A301893 Numbers m that set records for the ratio A010846(m)/A000005(m).

1, 6, 10, 18, 22, 30, 42, 66, 78, 102, 114, 138, 150, 174, 210, 330, 390, 510, 570, 690, 870, 1110, 1230, 1290, 1410, 1590, 1770, 1830, 2010, 2130, 2190, 2310, 2730, 3570, 3990, 4830, 6090, 6510, 7770, 8610, 9030, 9870, 11130, 12390, 12810, 14070, 14910, 15330

Offset: 1

Author

Michael De Vlieger, Mar 28 2018

Keywords

• nonn

Comments

We define an "n-regular" number as 1 <= m <= n such that m | n^e with integer e >= 0. The divisor d is a special case of regular number m such that d | n^e with e = 0 or e = 1. Regular numbers m can exceed n; we are concerned only with regulars m <= n herein.

Since divisors are a special case of regular numbers, we examine those numbers m that set records for the ratio of the "regular counting function" A010846(m) and the divisor counting function A000005(m).

There are 2 nonsquarefree terms {18, 150} less than 36,000,000.

The sequence contains no numbers with omega(m) = 1. This is because all regular m divide p^e, and since all the regulars of 1 also divide 1, no primes or prime powers greater than 1 appear in a(n).

The values of A000005(a(n)) are in A000079, i.e., powers 2^e except e = 1.

Aside from the 2 nonsquarefree terms, many terms m are products of A002110(i) * p_j, with j > i between some lower and upper bound outside of when m is in A002110. Example: 30 is in A002110; {42, 66, 78, 102, 114, 138, 174} are A002110(3) * p_j with 2 <= j <= 8.

There are a few terms of the form A002110(i) * p_j * p_k, with i + 1 < j < k. In other words, there is a gap in the indices of the prime divisors between the 3rd and 2nd largest prime divisors, as well as one potentially between the 2nd and largest prime divisors. The smallest m of this type is 46410 = 2 * 3 * 5 * 7 * 13 * 17, followed by 51870 = 2 * 3 * 5 * 7 * 13 * 19.

Conjectures:

1. The only nonsquarefree terms are 18 and 150.

2. Primorials A002110(i) for i = 0 and i > 2 are in the sequence.

Examples

			The number 1 sets a record as it is the first term; the ratio A010846(1)/A000005(1) = 1. Since 2 <= m <= 5 have omega(m) = 1, they too have ratio = 1 and do not appear.
6 is the next term since those numbers 1 <= k <= 6 that divide some nonnegative integer power of 6 are {1, 2, 3, 4, 6}; of these, 4 are divisors, thus we have the ratio 5/4. This exceeds 1, so 6 follows 1 in the sequence. The numbers 7 <= m <= 9 have omega(m) = 1.
10 appears next since the regular m of 10 are {1, 2, 4, 5, 8, 10}; of these 4 divide 10, thus we have ratio 6/4 which exceeds that of 6, so 10 follows 6.
12 is not in the sequence since the regular m of 12 are {1, 2, 3, 4, 6, 8, 9, 12} and 6 of these divide 12, giving us the ratio 8/6 which is less than the 6/4 of 10.
		

Crossrefs

Cf. A000005, A010846, A002110, A002182, A244052.

Programs

Mathematica

With[{s = Table[Count[Range@ n, _?(PowerMod[n, Floor@ Log2@ n, #] == 0 &)]/DivisorSigma[0, n], {n, 3000}]}, Map[FirstPosition[s, #][[1]] &, Union@ FoldList[Max, s]]]

PARI

a000005(n) = if(n==0, 0, numdiv(n)) \\ after Michael Somos in A000005
a010846(n) = sum(k=1, n, if(gcd(n, k)-1, 0, moebius(k)*(n\k))) \\ after Benoit Cloitre in A010846
r=0; for(m=1, oo, if(a010846(m)/a000005(m) > r, print1(m, ", "); r=a010846(m)/a000005(m))) \\ Felix Fröhlich, Mar 30 2018