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A301893
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Numbers m that set records for the ratio A010846(m)/A000005(m).
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0
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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]]]
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PROG
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r=0; for(m=1, oo, if(a010846(m)/a000005(m) > r, print1(m, ", "); r=a010846(m)/a000005(m))) \\ Felix Fröhlich, Mar 30 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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