A277289 -id:A277289 - cp's OEIS frontend

A015973 Positive integers n such that n | (3^n + 2).

1, 5, 77, 278377, 3697489, 219596687717, 56865169816619

Offset: 1

No other terms below 10^15. Some larger term: 3142423971953435020522506484187. - Max Alekseyev, Aug 04 2011

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), this sequence (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), A277126 (k=7), A277274 (k=11).

a(1)=1 prepended and a(6)-a(7) added by Max Alekseyev, Aug 04 2011

A276740 Numbers n such that 3^n == 5 (mod n).

Original entry on oeis.org

1, 2, 4, 76, 418, 1102, 4687, 7637, 139183, 2543923, 1614895738, 9083990938, 23149317409, 497240757797, 4447730232523, 16000967516764, 65262766108619, 141644055557882

Offset: 1

Dmitry Ezhov, Sep 16 2016

No other terms below 10^15. Some larger terms: 194995887252090239, 2185052151122686482926861593785262. - Max Alekseyev, Oct 13 2016

			3 == 5 (mod 1), so 1 is a term;
9 == 5 (mod 2), so 2 is a term.

Cf. A066601.

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), this sequence (k=5), A277628 (k=6), A277126 (k=7), A277630 (k=8), A277274 (k=11).

Select[Range[10^7], PowerMod[3, #, #] == Mod[5, #] &] (* Michael De Vlieger, Sep 26 2016 *)

isok(n) = Mod(3, n)^n == Mod(5, n); \\ Michel Marcus, Sep 17 2016

A276740_list = [1,2,4]+[n for n in range(5,10**6) if pow(3,n,n) == 5] # Chai Wah Wu, Oct 04 2016

a(11)-a(13) from Chai Wah Wu, Oct 05 2016
a(14) from Lars Blomberg, Oct 12 2016
a(15)-a(18) from Max Alekseyev, Oct 13 2016
a(12) was missing Robert G. Wilson v, Oct 19 2016

A277126 Positive integers n such that 3^n == 7 (mod n).

Original entry on oeis.org

1, 2, 295, 883438, 252027511, 7469046275, 26782373099, 53191768475, 55246802458, 819613658855, 893727887879978

Offset: 1

Seiichi Manyama, Oct 06 2016

No other terms below 10^15. A larger term: 9135884036634915191945452485106476242. - Max Alekseyev, Oct 12 2016

Terms are not divisible by 127 (Alekseyev 2016).

			3 == 7 mod 1, so 1 is a term;
9 == 7 mod 2, so 2 is a term.

M. A. Alekseyev. "Problem 4101". Crux Mathematicorum 42:1 (2016), 28.

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), this sequence (k=7), A277274 (k=11).

isok(n) = Mod(3, n)^n == 7; \\ Michel Marcus, Oct 06 2016

a(5) from Joerg Arndt, Oct 06 2016

a(6)-a(11) from Max Alekseyev, Oct 12 2016

A277288 Positive integers k such that k divides 3^k + 5.

Original entry on oeis.org

1, 2, 14, 1978, 38209, 4782974, 9581014, 244330711, 365496202, 1661392258, 116084432414, 288504187458218, 490179448388654, 802245996685561

Offset: 1

Seiichi Manyama, Oct 09 2016

No other terms below 10^15. Some larger terms: 79854828136468902206, 3518556634988844968631084847788071912030455376274045370172567094578. - Max Alekseyev, Oct 14 2016

			3^14 + 5 = 4782974 = 14 * 341641, so 14 is a term.

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), this sequence (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), A277126 (k=7), A277274 (k=11).

is(n)=Mod(3,n)^n==-5; \\ Joerg Arndt, Oct 09 2016

A277288_list = [1,2]+[n for n in range(3,10**6) if pow(3,n,n)==n-5] # Chai Wah Wu, Oct 09 2016

def A277288_list(search_limit):
    n, t, r = 1, Integer(3), [1]
    while n < search_limit:
        n += 1
        t *= 3
        if n.divides(t+5): r.append(n)
    return r # Peter Luschny, Oct 10 2016

a(9) from Joerg Arndt, Oct 09 2016
a(10) from Chai Wah Wu, Oct 09 2016
a(11)-a(14) from Max Alekseyev, Oct 14 2016

A277274 Positive integers n such that 3^n == 11 (mod n).

Original entry on oeis.org

1, 2, 1162, 1692934, 3851999, 274422823, 14543645261, 492230729674, 773046873382, 13010754158393, 31446154470014, 583396812890467, 598371102650063

Offset: 1

Seiichi Manyama, Oct 08 2016

No other terms below 10^15. Some larger terms: 38726095838775708310162, 2682806839696008709567739369. - Max Alekseyev, Oct 12 2016

			3 == 11 mod 1, so 1 is a term.
9 == 11 mod 2, so 2 is a term.

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), A277126 (k=7), this sequence (k=11).

k = 3; lst = {1, 2}; While[k < 12000000001, If[ PowerMod[3, k, k] == 11, AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Oct 08 2016 *)

a(7)-a(13) from Max Alekseyev, Oct 12 2016

A277340 Positive integers n such that n | (3^n + 11).

Original entry on oeis.org

1, 2, 4, 7, 10, 92, 1099, 29530, 281473, 657892, 3313964, 9816013, 18669155396, 94849225930, 358676424226, 957439868543, 1586504109310, 41431374800470, 241469610359708, 256165266592379

Offset: 1

Seiichi Manyama, Oct 09 2016

nonn

No other terms below 10^15. Some larger terms: 9151612250553176993, 1401778935853533028413047652833, 5645122353966835994338815444821661584288016927879134, 313*(3^626+11)/6562567821545333606830 (280 digits). - Max Alekseyev, Oct 14 2016

			3^10 + 11 = 59060 = 10 * 5906, so 10 is a term.

Solutions to 3^n == k (mod n): this sequence (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), A277126 (k=7), A277274 (k=11).

is(n)=Mod(3,n)^n==-11; \\ Joerg Arndt, Oct 10 2016

A277340_list = [1,2,4,7,10]+[n for n in range(11,10**6) if pow(3,n,n)==n-11] # Chai Wah Wu, Oct 11 2016

a(13)-a(14) from Chai Wah Wu, Oct 12 2016

a(15)-a(20) from Max Alekseyev, Oct 14 2016

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

A277628 Positive integers n such that 3^n == 6 (mod n).

Original entry on oeis.org

1, 3, 21, 936340943, 10460353197, 9374251222371, 23326283250291, 615790788171551

Offset: 1

Dmitry Ezhov, Oct 24 2016

No other terms below 10^15. - Max Alekseyev, Sep 12 2017

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), this sequence (k=6), A277126 (k=7), A277630 (k=8), A277274 (k=11).

PARI
```
isok(n) = Mod(3, n)^n == Mod(6, n);
```

a(6)-a(8) from Max Alekseyev, Sep 12 2017

A277630 Positive integers n such that 3^n == 8 (mod n).

Original entry on oeis.org

1, 5, 2352527, 193841707, 17126009179703, 380211619942943

Offset: 1

Dmitry Ezhov, Oct 24 2016

No other terms below 10^15. - Max Alekseyev, Sep 13 2017

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), A277628 (k=6), A277126 (k=7), this sequence (k=8), A277274 (k=11).

PARI
```
isok(n) = Mod(3, n)^n == Mod(8, n);
```

a(5)-a(6) established by Max Alekseyev, Sep 13 2017

cp's OEIS Frontend

A015973 Positive integers n such that n | (3^n + 2).

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A276740 Numbers n such that 3^n == 5 (mod n).

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A277126 Positive integers n such that 3^n == 7 (mod n).

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A277288 Positive integers k such that k divides 3^k + 5.

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A277274 Positive integers n such that 3^n == 11 (mod n).

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A277340 Positive integers n such that n | (3^n + 11).

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

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A277628 Positive integers n such that 3^n == 6 (mod n).

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