sum{k=1 to m} 1/k^n

[hv at crypt.org]

Leroy Quet <qq-quet at mindspring.com> wrote:
:Consider the sequence where each term is a numerator of at least one 
:reduced rational
:sum{k=1 to m} 1/k^n,
:taken over all positive integers m and n.
:
:So, unless I carelessly erred, the sequence begins:
:
:1, 3, 5, 9, 11, 17, 25, 33, 49,...
:
:
:(Not in EIS.)
:
:3 things:
:
:1) Except for 1 and 49, do any of the terms of this sequence correspond
:with more than one {m,n}?
:
:(49 is numerator when m = 6 and n = 1, and when m = 3 and n =2.)

I wrote a quick script for the sequence, and I get these results:
1 3 5 9 11 17 25 33 49 65 129 137 205 251 257 363 513 761 1025 1393 2035 2049
4097 5269 5369 7129 7381 8051 8193 16385 22369 28567 32769 47449 65537 83711
86021 131073 256103 257875 262145 266681 282251 524289 1048577 1077749 1145993
1171733 1195757 1686433 1968329 2097153 2436559 3037465 4194305 8388609
9778141 9822481 10097891 14001361 14011361 14274301 16777217 18858053 19093197
33554433 36130315 42142223 55835135 60526249 67108865 78708473 134217729
239437889 240505109 268435457 268736069 275295799 362976251 431733409
444316699 536870913 806108207 1073741825 1347822955 2147483649 2177317873
4294967297 5170139875 8589934593 13062296531 17179869185 19148110939
19164113947 33654237761 34052522467 34359738369 34395742267 40799043101
40931552621 47463376609 47464376609 61978938025 68719476737 78368963449
137438953473 205234915681 274877906945 312536252003 315404588903 470199366251
538589354801 549755813889 743375539195 822968714749 940908897061 ...

Given the sparseness of the list, further coincidence seems possible but
unlikely.

The list for denominators seems much more likely to have repeats due to the
divisibility constraints. I see repeats for 2520, 3600, 27720, 360360 (twice),
5173168, 1296000, 153679680, 8923714800, 16003008000 ...

I've coped the perl I used below, in case it is of interest; note that it
assumes that the result for (m=1, n=p) is always greater than that for
(m=1, n<p) for all prime p; I'm not sure how justified that is.

Hugo van der Sanden
---
#!/usr/bin/perl -w
use Math::BigRat;
my @index = (undef, undef, 0);
my @p = (2, 3);
my @calc;
$| = 1;
print "1 ";
print nextn(), " " while 1;

sub nextn {
	my $min;
	for (2 .. $#index) {
		$calc[$_] ||= calc($_, $index[$_] ||= 1);
		$min = $_ if !$min || $calc[$min] > $calc[$_];
	}
	$#index = nextprime() - 1 if $min == $#index && $index[$min] == 1;
	++$index[$min];
	my $ret = $calc[$min];
	$calc[$min] = 0;
	return $ret;
}

sub calc {
	my($m, $n) = @_;
	my $sum = Math::BigRat->new(1);
	$sum += Math::BigRat->new("1/$_") ** $n for 2 .. $m;
	return $sum->numerator;
}

sub nextprime {
	my $p = $p[$#p];
	DIV: while (1) {
		$p += 2;
		my $pc = 0;
		while (1) {
			my $d = $p[$pc++];
			last if $d * $d > $p;
			next DIV unless $p % $d;
		}
		$p[@p] = $p;
		return $p;
	}
}