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A018894
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Numbers n such that sigma(n)/phi(n) sets a new record.
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9
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1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 360, 420, 840, 1260, 1680, 2520, 4620, 9240, 13860, 18480, 27720, 55440, 110880, 120120, 180180, 240240, 360360, 720720, 1441440, 2162160, 3603600, 4084080, 4324320, 6126120, 12252240, 24504480, 36756720, 61261200
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OFFSET
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1,2
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COMMENTS
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Is a(n+1) <= 2*a(n)? Is a(n) divisible by the primorial p# where p is the largest prime divisor of a(n)? Is a(k) divisible by p# for all k > n + 1? (Cf. A002110.) - David A. Corneth, May 22 2016
Yes, a(n+1) <= 2*a(n) -- if m is odd, phi(2m) = phi(m) and sigma(2m) = 3*sigma(m).
If m is even then phi(2m) = 2*phi(m) and sigma(2m) > 2*sigma(m).
So sigma(2m)/phi(2m) > sigma(m)/phi(m). (end)
Let prime(n)# be the product of the first n primes. Then the LCM of the terms <= 10^40 is 89# * 7# * 5# * (3#)^2 * (2#)^4.
We can assume a larger LCM for terms <= 10^60 namely P# * (13#)^3 * (11#) * (5#) * (3#)^2 * (2#)^4. This gives a total of 466 terms <= 10^75 where P is an arbitrary large prime such that P# <= 10^75.
The LCM of these found terms is a proper divisor and for all primes p <= 13 the exponent is less than the assumed prime. Conjecture: These 466 terms are the terms <= 10^75.
For all 240 terms 1 < t <= 10^40 the following holds: there exists a p|t such that t/p is a term. Conjecture: This holds for all terms t > 1.
Using this technique to find terms I get 6522 terms <= 10^1000 and no conflict with terms found above.
See attached file with terms assuming these conjectures. (End)
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LINKS
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MATHEMATICA
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Flatten@ Function[k, FirstPosition[k, #] & /@ Union@ Rest@ FoldList[Max, 0, k]]@ Array[DivisorSigma[1, #]/EulerPhi@ # &, 10^7] (* Michael De Vlieger, May 27 2016, Version 10 *)
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PROG
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(PARI) lista(nn) = {mse = 0; for (n=1, nn, se = sigma(n)/eulerphi(n); if (se > mse, print1(n, ", "); mse = se); ); } \\ Michel Marcus, Jul 10 2015
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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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