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Revision History for A143027 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Sturdy prime numbers: p such that in binary notation k*p has at least as many 1-bits as p for all k > 0.
(history; published version)
#36 by N. J. A. Sloane at Sat Apr 20 10:01:30 EDT 2024
STATUS

proposed

approved

#35 by Jason Yuen at Sat Mar 30 21:24:35 EDT 2024
STATUS

editing

proposed

#34 by Jason Yuen at Sat Mar 30 21:13:53 EDT 2024
CROSSREFS
Discussion
Sat Mar 30
21:23
Jason Yuen: For the base 10 version, see my draft edits to A181863.
#33 by Jason Yuen at Sat Mar 30 21:10:57 EDT 2024
DATA

2, 3, 5, 7, 17, 31, 73, 89, 127, 257, 1801, 2089, 8191, 65537, 131071, 178481, 262657, 524287, 2099863, 616318177, 2147483647, 4432676798593

COMMENTS

From Jason Yuen, Mar 30 2024: (Start)

For all x>log_2(p), 1+A000120(p-(2^x mod p)) >= A000120(p). This follows from the fact that 2^x+p-(2^x mod p) is a multiple of p.

a(23) > 5*10^12. See a143027_5e12.txt for more details. (End)

LINKS

Jason Yuen, <a href="/A143027/a143027.txt">a143027_5e12.txt</a>. This file shows that a(23) > 5*10^12.

EXTENSIONS

4432676798593 added by Jason Yuen, Mar 30 2024

STATUS

approved

editing

#32 by Michael De Vlieger at Sat Sep 02 17:00:14 EDT 2023
STATUS

reviewed

approved

#31 by Andrew Howroyd at Sat Sep 02 14:09:08 EDT 2023
STATUS

proposed

reviewed

#30 by Jon E. Schoenfield at Sat Sep 02 13:44:08 EDT 2023
STATUS

editing

proposed

#29 by Jon E. Schoenfield at Sat Sep 02 13:43:44 EDT 2023
NAME

Sturdy prime numbers: p such that in binary notation k*p has at least as many 1-bits as p for all k > 0.

COMMENTS

The primes in A125121. This sequence includes the Fermat primes (A019434), Mersenne primes (A000668) and the three known primes in A051154, . It appears that almost all primes are flimsy numbers, A005360.

Clokie et al. verify that the next two sturdy primes after 2099863 are 616318177 and 2147483647. These are all up to 2^{32}. Two additional sturdy primes are 57912614113275649087721 = (2^{83} - 1)/167 and 10350794431055162386718619237468234569 = (2^{131} - 1)/263, but probably there are some in between these and 2147483647. Jeffrey Shallit, Feb 10 2020

STATUS

approved

editing

#28 by Bruno Berselli at Mon Feb 10 06:16:15 EST 2020
STATUS

reviewed

approved

#27 by Joerg Arndt at Mon Feb 10 05:24:59 EST 2020
STATUS

proposed

reviewed