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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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4
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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
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refs;
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OFFSET
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1,2
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COMMENTS
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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.
Changing the definition to use "less than sqrt(n)" doesn't change the sequence. - Stewart Gordon, Sep 27 2011
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LINKS
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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]
cdQ[n_]:=Module[{d=Union[Differences[Select[Divisors[n], #<=Sqrt[n]&]]]}, d=={}||d=={1}]; Select[Range[200], cdQ] (* Harvey P. Dale, Feb 12 2017 *)
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PROG
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(PARI) { n=0; for (m=1, 10^10, d=divisors(m); b=1; for (i=2, ceil(length(d)/2), if (d[i] - d[i-1] > 1, b=0; break)); if (b, write("b066522.txt", n++, " ", m); if (n==1000, return)) ) } \\ Harry J. Smith, Feb 21 2010
(Haskell)
import Data.List (genericLength)
a066522 n = a066522_list !! (n-1)
a066522_list = filter f [1..] where
f x = genericLength ds == maximum ds where ds = a161906_row x
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CROSSREFS
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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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STATUS
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approved
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