A359028 - OEIS

%I #27 Dec 17 2022 20:02:11

%S 1,2,3,5,7,8,9,11,13,14,15,17,19,20,21,23,25,26,27,29,31,32,33,34,35,

%T 37,38,39,41,43,44,47,49,51,53,55,59,62,63,65,67,69,71,74,75,77,79,80,

%U 83,87,89,91,95,97,98,99,101,103,104,107,109,111,113,115,116,119,123,125,127,129

%N Integers m such that A006218(m+1)/(m+1) > A006218(m)/m.

%C Equivalently: Indices m such that f(m + 1) > f(m) where f(m) = Sum_{k=1..m} d(k) / m, where d(k) is the number of divisors of k (A000005).

%C This sequence comes from a problem proposed by South Africa during the 47th International Mathematical Olympiad, in 2006 at Ljubljana, Slovenia, but not used for the competition (see link).

%C In fact, the problem asked for a proof that, for the sequence {f(m)} defined by f(m) = (1/m) * ([m/1] + [m/2] + ... + [m/m]), where [x] denotes the integer part of x,

%C (a) f(m + 1) > f(m) occurs infinitely often (these are the terms m of this sequence),

%C (b) f(m + 1) < f(m) occurs infinitely often (see A359029).

%C Differs from A047255 when a(24) = 34 while A047255(24) = 35.

%C Some results:

%C 1. For every m, f(m) = (1/m) * ([m/1] + [m/2] + ... + [m/m]) proposed in the problem is the arithmetic mean of d(1), d(2), ..., d(m) = A006218(m)/m.

%C 2. f(m + 1) > f(m) is equivalent to d(m + 1) > f(m).

%C 3. Each m = c - 1, where c is a highly composite number (A002182) is a term.

%C Proof: in this case, d(m+1) = d(c) > max{d(1), ..., d(m)}; as f(m) = (d(1)+...+d(m)) / m < m*d(c)/m = d(c), it follows that d(m+1) = d(c) > f(m).

%C 4. As there are infinitely many highly composite numbers, that also proves that f(m + 1) > f(m) occurs infinitely often, answer to IMO problem (a).

%C 5. There exist other terms not of the form A002182 - 1: 2, 7, 8, 9, 13, 14, 15, ...

%C Note that f(m) = f(m+1) is possible iff f(m) = tau(m+1), so f(m) must be an integer (A050226) but this is not sufficient. The only known term such that f(m) = f(m+1) is at m=4, with f(4) = 2 and f(5) = tau(5) = 2.

%H 47th International Mathematical Olympiad, Slovenia, 2006, <a href="http://www.imo-official.org/problems/IMO2006SL.pdf">Problem N3, page 57</a>, Shortlisted problems with solutions.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%e f(7) = (d(1)+d(2)+d(3)+d(4)+d(5)+d(6)+d(7)) / 7 = (1+2+2+3+2+4+2) / 7 = 16/7 < f(6) = (d(1)+d(2)+d(3)+d(4)+d(5)+d(6)) / 6 = (1+2+2+3+2+4) / 6 = 14/6 = 7/3, so 6 is a term.

%p with(numtheory):

%p for n from 1 to 100 do

%p m := (1/(n+1))*sum(tau(k),k=1..n+1) - (1/n)*sum(tau(k),k=1..n);

%p if m>0 then print(n); else fi; od:

%t With[{m = 130}, Position[Differences[Accumulate[DivisorSigma[0, Range[m]]]/Range[m]], _?(# > 0 &)] // Flatten] (* _Amiram Eldar_, Dec 12 2022 *)

%Y Cf. A002182, A006218, A047255, A050226, A359029.

%K nonn

%O 1,2

%A _Bernard Schott_, Dec 12 2022