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A307640
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Least number k such that n divides gcd(sigma(k), phi(k), tau(k)).
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0
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1, 3, 18, 15, 3344, 45, 24128, 30, 882, 3344, 1012736, 126, 1953792, 24128, 16200, 168, 452263936, 2016, 1852571648, 3344, 40768, 1012736, 27007123456, 420, 1490000, 1953792, 103968, 24128, 2739920699392, 30096, 8348342681600, 840, 9114624, 452263936, 6163776, 2016
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OFFSET
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1,2
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COMMENTS
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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.
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REFERENCES
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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).
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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Array[Block[{i = 1}, While[Mod[GCD[DivisorSigma[1, i], EulerPhi@ i, DivisorSigma[0, i]], #] != 0, i++]; i] &, 16] (*Adaptation after A222713*)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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