A296369 -id:A296369 - cp's OEIS frontend

A055685 Numbers k such that 2^k == -1 (mod k-1).

2, 6, 66, 378, 1190, 1470, 25806, 58590, 134946, 137346, 170586, 272610, 285390, 420210, 538734, 592410, 618450, 680706, 778806, 1163066, 1520442, 1700946, 2099202, 2831010, 4020030, 4174170, 4516110, 5059890, 5215770

Offset: 1

Robert G. Wilson v, Jun 09 2000

nonn

The sequence is infinite. In fact, even its intersection with A006517 (given by A219037) is infinite.

Chai Wah Wu, Table of n, a(n) for n = 1..1192
Kin Y. Li et al., Solution to Problem 323, Mathematical Excalibur 14(2), 2009, p. 3.

Cf. A064935, A006517, A219037, A296369.

Do[If[PowerMod[2, n, n-1]==n-2, Print[n]], {n, 2, 12900000}]

A055685_list = [n for n in range(2,10**6) if pow(2,n,n-1) == n-2] # Chai Wah Wu, Nov 04 2019

a(n) = A296369(n) + 1.

Edited by Max Alekseyev, Oct 11 2012

Incorrect term 4285390 removed by Chai Wah Wu, Nov 04 2019

A128122 Numbers m such that 2^m == 6 (mod m).

Original entry on oeis.org

1, 2, 10669, 6611474, 43070220513807782

Offset: 1

Alexander Adamchuk, Feb 15 2007

No other terms below 10^17. - Max Alekseyev, Nov 18 2022

A large term: 862*(2^861-3)/281437921287063162726198552345362315020202285185118249390789 (203 digits). - Max Alekseyev, Sep 24 2016

			2 == 6 (mod 1), so 1 is a term;
4 == 6 (mod 2), so 2 is a term.

Joe K. Crump, 2^n mod n
OEIS Wiki, 2^n mod n

Solutions to 2^m == k (mod m): A000079 (k=0),A187787 (k=1/2), A296369 (k=-1/2), A006521 (k=-1), A296370 (k=3/2), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), this sequence (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)

Cf. A015910, A036236.

m = 6; Join[Select[Range[m], Divisible[2^# - m, #] &],
Select[Range[m + 1, 10^6], PowerMod[2, #, #] == m &]] (* Robert Price, Oct 08 2018 *)

1 and 2 added by N. J. A. Sloane, Apr 23 2007

a(5) from Max Alekseyev, Nov 18 2022

A296370 Numbers m such that 2^m == 3/2 (mod m).

Original entry on oeis.org

1, 111481, 465793, 79036177, 1781269903307, 250369632905747, 708229497085909, 15673900819204067

Offset: 1

Max Alekseyev, Dec 11 2017

Equivalently, 2^(m+1) == 3 (mod m).
Also, numbers m such that 2^(m+1) - 2 is a Fermat pseudoprime base 2, i.e., 2^(m+1) - 2 belongs to A015919 and A006935.
Some larger terms (may be not in order): 2338990834231272653581, 341569682872976768698011746141903924998969680637.

OEIS Wiki, 2^n mod n

Solutions to 2^m == k (mod m): this sequence (k=3/2), A187787 (k=1/2), A296369 (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)

Select[Range[10^6], Divisible[2^(# + 1) - 3, #] &] (* Robert Price, Oct 11 2018 *)

a(n) = A296104(n) - 1.

A328230 Numbers m that divide 3^(m + 1) + 1.

Original entry on oeis.org

1, 2, 4, 5, 14, 244, 365, 434, 854, 2294, 3794, 5966, 7874, 10877, 26474, 33914, 117614, 188774, 231434, 284354, 487634, 501038, 589154, 593774, 621674, 755594, 1255814, 1306934, 1642094, 1911194, 2193124, 2434754, 2484674, 2507834, 2621654, 2643494, 3512114, 3759854, 3997574, 4082246

Offset: 1

Juri-Stepan Gerasimov, Oct 08 2019

nonn

Conjecture: For k > 2, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.

Chai Wah Wu, Table of n, a(n) for n = 1..1528 (n = 1..100 from Robert Israel)

Cf. A055685, A277289, A296369.

[n+1: n in [0..5000000] | Modexp(3,n+2,n+1) eq n];

filter:= m -> 3 &^ (m+1) + 1 mod m = 0:
select(filter, [$1..10^7]); # Robert Israel, Oct 30 2019

isok(m) = Mod(3, m)^(m+1) == -1; \\ Michel Marcus, Oct 10 2019

A329168 Numbers m that divide 4^(m + 1) + 1.

Original entry on oeis.org

1, 17, 4097, 7361, 85073, 658529, 3999137, 72281281, 285143057, 628944689, 854112113, 1423081169, 2561019281, 3111576929, 4298117633, 5921265041, 14224884929, 21336998129, 34317377233, 50723421713, 63797137889, 144269032049, 163834314353, 187397322209, 212565453281

Offset: 1

Juri-Stepan Gerasimov, Nov 06 2019

nonn

Conjecture: For k > 1, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.

Conjecture: For k > 1, if f(1) = p(1) equals one of the prime factors of k^2 + 1, p(i+1) equals one of the prime factors of k^(f(i)+1) + 1 greater than p(i), f(i+1) = f(i)*p(i+1), then k^(f(i) + 1) == -1 (mod f(i)) for all integers i. (Especially in this sequence, k = 4, so {f(i)} can be 17, 4097, 4298117633, ...) - Jinyuan Wang, Nov 16 2019

Cf. A055685.

Solutions to k^(m + 1) + 1 == -1 (mod m): A296369 (k = 2), A328230 (k = 3).

[n + 1: n in [0..5000000] | Modexp(4, n + 2, n + 1) eq n ];

isok(m) = Mod(4, m)^(m+1) == -1; \\ Jinyuan Wang, Nov 16 2019

a(13)-a(25) from Giovanni Resta, Nov 08 2019

A319222 Numbers k such that k divides 2^(2k+1) + 1.

Original entry on oeis.org

1, 3, 129, 2537, 51889, 101617, 226873, 270427, 653467, 945667, 1740979, 5819937, 6520987, 9828587, 15452867, 24950857, 51377539, 89519449, 108627601, 135776371, 160126609, 296338873, 310026163, 400431289, 641706243, 643359937, 678257563, 803419697, 902661523, 952431331, 1004273987, 1243893697, 1796055907

Offset: 1

Altug Alkan, Sep 13 2018

nonn

Also, numbers k such that 4^k == -1/2 (mod k) (cf. A296369). - Max Alekseyev, Sep 15 2018

If k is in the sequence, and m is another divisor of 2^(2*k+1)+1 and is coprime to k, then m*k is in the sequence. - Robert Israel, Sep 14 2018

Cf. A006517, A006521, A015950.

filter:= n -> 2 &^ (2*n+1)+1 mod n = 0:
select(filter, [$1..10^7]); # Robert Israel, Sep 14 2018

PARI
```
is_A319222(n) = Mod(2, n)^(2*n+1)==-1;
```

A329222 Numbers m that divide 5^(m + 1) + 1.

Original entry on oeis.org

1, 2, 6, 13, 14, 174, 854, 2694, 78126, 103973, 106694, 121974, 420209, 487374, 1299374, 2174654, 3895094, 4151454, 5842214, 5951129, 6508334, 10637054, 20117894, 24482957, 31999694, 32282053, 32620202, 32872454, 34258454, 52657397, 56114618, 57679082, 65538437, 70782774, 71899526

Offset: 1

Juri-Stepan Gerasimov, Nov 08 2019

nonn

Conjecture: For k > 1, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.

Cf. A055685.

Solutions to k^(m + 1) == -1 (mod m): A296369 (k=2), A328230 (k=3), A329168 (k=4), this sequence (k=5), A329226 (k=6).

[n + 1: n in [0..2000000] | Modexp(5, n + 2, n + 1) eq n];

Select[Range[719*10^5],PowerMod[5,#+1,#]==#-1&] (* Harvey P. Dale, Jul 03 2020 *)

isok(m) = Mod(5, m)^(m+1) == -1; \\ Jinyuan Wang, Nov 16 2019

A329226 Numbers m that divide 6^(m + 1) + 1.

Original entry on oeis.org

1, 37, 16987849, 2416266949, 5995229029, 7193673829, 11465419549, 17783484529, 72155530501, 142013229529, 174523785589, 189282539137, 294183810997, 302690164297, 354613312129, 774557575609, 933821938789, 1407294504937, 1974020768389, 2112969494569, 2878251281401

Offset: 1

Juri-Stepan Gerasimov, Nov 08 2019

nonn

Conjecture: For k > 1, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.

Terms cannot be a multiple of the following primes below 100: 2, 3, 5, 7, 11, 19, 23, 29, 31, 43, 47, 53, 59, 67, 71, 79, 83. - Giovanni Resta, Nov 09 2019

Cf. A055685.

Solutions to k^(m + 1) == -1 (mod m): A296369 (k=2), A328230 (k=3), A329168 (k=4), A329222 (k=5).

[n + 1: n in [0..20000000] | Modexp(6, n + 2, n + 1) eq n];

isok(m) = Mod(6, m)^(m+1) == -1; \\ Jinyuan Wang, Nov 16 2019

a(5)-a(21) from Giovanni Resta, Nov 09 2019

A319538 Numbers k such that k divides 2^(2*k+1) - 1.

Original entry on oeis.org

1, 7, 217, 1057, 3937, 10447, 24601, 32767, 91657, 145337, 279527, 666967, 1412113, 2484247, 2874847, 3124327, 4169137, 4472167, 9526207, 12021439, 16539337, 16646017, 19384207, 20139367, 24639727, 28127137, 28940887, 30583087, 66131279, 68068777, 70694167, 72299857, 72903847, 73498471, 87507049

Offset: 1

Altug Alkan, Sep 22 2018

nonn

Sequence is infinite because 2^A187787(t) - 1 is a term for all t >= 1.

Cf. A187787, A296369, A319222.

Select[Range[10^5], Mod[2^(2 # + 1) - 1, #] == 0 &] (* Michael De Vlieger, Sep 24 2018 *)

PARI
```
isok(n) = Mod(2, n)^(2*n+1)==1;
```

A334634 Numbers m that divide 2^m + 11.

Original entry on oeis.org

1, 13, 16043199041, 91118493923, 28047837698634913

Offset: 1

Max Alekseyev, Sep 10 2020

Equivalently, numbers m such that 2^m == -11 (mod m).

No other terms below 10^17.

OEIS Wiki, 2^n mod n

Solutions to 2^n == k (mod n): A296370 (k=3/2), A187787 (k=1/2), A296369 (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), this sequence (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).

a(5) from Sergey Paramonov, Oct 10 2021

cp's OEIS Frontend

A055685 Numbers k such that 2^k == -1 (mod k-1).

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A128122 Numbers m such that 2^m == 6 (mod m).

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A296370 Numbers m such that 2^m == 3/2 (mod m).

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A328230 Numbers m that divide 3^(m + 1) + 1.

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A329168 Numbers m that divide 4^(m + 1) + 1.

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A319222 Numbers k such that k divides 2^(2k+1) + 1.

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A329222 Numbers m that divide 5^(m + 1) + 1.

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A329226 Numbers m that divide 6^(m + 1) + 1.

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