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#32 by Joerg Arndt at Mon Feb 19 01:57:54 EST 2024
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#31 by Paolo P. Lava at Sun Feb 18 15:22:31 EST 2024
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| COMMENTS
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Numerator(B_m) mod denominator(B_m) = 5. - Paolo P. Lava, Mar 30 2015
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approved
editing
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#30 by Michel Marcus at Sun Jun 07 01:24:36 EDT 2020
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#29 by Joerg Arndt at Sun Jun 07 01:19:13 EDT 2020
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#28 by Petros Hadjicostas at Sat Jun 06 17:10:19 EDT 2020
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#27 by Petros Hadjicostas at Sat Jun 06 17:08:40 EDT 2020
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#26 by Petros Hadjicostas at Sat Jun 06 17:07:57 EDT 2020
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| EXAMPLE
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The numbers m = 10, 50 belong to the list because B_10 = 5/66 and B_50 = 495057205241079648212477525/66. - Petros Hadjicostas, Jun 06 2020
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#25 by Petros Hadjicostas at Sat Jun 06 17:01:27 EDT 2020
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Numbers m such that the Bernoulli number B_{n} _m has denominator 66.
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| COMMENTS
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From the von Staudt-Clausen theorem, denominator(B_2n) = _{2*m}) = product of primes p such that (p-1)|2n2*m.
Numerator(B_{n}) _m) mod denominator(B_{n}) = _m) = 5. - Paolo P. Lava, Mar 30 2015
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| LINKS
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Wikipedia, <a href="https://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem">Von Staudt-Clausen theorem</a>.
<a href="/index/Be#Bernoulli">Index entries for sequences related to Bernoulli numbers.</</a>>.
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| CROSSREFS
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Cf. A045979, A051222, A051225-A051229. Equals 2*A051229, A051226, A051227, A051228.
Equals 2*A051229.
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approved
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#24 by Jon E. Schoenfield at Mon Jan 28 23:53:36 EST 2019
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#23 by Jon E. Schoenfield at Mon Jan 28 23:53:32 EST 2019
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From the Vonvon Staudt-Clausen theorem, denominator(B_2n) = product of primes p such that (p-1)|2n.
Numerator(B_{n}) mod Denominatordenominator(B_{n}) = 5. - Paolo P. Lava, Mar 30 2015
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| LINKS
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T. D. Noe, <a href="/A051230/b051230.txt">Table of n, a(n) for n= = 1..1000</a>
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approved
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