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Numbers m such that tau(sigma(m)) divides m, where tau(m) is the number of divisors function (A000005) and sigma(m) is the sum of divisors function (A000203).
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%I #17 May 15 2026 16:21:43

%S 1,2,3,4,6,8,12,16,24,36,48,64,72,80,81,84,100,112,120,128,140,144,

%T 156,160,162,168,192,198,200,208,210,216,240,256,270,288,300,320,324,

%U 336,357,360,368,384,390,420,432,448,464,468,480,512,560,576,592,600,624,630

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

%C Two subsets of terms:

%C 1) If 2^p - 1 is a Mersenne prime (p is in A000043 and 2^p-1 is in A000668), then m = 2^(p-1) is a term that belongs to A019279: the even superperfect numbers (2, 4, 16, 64, 4096, ...). Proof: sigma(m) = 1+2+...+2^(p-1) = 2^p - 1 that is a Mersenne prime so tau(2^p-1) = 2 that divides m = 2^(p-1); indeed, m/tau(sigma(m)) = 2^(p-2).

%C 2) If m = 2^(p-1) is a term as above, then 3*m is another term (see example) with 3*m/tau(sigma(3*m)) = 2^(p-2).

%H Robert Israel, <a href="/A336613/b336613.txt">Table of n, a(n) for n = 1..10000</a>

%e 48 = 2^4 * 3, so, sigma(48) = sigma(2^4) * sigma(3) = (2^5 - 1) * (1+3) = 31 * 4 = 124; then, tau(2^2 * 31) = tau(4) * tau(31) = 3 * 2 = 6, and 48/6 = 8 = 2^3, hence 48 is a term.

%p with(numtheory); filter:= m -> m/tau(sigma(m)) = floor(m/tau(sigma(m))) : select(filter, [$1..650]);

%t Select[Range[630], Divisible[#, DivisorSigma[0, DivisorSigma[1, #]]] &] (* _Amiram Eldar_, Jul 30 2020 *)

%o (PARI) isok(m) = !(m % numdiv(sigma(m))); \\ _Michel Marcus_, Jul 30 2020

%Y Cf. A000005, A000203, A062068.

%Y Cf. A019279 (subsequence), A336612 (sigma(tau(m)) divides m).

%K nonn

%O 1,2

%A _Bernard Schott_, Jul 29 2020