OFFSET
1,1
COMMENTS
Equivalently: numbers with only one proper divisor > sqrt(n).
Also: numbers with only one nontrivial divisor d with 1 < d < sqrt(n).
Four subsequences (see examples):
1) Squarefree semiprimes (A006881) p*q with p < q, then this unique divisor is q.
2) Cube of primes p^3 (A030078), then this unique divisor is p^2.
3) Primes^4 (A030514), then this unique divisor is p^3.
For n = 1 to n = 21, we have a(n) = A319238(n) = A331231(n) but a(22) = 65 <> A319238(22) = A331231(22) = 64.
From Marius A. Burtea, May 07 2020: (Start)
The sequence contains terms that are consecutive numbers.
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), ...
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)
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.
EXAMPLE
The divisors of 15 are {1, 3, 5, 15} and only 5 satisfies sqrt(15) < 5 < 15, hence 15 is a term.
The divisors of 27 are {1, 3, 9, 27} and only 9 satisfies sqrt(27) < 9 < 27, hence 27 is a term.
The divisors of 16 are {1, 2, 4, 8, 16} and only 8 satisfies sqrt(16) < 8 < 16, hence 16 is a term.
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.
MATHEMATICA
Select[Range[200], MemberQ[{4, 5}, DivisorSigma[0, #]] &] (* Amiram Eldar, May 04 2020 *)
PROG
(PARI) isok(m) = #select(x->(x^2 > m), divisors(m)) == 2; \\ Michel Marcus, May 05 2020
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, May 04 2020
STATUS
approved