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Search: id:A066522
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| A066522 |
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Numbers n whose divisors less than or equal to sqrt(n) are consecutive, from 1 up to some number k. |
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+0
2
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| 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 18, 19, 22, 23, 24, 26, 29, 31, 34, 37, 38, 41, 43, 46, 47, 53, 58, 59, 60, 61, 62, 67, 71, 73, 74, 79, 82, 83, 86, 89, 94, 97, 101, 103, 106, 107, 109, 113, 118, 122, 127, 131, 134, 137, 139, 142, 146, 149, 151, 157
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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The sequence consists of all numbers of the form p or 2p with p prime, along with 1, 8, 12, 18, 24 and 60. Sketch of proof: If k<=2 then n=1 or 8 or p or 2p. If k>2, then one of the numbers k+1, ..., k+4 is == 2 (mod 4); call it m. Then m/2 is an odd number <= k, so m = 2 * (m/2) divides n. Since m is not among 1,2,...,k, it must be greater than sqrt(n), so sqrt(n) < m <= k+4. Also, n is divisible by all positive integers <= k, including k, k-1 and k-2, whose least common multiple is their product divided by 1 or 2. So n >= k(k-1)(k-2)/2. Combining these inequalities implies k<=7 and n<=120.
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LINKS
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J. G. van der Galien, The Dawn of Science.
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EXAMPLE
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60 = 1*60 = 2*30 = 3*20 = 4*15 = 5*12 = 6*10.
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MATHEMATICA
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test[n_] := Module[{}, d=Divisors[n]; d=Take[d, Ceiling[Length[d]/2]]; Last[d]==Length[d]]; Select[Range[1, 200], test]
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CROSSREFS
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Adjacent sequences: A066519 A066520 A066521 this_sequence A066523 A066524 A066525
Sequence in context: A123345 A093641 A096157 this_sequence A007298 A127033 A028826
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KEYWORD
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nonn,nice,easy
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AUTHOR
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Johan G. van der Galien (galien8(AT)zonnet.nl), Jan 05 2002
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EXTENSIONS
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Edited by Dean Hickerson (dean(AT)math.ucdavis.edu), Jan 07, 2002.
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