1,2

For each n >= 1 there are infinitely many numbers s such that n divides sigma(s), phi(s) and tau(s).

From Dirichlet's theorem there are infinitely many numbers m for which the numbers p = n*m + 1 are prime. Then sigma(p^(n-1)), phi(p^(n-1)) and tau(p^(n-1)) numbers are divisible by n.

Laurențiu Panaitopol, Alexandru Gica, Arithmetic problems and number theory. Ideas and methods of solving, Ed. Gil, Zalău, 2006, ch. 13, p. 79, pr. 18. (in Romanian).

For n = 2, sigma(3) = 4, phi(3) = 2, tau(3) = 4 are divisible by 2.

For n = 5, sigma(3344) = 7440, phi (3344) = 1440, tau (3344) = 20 are divisible by 5 and by 10.

For n = 11, sigma(1012736) = 2161632 = 11 * 196512, phi(1012736) = 11 * 43008, tau(1012736) = 11 * 4 are divisible by 11.

Array[Block[{i = 1}, While[Mod[GCD[DivisorSigma[1, i], EulerPhi@ i, DivisorSigma[0, i]], #] != 0, i++]; i] &, 16] (*Adaptation after A222713*)

(Magma) for m in [1..16] do

for n in [1..2000000] do

if IsIntegral(SumOfDivisors(n)/m) and IsIntegral(EulerPhi(n)/m) and IsIntegral(NumberOfDivisors(n)/m) then

m, n;

break;

end if;

end for;

end for;

(PARI) isok(n, k) = ! frac(gcd(sigma(k), gcd(eulerphi(k), numdiv(k)))/n);

a(n) = my(k=1); while(!isok(n, k), k++); k; \\ Michel Marcus, Apr 20 2019

(PARI) a(n) = {if(n==1, return(1)); my(res = oo, f = factor(n), hpf = f[#f~, 1]); forprime(p = 2, oo, if(p ^ (hpf - 1) > res, return(res)); forstep(i = p ^ (hpf - 1), res, p ^ (hpf - 1), if(isok(n, i), res = min(res, i); next(2) ) ) ) } \\ uses isok from above \\ David A. Corneth, Apr 22 2019

Cf. A000005, A000010, A000203, A009223, A222713.

Sequence in context: A077104 A073815 A212994 * A339631 A195998 A291167

Adjacent sequences: A307637 A307638 A307639 * A307641 A307642 A307643

Marius A. Burtea, Apr 19 2019

More terms from David A. Corneth, Apr 21 2019

approved