login

Revision History for A033981

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

Showing entries 1-10 | older changes
Integers k such that 2^k == 7 (mod k).
(history; published version)
#25 by Michel Marcus at Sun Sep 30 03:13:08 EDT 2018
STATUS

reviewed

approved

#24 by Joerg Arndt at Sun Sep 30 02:49:35 EDT 2018
STATUS

proposed

reviewed

#23 by Joerg Arndt at Sun Sep 30 02:49:32 EDT 2018
STATUS

editing

proposed

#22 by Joerg Arndt at Sun Sep 30 02:49:30 EDT 2018
MATHEMATICA

Select[Range[m + 1, 10^7], PowerMod[2, #, #] == m &]] (* Robert Price, Sep 26 2018 *)

EXTENSIONS

Improved Mathematica program by Robert Price, Sep 29 2018

STATUS

proposed

editing

#21 by Robert Price at Sat Sep 29 21:07:34 EDT 2018
STATUS

editing

proposed

#20 by Robert Price at Sat Sep 29 21:07:16 EDT 2018
MATHEMATICA

m = 7; Join[Select[Range[m], Divisible[2^# - m, #] &],

Select[Range[m + 1, 10^7], PowerMod[2, #, #] == 7 m &] ] (* Robert Price, Sep 26 2018 *)

EXTENSIONS

Improved Mathematica program by Robert Price, Sep 29 2018

STATUS

approved

editing

#19 by N. J. A. Sloane at Fri Sep 28 22:36:02 EDT 2018
STATUS

proposed

approved

#18 by Michel Marcus at Thu Sep 27 00:22:44 EDT 2018
STATUS

editing

proposed

Discussion
Thu Sep 27
12:00
Robert Price: Michel, I must be missing something basic here.
For k=1: 2^1 = 2; Mod[2,1]=0.
For k=5: 2^5=32; Mod[32,5]=2.
For k=25; 2^25=33554432; Mod[33554432,25]=7.

For integers m and n, Mod[m,n] lies between 0 and n-1. 
So, for n<7, Mod[m,n] < 6, and cannot be 7.

Where am I going wrong?
Fri Sep 28
22:35
N. J. A. Sloane: Bob,  remember that x == y (mod z) means precisely that z divides x-y.  So for any integers p and q, p == q mod 1 since 1 always divides p-q.  You were probably thinking of a different condition, namely  x (mod y) = u (mod v)
#17 by Michel Marcus at Thu Sep 27 00:22:09 EDT 2018
PROG

(PARI) isok(n) = Mod(2, n)^n == 7; \\ Michel Marcus, Sep 27 2018

STATUS

proposed

editing

Discussion
Thu Sep 27
00:22
Michel Marcus: with this pari script , I do get 1, 5, 25, 1727,
#16 by Jon E. Schoenfield at Wed Sep 26 23:39:14 EDT 2018
STATUS

editing

proposed