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Numbers m with only one divisor d such that sqrt(m) < d < m.
4

%I #58 Sep 08 2022 08:46:25

%S 6,8,10,14,15,16,21,22,26,27,33,34,35,38,39,46,51,55,57,58,62,65,69,

%T 74,77,81,82,85,86,87,91,93,94,95,106,111,115,118,119,122,123,125,129,

%U 133,134,141,142,143,145,146,155,158,159,161,166,177,178,183,185,187

%N Numbers m with only one divisor d such that sqrt(m) < d < m.

%C Equivalently: numbers with only one proper divisor > sqrt(n).

%C Also: numbers with only one nontrivial divisor d with 1 < d < sqrt(n).

%C Four subsequences (see examples):

%C 1) Squarefree semiprimes (A006881) p*q with p < q, then this unique divisor is q.

%C 2) Cube of primes p^3 (A030078), then this unique divisor is p^2.

%C 3) Primes^4 (A030514), then this unique divisor is p^3.

%C 4) Numbers with 4 divisors: A030513 = A006881 Union A030078.

%C For n = 1 to n = 21, we have a(n) = A319238(n) = A331231(n) but a(22) = 65 <> A319238(22) = A331231(22) = 64.

%C From _Marius A. Burtea_, May 07 2020: (Start)

%C The sequence contains terms that are consecutive numbers.

%C If the numbers 4*k + 1 and 6*k + 1, k >= 1, are prime numbers, then the numbers 12*k + 2 and 12*k + 3 are terms. Examples: (14, 15), (38, 39), (86, 87), (122, 123), (158, 159), (218, 219), (302, 303), ...

%C If the numbers 6*m + 1, 10*m + 1 and 15*m + 2, m >= 1, are prime numbers, then the numbers 30*m + 3, 30*m + 4 and 30*m + 5 are terms. Examples: (33, 34, 35), (93, 94, 95), (213, 214, 215), (393, 394, 395), (633, 634, 635), ... (End)

%C There are never more than 3 consecutive terms because one of them would be divisible by 4, and neither 8 nor 16 belong to such a string of 4 consecutive terms.

%F m is a term iff tau(m) - A038548(m) = 2 where tau = A000005.

%e The divisors of 15 are {1, 3, 5, 15} and only 5 satisfies sqrt(15) < 5 < 15, hence 15 is a term.

%e The divisors of 27 are {1, 3, 9, 27} and only 9 satisfies sqrt(27) < 9 < 27, hence 27 is a term.

%e The divisors of 16 are {1, 2, 4, 8, 16} and only 8 satisfies sqrt(16) < 8 < 16, hence 16 is a term.

%e The divisors of 28 are {1, 2, 4, 7, 14, 28} but 7 and 14 satisfy sqrt(28) < 7 < 14 < 28, hence 28 is not a term.

%t Select[Range[200], MemberQ[{4, 5}, DivisorSigma[0, #]] &] (* _Amiram Eldar_, May 04 2020 *)

%o (PARI) isok(m) = #select(x->(x^2 > m), divisors(m)) == 2; \\ _Michel Marcus_, May 05 2020

%o (Magma) [k:k in [1..200]|#[d:d in Divisors(k)|d gt Sqrt(k) and d lt k] eq 1]; // _Marius A. Burtea_, May 07 2020

%Y Disjoint union of A006881, A030078, and A030514.

%Y Disjoint union of A030513 and A030514.

%Y Cf. A000005, A038548.

%K nonn

%O 1,1

%A _Bernard Schott_, May 04 2020