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A359029 Integers m such that A006218(m+1)/(m+1) < A006218(m)/m.

%I A359029 #18 Dec 23 2022 13:07:26
%S A359029 6,10,12,16,18,22,24,28,30,36,40,42,45,46,48,50,52,54,56,57,58,60,61,
%T A359029 64,66,68,70,72,73,76,78,81,82,84,85,86,88,90,92,93,94,96,100,102,105,
%U A359029 106,108,110,112,114,117,118,120,121,122,124,126,128,130,132,133,136
%N A359029 Integers m such that A006218(m+1)/(m+1) < A006218(m)/m.
%C A359029 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 A359029 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 A359029 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 A359029 (a) f(m + 1) > f(m) occurs infinitely often (see A359028),
%C A359029 (b) f(m + 1) < f(m) occurs infinitely often (this sequence).
%C A359029 Some results:
%C A359029 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 A359029 2. f(m + 1) < f(m) is equivalent to d(m + 1) < f(m).
%C A359029 3. Each m = p - 1, p prime >= 7 is a term, so A006093 \ {1,2,4} is a subsequence.
%C A359029 Proof: in this case, d(m+1) = 2 < f(6) = 7/3. Since f(6) > 2, it follows that f(m) > 2 holds for all m >= 6; so for m = p - 1, p prime >= 7, d(m+1) = 2 < f(m) because when m >= 6, f(m) is > 2.
%C A359029 4. As there are infinitely many primes, that also proves that f(m + 1) < f(m) occurs infinitely often, which answers IMO problem (b).
%C A359029 5. There exist other terms not belonging to A006093: 24, 45, 48, 50, 54, ...
%C A359029 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 A359029 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 A359029 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%e A359029 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 A359029 with(numtheory):
%p A359029 for n from 1 to 150 do
%p A359029 m:= (1/(n+1))*sum(tau(k),k=1..n+1) - (1/n)*sum(tau(k),k=1..n);
%p A359029 if m<0 then print(n); else fi; od:
%t A359029 With[{m = 140}, Position[Differences[Accumulate[DivisorSigma[0, Range[m]]]/Range[m]], _?(# < 0 &)] // Flatten] (* _Amiram Eldar_, Dec 18 2022 *)
%o A359029 (PARI) f(n) = sum(k=1, n, n\k); \\ A006218
%o A359029 isok(m) = f(m+1)/(m+1) < f(m)/m; \\ _Michel Marcus_, Dec 19 2022
%Y A359029 Cf. A006093, A006218, A050226, A359028.
%K A359029 nonn
%O A359029 1,1
%A A359029 _Bernard Schott_, Dec 18 2022