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A334420
Numbers m such that sigma(d)/tau(d) is an integer for all divisors d of m.
6
1, 3, 5, 7, 11, 13, 15, 17, 19, 21, 23, 29, 31, 33, 35, 37, 39, 41, 43, 47, 49, 51, 53, 55, 57, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 93, 95, 97, 101, 103, 105, 107, 109, 111, 113, 115, 119, 123, 127, 129, 131, 133, 137, 139, 141, 143, 145
OFFSET
1,2
COMMENTS
Sequences of numbers m from this sequence with k such divisors for 1 < k < 6:
k = 2: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... (A065091 - odd primes).
k = 3: 49, 169, 361, 961, 1369, 1849, 3721, 4489, 5329, 6241, 9409, ...
k = 4: 15, 21, 33, 35, 39, 51, 55, 57, 65, 69, 77, 85, 87, 91, 93, ...
k = 5: 923521, 13845841, 519885601, 1073283121, 1982119441, ...
See A334421 for sequence of the smallest numbers m with k such divisors.
All divisors of a member of the sequence are members of the sequence. - Robert Israel, May 01 2020
Numbers for which all divisors are in A003601. - Michel Marcus, May 02 2020
LINKS
FORMULA
A324500(a(n)) = 1.
EXAMPLE
Number 15 with divisors 1, 3, 5 and 15 is a term because sigma(1)/tau(1) = 1/1 = 1, sigma(3)/tau(3) = 4/2 = 2, sigma(5)/tau(5) = 6/2 = 3, sigma(15)/tau(15) = 24/4 = 6.
MAPLE
filter:= n -> andmap(d -> numtheory:-sigma(d) mod numtheory:-tau(d)=0, numtheory:-divisors(n)):
select(filter, [$1..200]); # Robert Israel, May 01 2020
MATHEMATICA
divQ[n_] := Divisible[DivisorSigma[1, n], DivisorSigma[0, n]]; Select[Range[150], AllTrue[Divisors[#], divQ] &] (* Amiram Eldar, Apr 29 2020 *)
PROG
(Magma) [m: m in [1..10^6] | &+[SumOfDivisors(d) mod NumberOfDivisors(d): d in Divisors(m)] eq 0]
(PARI) isok(m) = fordiv(m, d, if (sigma(d) % numdiv(d), return (0))); return(1); \\ Michel Marcus, Apr 29 2020
CROSSREFS
Subsequence of A306639.
Cf. A000005 (tau), A000203 (sigma), A003601, A324499, A324500, A334421.
Includes A056911.
Sequence in context: A376218 A247424 A305635 * A342144 A357014 A056911
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Apr 29 2020
STATUS
approved