A359029 -id:A359029 - cp's OEIS frontend

A359028 Integers m such that A006218(m+1)/(m+1) > A006218(m)/m.

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, 37, 38, 39, 41, 43, 44, 47, 49, 51, 53, 55, 59, 62, 63, 65, 67, 69, 71, 74, 75, 77, 79, 80, 83, 87, 89, 91, 95, 97, 98, 99, 101, 103, 104, 107, 109, 111, 113, 115, 116, 119, 123, 125, 127, 129

Offset: 1

Bernard Schott, Dec 12 2022

nonn

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).
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).
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,
(a) f(m + 1) > f(m) occurs infinitely often (these are the terms m of this sequence),
(b) f(m + 1) < f(m) occurs infinitely often (see A359029).
Differs from A047255 when a(24) = 34 while A047255(24) = 35.
Some results:
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.
2. f(m + 1) > f(m) is equivalent to d(m + 1) > f(m).
3. Each m = c - 1, where c is a highly composite number (A002182) is a term.
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).
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).
5. There exist other terms not of the form A002182 - 1: 2, 7, 8, 9, 13, 14, 15, ...
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.

			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.

47th International Mathematical Olympiad, Slovenia, 2006, Problem N3, page 57, Shortlisted problems with solutions.
Index to sequences related to Olympiads.

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

with(numtheory):
for n from 1 to 100 do
m := (1/(n+1))*sum(tau(k),k=1..n+1) - (1/n)*sum(tau(k),k=1..n);
if m>0 then print(n); else fi; od:

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

cp's OEIS Frontend

A359028 Integers m such that A006218(m+1)/(m+1) > A006218(m)/m.

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