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A019505
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a(1)=1; for n > 1, a(n) is the smallest number with the same number of divisors as 2*a(n-1).
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12
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1, 2, 4, 6, 12, 24, 48, 60, 120, 240, 360, 720, 1260, 2520, 5040, 10080, 20160, 27720, 55440, 110880, 221760, 332640, 665280, 1081080, 2162160, 4324320, 8648640, 17297280, 21621600, 43243200, 73513440, 147026880, 294053760, 367567200, 735134400, 1396755360, 2793510720
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OFFSET
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1,2
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COMMENTS
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From J. Lowell, Mar 19 2012 and Apr 05 2012: (Start)
Conjectures:
Subsequence of A002182. [This conjecture is false. The 64th term is 97039187544499200, which has exactly 63360 divisors, but is NOT in A002182; which has the smaller number 74801040398884800, which has 64512 divisors. - J. Lowell, Nov 29 2021]
In order for n to be followed by a number less than 2n, a requirement is that the number of 2's in the prime factorization of n must not be of the form p-2 where p is a prime.
There are infinitely many values where n, 2n, and 3n are all in this sequence. (It can be proved that n, 2n, 3n, and 4n can never all be in this sequence.)
In any group of 3 consecutive terms of this sequence a,b,c at most one of the following statements is true:
The value of b is less than twice a.
The value of c is less than twice b. [This conjecture is false. Terms 121-123 are 9363553722094352358689983872000, 14258138622280036546187020896000, and 26139920807513400334676204976000. - J. Lowell, Jan 28 2022]
There are terms divisible by 2^k no matter how large k is.
The primes and the powers of 3 are the only numbers that never "come and go" from the lists of divisors of the numbers in the sequence as the terms advance. [This conjecture is false. The 122nd term, 14258138622280036546187020896000, is a multiple of 3^5=243, but the 123rd term, 26139920807513400334676204976000, is not. - J. Lowell, Jan 28 2022]
(End)
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LINKS
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FORMULA
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EXAMPLE
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After a(3)=4 we argue as follows: 2*4 = 8 has 4 factors (1,2,4,8), but smallest number with 4 factors is 6, so a(4)=6.
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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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