RE: [math-fun] First time harmonic series exceeds 100
The harmonic number H(n) := 1 + 1/2 + 1/3 + . . . + 1/n has the interesting property that although it -> oo as n does, it's never an integer. So let H(n) = p(n)/q(n) in lowest terms. (I have full confidence that p and q are already in the OEIS.) What is known about the factorizations of the p(n)'s and q(n)'s ? In particular, are their factorizations known or believed to have statistical properties atypical of numbers of the same size (re the number, size, and exponents of their prime factors) ? Do the factors of p(n) (or q(n)) have special number-theoretical properties? (I.e., what is known about the primes -- if any -- that *never* occur as a factor of any p(n) (q(n)) ? --Dan
The numerators are A001008, and the denominators are A002805. The denominators are the easy part, so I'll mostly address that. The q(n)'s definitely have atypical factorizations; in particular, q(n) divides n!. (In fact, it divides lcm(1,2,3,...,n).) Note that if p is prime, q(p^k) is always divisible by p^k (in fact, q(n) is divisible by p^k for p^k <= n < 2*p^k - you can improve this result further for p != 3). The numerators are also divisible by every p (except 2): see http://mathworld.wolfram.com/WolstenholmesTheorem.html/. Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net The harmonic number H(n) := 1 + 1/2 + 1/3 + . . . + 1/n has the interesting property that although it -> oo as n does, it's never an integer. So let H(n) = p(n)/q(n) in lowest terms. (I have full confidence that p and q are already in the OEIS.) What is known about the factorizations of the p(n)'s and q(n)'s ? In particular, are their factorizations known or believed to have statistical properties atypical of numbers of the same size (re the number, size, and exponents of their prime factors) ? Do the factors of p(n) (or q(n)) have special number-theoretical properties? (I.e., what is known about the primes -- if any -- that *never* occur as a factor of any p(n) (q(n)) ? --Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
I always wondered how fast the denominators of the harmonic numbers grow. This is answered in: http://mathworld.wolfram.com/HarmonicNumber.html It says that for n = 0, 1, 2, 3, ..., the number of digits in the denominator of HarmonicNumber[10^n] is given by 1, 4, 40, 433, 4345, 43450, 434110, 4342302, 43428678, ... (Sloane's A114468). These digits converge to what appears to be the decimal digits of log_(10)e==0.43429448... I don't know about anyone else, but I think this is neat! (Is it obvious?) Bob --- franktaw@netscape.net wrote:
The numerators are A001008, and the denominators are A002805.
The denominators are the easy part, so I'll mostly address that.
The q(n)'s definitely have atypical factorizations; in particular, q(n) divides n!. (In fact, it divides lcm(1,2,3,...,n).) Note that if p is prime, q(p^k) is always divisible by p^k (in fact, q(n) is divisible by p^k for p^k <= n < 2*p^k - you can improve this result further for p != 3).
The numerators are also divisible by every p (except 2): see http://mathworld.wolfram.com/WolstenholmesTheorem.html/.
Franklin T. Adams-Watters
-----Original Message----- From: dasimov@earthlink.net
The harmonic number H(n) := 1 + 1/2 + 1/3 + . . . + 1/n has the interesting property that although it -> oo as n does, it's never an integer.
So let H(n) = p(n)/q(n) in lowest terms. (I have full confidence that p and q are already in the OEIS.)
What is known about the factorizations of the p(n)'s and q(n)'s ?
In particular, are their factorizations known or believed to have statistical properties atypical of numbers of the same size (re the number, size, and exponents of their prime factors) ?
Do the factors of p(n) (or q(n)) have special number-theoretical properties? (I.e., what is known about the primes -- if any -- that *never* occur as a factor of any p(n) (q(n)) ?
--Dan
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I always wondered how fast the denominators of the harmonic numbers grow.
This is answered in: http://mathworld.wolfram.com/HarmonicNumber.html
It says that for n = 0, 1, 2, 3, ..., the number of digits in the denominator of HarmonicNumber[10^n] is given by 1, 4, 40, 433, 4345, 43450, 434110, 4342302, 43428678, ... (Sloane's A114468). These digits converge to what appears to be the decimal digits of log_(10)e==0.43429448...
I don't know about anyone else, but I think this is neat! (Is it obvious?)
Well, let's see. Call the denominator D(n). The number of digits is basically log10 D(n), and the claim is that log10 D(10^n) ~ 10^n log10 e, which (dividing through by log10 e) is the same as log D(10^n) ~ 10^n or log D(N) ~ N. Now, naively we'd expect D(N) to be something like gcd(1,2,...,N), whose exponent of p is approximately log N / log p so that log D(N) ~ sum over p of log N ~ (N/log N) log N ~ N. Of course I'm using "~" in a somewhat looser sense than is traditional and I haven't proved any of the potentially tricky bits. :-) -- g
Looking at A003418, and fixing Gareth's typo, we have lcm(1,2,...,N) = e^(N(1+o(1)), which is the result we want. Note that we haven't proved that the numerators and denominators act like lcm(1,2,...,N); but it shouldn't be too hard to do so. We don't need a very close approximation. Franklin T. Adams-Watters -----Original Message----- From: gareth.mccaughan@pobox.com
I always wondered how fast the denominators of the harmonic numbers grow.
This is answered in: http://mathworld.wolfram.com/HarmonicNumber.html
It says that for n = 0, 1, 2, 3, ..., the number of digits in the denominator of HarmonicNumber[10^n] is given by 1, 4, 40, 433, 4345, 43450, 434110, 4342302, 43428678, ... (Sloane's A114468). These digits converge to what appears to be the decimal digits of log_(10)e==0.43429448...
I don't know about anyone else, but I think this is neat! (Is it obvious?)
Well, let's see. Call the denominator D(n). The number of digits is basically log10 D(n), and the claim is that log10 D(10^n) ~ 10^n log10 e, which (dividing through by log10 e) is the same as log D(10^n) ~ 10^n or log D(N) ~ N. Now, naively we'd expect D(N) to be something like gcd(1,2,...,N), whose exponent of p is approximately log N / log p so that log D(N) ~ sum over p of log N ~ (N/log N) log N ~ N. Of course I'm using "~" in a somewhat looser sense than is traditional and I haven't proved any of the potentially tricky bits. :-) -- g _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
participants (4)
-
Daniel Asimov -
franktaw＠netscape.net -
Gareth McCaughan -
Robert Baillie