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Search: id:A140110
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| A140110 |
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Numbers n such that for all divisors of n, ratios of 2 consecutive divisors of n will always reduce to lowest terms as a ratio of consecutive integers. |
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1
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| 1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 42, 48, 54, 64, 72, 96, 100, 108, 120, 128, 144, 156, 162, 168, 180, 192, 216, 240, 256, 272, 288, 294, 324, 342, 360, 384, 432, 480, 486, 500, 512, 576, 600, 648, 720, 768, 840, 900, 960, 972, 1024, 1080, 1152, 1176
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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All numbers in sequence greater than 2 must leave remainders of 0, 4, 6, or 8 when divided by 12.
Every term after the first appears to be a multiple of 4 or 6. - John W. Layman (layman(AT)math.vt.edu), Jul 03 2008
Proof of John W. Layman's conjecture that every term after the second is a multiple of 4 or 6. The first divisor of any number is always 1. Because the only divisor of 1 is 1, the second divisor of any member of this sequence greater than 1 is 2. Because the divisors of 2 are 1 and 2, the third divisor of any term of this sequence is always either 3 (proving it is a multiple of 6 because 6 is the LCM of 2 and 3) or 4. Therefore every term greater than 2 in this sequence is a multiple of at least one of 4 and 6. - J. Lowell (jhbubby(AT)mindspring.com), Jul 07 2008
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EXAMPLE
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Divisors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60. The "6,10" disqualifies 60 from being in this sequence because 6/10 = 3/5, or more generally, a fraction in lowest terms a/b with b-a greater than 1. Specifically, if 6 is a divisor of a member of this sequence, the next divisor must be 7, 8, 9, or 12.
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CROSSREFS
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Sequence in context: A068997 A067712 A060765 this_sequence A128397 A120383 A055932
Adjacent sequences: A140107 A140108 A140109 this_sequence A140111 A140112 A140113
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KEYWORD
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nonn
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AUTHOR
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J. Lowell (jhbubby(AT)mindspring.com), Jun 21 2008
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EXTENSIONS
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More terms from John W. Layman (layman(AT)math.vt.edu), Jul 03 2008
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