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#13 by Alois P. Heinz at Thu Jul 25 08:45:39 EDT 2019
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#12 by Amiram Eldar at Thu Jul 25 07:41:03 EDT 2019
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#11 by Amiram Eldar at Thu Jul 25 07:40:52 EDT 2019
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| LINKS
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Amiram Eldar, <a href="/A319745/b319745.txt">Table of n, a(n) for n = 1..10000</a>
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| STATUS
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approved
editing
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#10 by Bruno Berselli at Tue Nov 20 04:02:37 EST 2018
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#9 by Giovanni Resta at Tue Nov 20 04:00:31 EST 2018
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#8 by Michel Marcus at Sun Oct 28 06:35:59 EDT 2018
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#7 by Michel Marcus at Sun Oct 28 06:35:54 EDT 2018
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| PROG
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(PARI) hm(v) = #v/sum(k=1, #v, 1/v[k]);
vnud(n) = select(x->(gcd(x, n/x)!=1), divisors(n));
isok(n) = iferr(denominator(hm(vnud(n))) == 1, E, 0); \\ Michel Marcus, Oct 28 2018
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| STATUS
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proposed
editing
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#6 by Amiram Eldar at Thu Sep 27 08:57:33 EDT 2018
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#5 by Amiram Eldar at Thu Sep 27 08:56:11 EDT 2018
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| COMMENTS
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Ligh & Wall showed that if p, 2p-1 and 2^p-1 are distinct primes (A172461, except for 2), then the following numbers are in the sequence: 6*p^2, p^2*(2p-1), 6*p^2*(2p-1), 2^(p+1)*3*(2p2^p-1), 2^(p+1)*15*(2p2^p-1) and 2^(p+1)*(2p-1)*(2^p-1).
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#4 by Amiram Eldar at Thu Sep 27 08:50:52 EDT 2018
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| COMMENTS
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Ligh & Wall showed that if p, 2p-1 and 2^p-1 are distinct primes (A172461, except for 2), then the following numbers are in the sequence: 6*p^2, p^2*(2p-1), 6*p^2*(2p-1), 2^(p+1)*3*(2p-1), 2^(p+1)*15*(2p-1) and 2^(p+1)*(2p-1)*(2^p-1).
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| CROSSREFS
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Cf. A001599, A006086, A063947, A064591, A172461, A286325.
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