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#19 by Andrew Howroyd at Fri Jan 17 14:05:53 EST 2020
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#18 by Michel Marcus at Fri Jan 17 00:57:16 EST 2020
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#17 by Jon E. Schoenfield at Fri Jan 17 00:52:36 EST 2020
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#16 by Jon E. Schoenfield at Fri Jan 17 00:52:29 EST 2020
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| COMMENTS
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ItAt first, it lookslooked as if most terms were multiplemultiples of 5. The first exceptions are a({4, 11, 16}) = {539, 1127, 1421}. However, after the first 30 terms, almost every other term is not divisible by 5.
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approved
editing
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Discussion
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| Fri Jan 17
| 00:52
| Jon E. Schoenfield: Rewording okay?
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#15 by M. F. Hasler at Wed Nov 14 13:51:04 EST 2018
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#14 by M. F. Hasler at Wed Nov 14 13:50:36 EST 2018
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| NAME
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Numbers in A121707 (n^3 | > 1 divides Sum_{k=1..n-1} k^n = A121706(n)) ) which are not semiprimes.
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| COMMENTS
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Most terms of A121707 and its (conjectured) subsequence A267999 are semiprimes. This sequence lists the " exceptions"..
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#13 by M. F. Hasler at Wed Nov 14 13:49:00 EST 2018
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Numbers n > 1 such thatin A121707 (n^3 divides | Sum_{k=1..n-1} k^n = A121706(n) but n)) which isare not a semiprimesemiprimes.
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| COMMENTS
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Most terms of A121706A121707 and its (conjectured) subsequence A267999 are semiprimes. This sequence lists the "exceptions".
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| STATUS
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approved
editing
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#12 by N. J. A. Sloane at Wed Nov 14 13:39:08 EST 2018
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#11 by N. J. A. Sloane at Wed Nov 14 13:39:06 EST 2018
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NonsemiprimesNumbers n > 1 such that n^3 divides Sum_{k=1..n-1} k^n = A121706(n) but n). is not a semiprime.
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| STATUS
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proposed
editing
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#10 by M. F. Hasler at Sun Nov 11 09:06:43 EST 2018
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Discussion
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| Sun Nov 11
| 09:08
| M. F. Hasler: I think this is ugly, but if it has to be, OK.
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| Wed Nov 14
| 13:38
| N. J. A. Sloane: It can be made less ugly. Look!
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