The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A337157

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A337157

a(n) is the smallest m such that A054024(m) = prime(n), where A054024(m) is A000203(m) mod m, or -1 if there is no such m.
(history; published version)

#19 by Alois P. Heinz at Fri Jan 29 14:21:36 EST 2021

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proposed

approved

#18 by Michel Marcus at Fri Jan 29 12:54:33 EST 2021

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editing

proposed

#17 by Michel Marcus at Fri Jan 29 12:54:20 EST 2021

EXAMPLE

For prime(5) = 11, 11-1=3+7 with 3*7 = 21, so a(115) <= 21 and a(115) = 21.

For prime(6) = 13, 13-1=5+7 with 5*7 = 35, so a(136) <= 35 but sigma(27) = 40 == 13 (mod 27), hence a(136)=27.

#16 by Michel Marcus at Fri Jan 29 12:52:48 EST 2021

LINKS

Michel Marcus, <a href="/A337157/b337157.txt">Table of n, a(n) for n = 1..4000</a>

EXAMPLE

For prime(6) = 13, 13-1=5+7 with 5*7 = 35, so a(1113) <= 35 but sigma(27) = 40 == 13 (mod 27), hence a(13)=27.

STATUS

approved

editing

#15 by Alois P. Heinz at Fri Jan 29 05:17:26 EST 2021

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proposed

approved

#14 by Michel Marcus at Fri Jan 29 03:49:16 EST 2021

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editing

proposed

#13 by Michel Marcus at Fri Jan 29 03:49:13 EST 2021

EXAMPLE

For prime(5) = 11, by Goldbach conjecture 11-1=3+7 with 3*7 = 21, so a(11) <= 21 and a(11) = 21.

For prime(6) = 13, by Goldbach conjecture 13-1=5+7 with 5*7 = 35 , so a(11) <= 35 but sigma(27) = 40 == 13 (mod 27), hence a(13)=27.

STATUS

approved

editing

#12 by N. J. A. Sloane at Thu Jan 28 21:51:14 EST 2021

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proposed

approved

#11 by Bernard Schott at Thu Jan 28 10:49:51 EST 2021

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editing

proposed

Discussion

Thu Jan 28

11:02

Michel Marcus: by Goldbach conjecture 11-1=3+7   ?   rather by the magical power of addition ??

11:44

Bernard Schott: Je ne comprends pas, sorry.

13:32

Hugo Pfoertner: Long matches with A177900.

#10 by Bernard Schott at Thu Jan 28 10:48:59 EST 2021

COMMENTS

Now, when n = 3, not no k is known such that sigma(k) == 5 (mod k)? (StopEnd)

STATUS

proposed

editing

Discussion

Thu Jan 28

10:49

Bernard Schott: 10:32 both corrected, thanks.