7, 15, ...) having to contain a new prime as a factor.

2. Possible typo? In reading the article on Zsigmondy's theorem, I came across the following sentence in the last paragraph of the Generalizations section:

However, the result is ineffective in the sense that the proof does give an explicit upper bound for the largest element in \mathcal{Z}(W_n), although it is possible to give an effective upper bound for the number of elements in \mathcal{Z}(W_n).

In the above sentence, should it say "does not give" instead of "does give"?Bh12 (talk) 23:10, 20 May 2014 (UTC)[reply]

1. That's exactly what's meant by the term primitive prime divisor in the introduction to the Zsigmondy's theorem article - in your case, a=2 and b=1.

2. I think you're right, but the maths is beyond me. AlexTiefling (talk) 23:20, 20 May 2014 (UTC)[reply]

Computing harmonic numbers in PARI/GP

Is there an easy (and perhaps efficient) way to compute the n-th harmonic number in PARI/GP? I need to evaluate the expression ${H_{p-1} \over p^{2}}$ for primes p in a computation I want to perform. -- Toshio Yamaguchi 14:48, 20 May 2014 (UTC)[reply]

A bit more context: In this paper it is mentioned that for any prime p > 3 ${H_{p-1} \over p^{2}}\equiv -{B_{p-3} \over 3}{\pmod {p}}$ with B_n a Bernoulli number and H_n a harmonic number (see page 4). I want to do some computations related to that result. -- Toshio Yamaguchi 15:02, 20 May 2014 (UTC)[reply]

With an integral approximation? It really depends how much accuracy you want. 70.190.182.236 (talk) 05:17, 21 May 2014 (UTC)[reply]

Well, I need enough accuracy to check for a given p whether

{H_{p-1} \over p^{2}}\equiv -{B_{p-3} \over 3}{\pmod {p}}

holds or not, i.e. whether

{H_{p-1} \over p^{2}}{\pmod {p}}{=}-{B_{p-3} \over 3}

or

{H_{p-1} \over p^{2}}{\pmod {p}}\neq -{B_{p-3} \over 3}

. -- Toshio Yamaguchi 09:42, 22 May 2014 (UTC)[reply]