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#77 by Michael De Vlieger at Sun Oct 23 16:11:45 EDT 2022
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#76 by Jianing Song at Sun Oct 23 14:56:00 EDT 2022
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#75 by Jianing Song at Sun Oct 23 14:54:39 EDT 2022
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| COMMENTS
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a(n^4+n^2+1) = n^2-n+1: suppose that n^2-n+k divides n^4+n^2+1 = (n^2-n+k)*(n^2+n-k+2) - (k-1)*(2*n+1-k) for 2 <= k <= 2*n, then (k-1)*(2*n+1-k) >= n^2-n+k, or n^2 - (2*k-1)*n + (k^2-k+1) = (n-k+1/2)^2 + 3/4 < 0, which is impossible. Hence the next smallest divisor of n^4+n^2+1 than n^2-n+1 is at least n^2-n+(2*n+1) = n^2+n+1 > sqrt(n^4+n^2+1). - Jianing Song, Oct 23 2022
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| CROSSREFS
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The strictly inferior case is A060775. Cf. also A140271.
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| STATUS
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approved
editing
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#74 by Michel Marcus at Tue Mar 22 10:26:57 EDT 2022
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#73 by Joerg Arndt at Tue Mar 22 10:11:49 EDT 2022
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#72 by M. F. Hasler at Tue Mar 22 09:55:49 EDT 2022
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Discussion
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Tue Mar 22
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| Charles R Greathouse IV: I had just made a similar point over at Mathoverflow (for A060775, which is the same for non-squares): https://mathoverflow.net/q/418394/6043
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#71 by M. F. Hasler at Tue Mar 22 09:55:43 EDT 2022
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| CROSSREFS
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Indices of given values: A008578 (a(n) = 1: 1 together withand the prime numbers: a(n) = 1), A161344 (a(n) = 2), A161345 (a(n) = 3), A161424 (4), A161835 (5), A162526 (6), A162527 (7), A162528 (8), A162529 (9), A162530 (10), A162531 (11), A162532 (12), A282668 (indices of primes).
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| STATUS
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proposed
editing
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#70 by M. F. Hasler at Tue Mar 22 09:54:00 EDT 2022
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#69 by M. F. Hasler at Tue Mar 22 09:42:19 EDT 2022
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| FORMULA
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A033677(n) * a(n) = n / A033677(n.).
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| CROSSREFS
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Cf. A033677, (n/a(n)), A000196, (sqrt), A027750, (list of divisors), A056737, (n/a(n) - a(n)), A219695, A282668 (positionshalf of this for odd numbers), A207375 (list the central primes)divisor(s)).
From Omar E. Pol, Jul 05 2009: (Start)
SequenceIndices of given values: A008578 (a(n) = 1: 1 together with the correspondingprime numbers), A161344 (a(n) = 2), A161345 (a(n) = 3), A161424 (4), A161835 (5), A162526 (6), A162527 (7), A162528 (8), A162529 (9), A162530 (10), A162531 (11), A162532 (12), A282668 (indices: of primes).
.. 1 ..... A008578 (1 together with the prime numbers)
.. 2 ..... A161344
.. 3 ..... A161345
.. 4 ..... A161424
.. 5 ..... A161835
.. 6 ..... A162526
.. 7 ..... A162527
.. 8 ..... A162528
.. 9 ..... A162529
. 10 ..... A162530
. 11 ..... A162531
. 12 ..... A162532 (End)
Cf. A207375.
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| STATUS
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approved
editing
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Discussion
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Tue Mar 22
| 09:54
| M. F. Hasler: @ my comment from 2011: still no efficient algorithm! The given programs (PARI and others) first make the list of *all* divisors, there are e.g., 2^k for primorial(k).
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#68 by Susanna Cuyler at Mon Apr 05 13:43:29 EDT 2021
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