

A336687


Numbers m such that tau(sigma(m)) and sigma(tau(m)) both divide m, where tau(m) is the number of divisors function (A000005) and sigma(m) is the sum of divisors function (A000203).


0



1, 3, 4, 12, 64, 84, 140, 144, 162, 192, 336, 360, 420, 468, 480, 576, 600, 644, 720, 780, 1008, 1344, 1512, 1584, 1600, 1740, 1872, 2160, 2240, 2448, 2592, 2736, 2880, 2884, 3136, 3240, 3888, 4032, 4158, 4228, 4464, 4608, 4800, 5040, 5115, 5184, 5328, 5670, 6060, 6192, 6336
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OFFSET

1,2


COMMENTS

Conjecture: The only m such that m = tau(sigma(m))*sigma(tau(m)) are in {1,468,3240}. Verified for m up to 1*10^9.  Ivan N. Ianakiev, Aug 06 2020


LINKS



EXAMPLE

For 84: tau(84) = 12 and sigma(12) = 28 with 84/28 = 3. Also, sigma(84) = 224 and tau(224) = 12 with 84/12 = 7. Hence, 84 is a term.


MAPLE

with(numtheory):
filter:= m> irem(m, tau(sigma(m)))=0 and irem(m, sigma(tau(m)))=0:
select(filter, [$1..7000])[];


MATHEMATICA

Select[Range[6400], And @@ Divisible[#, {DivisorSigma[0, DivisorSigma[1, #]], DivisorSigma[1, DivisorSigma[0, #]]}] &] (* Amiram Eldar, Jul 31 2020 *)


PROG

(PARI) isok(m) = !(m % numdiv(sigma(m))) && !(m % sigma(numdiv(m))); \\ Michel Marcus, Aug 02 2020


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



