A296369 Numbers m such that 2^m == -1/2 (mod m).
1, 5, 65, 377, 1189, 1469, 25805, 58589, 134945, 137345, 170585, 272609, 285389, 420209, 538733, 592409, 618449, 680705, 778805, 1163065, 1520441, 1700945, 2099201, 2831009, 4020029, 4174169, 4516109, 5059889, 5215769
Offset: 1
Keywords
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..1192
- OEIS Wiki, 2^n mod n
Crossrefs
Solutions to 2^m == k (mod m): A296370 (k=3/2), A187787 (k=1/2), this sequence (k=-1/2), A000079 (k=0), A006521 (k=-1), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), A128122 (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12)
Programs
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Mathematica
Select[Range[10^5], Divisible[2^(# + 1) + 1, #] &] (* Robert Price, Oct 11 2018 *)
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Python
A296369_list = [n for n in range(1,10**6) if pow(2,n+1,n) == n-1] # Chai Wah Wu, Nov 04 2019
Formula
a(n) = A055685(n) - 1.
Extensions
Incorrect term 4285389 removed by Chai Wah Wu, Nov 04 2019
Comments