Q about LCM(1..n-1) <LCM(1..n)< LCM(1..n+1)

[Robert Israel]

lcm(1..n-1) < lcm(1..n) if and only if n is a prime power.
So your sequence consists of those n for which both n and n+1
are prime powers.  By Catalan's conjecture (proved by
Mihailescu), the only case where n and n+1 are both powers > 1
is n=8.  Otherwise, whichever of n and n+1 is even must be a
power of 2, and the other must be a prime: either a Mersenne
prime if n+1 is the power of 2, or a Fermat prime if n is the
power of 2.

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada

On Wed, 16 Jan 2008, zak seidov wrote:

> %I A000001
> %S A000001 2,3,4,7,8,16,31,127,256
> %N A000001 Numbers n such that
> lcm(1..(n-1))<lcm(1..n)<lcm(1..(n+1)).
> %C A000001 Or, numbers n such that
> A003418(n-1)<A003418(n)<A003418(n+1).
> Are all numbers of kind 2^m or 2^m-1?
> %t A000001 ta=Table[(LCM @@ Range[n]),{n,2000}];
> Do[If[ta[[i-1]]<ta[[i]]<ta[[i+1]],Print[i]],{i,2,1999}]
> %Y A000001 A003418
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov (zakseidov at yahoo.com), Jan 17
> 2008
>
>
>
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Earlier I wrote:
:I was prompted to look at A093407:
:  For p = prime(n), the least k such that p divides the numerator of
:  a sum 1/k + 1/x1 +...+ 1/xm, where x1,...,xm (for any m) are distinct
:  positive integers < k.
:by a reference in an email from Franklin T. Adams-Watters.
:
:>From this sequence we see that a(2 = pi(3)) = 2, since 1 + 1/2 == 0 (mod 3),
:and clearly no other entry in the sequence will take the value 2.
:
:Similarly, a(pi(2)) = a(pi(5)) = a(pi(11) = 3. Since the non-empty subsets
:of { 1, 1/2, 1/3 } sum to [2, 3, 5, 6, 8, 9, 11]/6, clearly we achieve
:a sum == 0 (mod p) only when p is one of {2, 3, 5, 11}.
:
:How hard is it to find a(n) = max({ k : A093407(pi(k)) == n })?
:The above shows a(2) = 3, a(3) = 11. I think knowing a few more terms of
:this sequence would enable some further optimisations in the calculation
:of new terms of A101877.
:
:I don't know, though, whether there is an easier way to determine a(n)
:than constructing the 2^n-1 subset sums. How tight are the theoretical
:limits we can impose on a(n)? It is easy to show a(n) <= n * A3418(n),
:but that's a pretty loose upper bound.

Ah, we have a tight upper bound of a(n) <= H_n * A3418(n); this becomes
equality whenever the RHS is prime. (And so by results on H_n we also
know the inequality is strict when n+1 is prime.) I don't know about
lower bounds though: it is possible an appeal to Bertrand's postulate
(er, Chebyshev's theorem) would get somewhere.

Calculating by hand, I find the sequence starts:
with offset 2, and that helps me find the related sequence A075226:

It follows from the definitions that a(n) >= A075226(n) whenever
A075226(n) != A075226(n-1), and it is likely to be equal in most
if not all of those cases.

(Interestingly, a(8) is also my current upper bound and conjectured
actual value for A101877(7). I don't know whether that is more than
coincidence, but I do know no similar coincidence occurs for A101877(8)
or A101877(9) by virtue of the bounds I've established for those.)

Hugo