A015973 -id:A015973 - cp's OEIS frontend

A006521 Numbers n such that n divides 2^n + 1.

1, 3, 9, 27, 81, 171, 243, 513, 729, 1539, 2187, 3249, 4617, 6561, 9747, 13203, 13851, 19683, 29241, 39609, 41553, 59049, 61731, 87723, 97641, 118827, 124659, 177147, 185193, 250857, 263169, 292923, 354537, 356481, 373977, 531441, 555579, 752571

Offset: 1

N. J. A. Sloane

nonn

Closed under multiplication: if x and y are terms then so is x*y.
More is true: 1. If n is in the sequence then so is any multiple of n having the same prime factors as n. 2. If n and m are in the sequence then so is lcm(n,m). For a proof see the Bailey-Smyth reference. Elements of the sequence that cannot be generated from smaller elements of the sequence using either of these rules are called *primitive*. The sequence of primitive solutions of n|2^n+1 is A136473. 3. The sequence satisfies various congruences, which enable it to be generated quickly. For instance, every element of this sequence not a power of 3 is divisible either by 171 or 243 or 13203 or 2354697 or 10970073 or 22032887841. See the Bailey-Smyth reference. - Toby Bailey and Christopher J. Smyth, Jan 13 2008
A000051(a(n)) mod a(n) = 0. - Reinhard Zumkeller, Jul 17 2014
The number of terms < 10^n: 3, 5, 9, 15, 25, 40, 68, 114, 188, 309, 518, 851, .... - Robert G. Wilson v, May 03 2015
Also known as Novák numbers after Břetislav Novák who was apparently the first to study this sequence. - Charles R Greathouse IV, Nov 03 2016
Conjecture: if n divides 2^n+1, then (2^n+1)/n is squarefree. Cf. A272361. - Thomas Ordowski, Dec 13 2018
Conjecture: For k > 1, k^m == 1 - k (mod m) has an infinite number of positive solutions. - Juri-Stepan Gerasimov, Sep 29 2019

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 243, p. 68, Ellipses, Paris 2008.
R. Honsberger, Mathematical Gems, M.A.A., 1973, p. 142.
W. Sierpiński, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #16.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Robert G. Wilson v, Table of n, a(n) for n = 1..1064
Toby Bailey and Chris Smyth, Primitive solutions of n|2^n+1.
Alexander Kalmynin, On Novák numbers, arXiv:1611.00417 [math.NT], 2016.
V. Meally, Letter to N. J. A. Sloane, May 1975
C. Smyth, The terms in Lucas Sequences divisible by their indices, JIS 13 (2010) #10.2.4.

Subsequence of A014945.
Cf. A057719 (prime factors), A136473 (primitive n such that n divides 2^n+1).
Cf. A000051, A006517.
Cf. A066807 (the corresponding quotients).
Solutions to k^m == k-1 (mod m): 1 (k = 1), this sequence (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), A328033 (k = 7).
Column k=2 of A333429.

a006521 n = a006521_list !! (n-1)
a006521_list = filter (\x -> a000051 x `mod` x == 0) [1..]
-- Reinhard Zumkeller, Jul 17 2014

[n: n in [1..6*10^5] | (2^n+1) mod n eq 0 ]; // Vincenzo Librandi, Dec 14 2018

for n from 1 to 1000 do if 2^n +1 mod n = 0 then lprint(n); fi; od;
S:=1,3,9,27,81:C:={171,243,13203,2354697,10970073,22032887841}: for c in C do for j from c to 10^8 by 2*c do if 2&^j+1 mod j = 0 then S:=S, j;fi;od;od; S:=op(sort([op({S})])); # Toby Bailey and Christopher J. Smyth, Jan 13 2008

Do[If[PowerMod[2, n, n] + 1 == n, Print[n]], {n, 1, 10^6}]
k = 9; lst = {1, 3}; While[k < 1000000, a = PowerMod[2, k, k]; If[a + 1 == k, AppendTo[lst, k]]; k += 18]; lst (* Robert G. Wilson v, Jul 06 2009 *)
Select[Range[10^5], Divisible[2^# + 1, #] &] (* Robert Price, Oct 11 2018 *)

for(n=1,10^6,if(Mod(2,n)^n==-1,print1(n,", "))); \\ Joerg Arndt, Nov 30 2014

A006521_list = [n for n in range(1,10**6) if pow(2,n,n) == n-1] # Chai Wah Wu, Jul 25 2017

More terms from David W. Wilson, Jul 06 2009

A276671 Positive integers k such that 3^k == 2 (mod k).

Original entry on oeis.org

1, 2929, 9742277641, 23341869101, 15092205901438895, 16311037042239935

Offset: 1

Max Alekseyev, Oct 05 2016

No other terms below 2*10^16. A larger term: 31744873758348589012852097851.

Cf. A015973, A067945, A015949, A055686, A242865, A234535, A234536, A050259.

Join[{1}, Select[Range[10000], PowerMod[3, #, #] == 2 &]] (* Alonso del Arte, Oct 11 2016 *)

isok(n) = Mod(3, n)^n == Mod(2, n); \\ Dmitry Ezhov, Sep 28 2016

Order of terms corrected by Felix Fröhlich, Oct 06 2016

a(5)-a(6) from Sergey Paramonov, Oct 03 2021

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

A277289 Positive integers n such that n | (3^n + 7).

Original entry on oeis.org

1, 2, 4, 5, 8, 25, 44, 4664, 6568, 1353025, 2919526, 5709589, 7827725, 64661225, 85132756, 153872408, 743947534, 34304296003, 38832409867, 40263727492, 1946603375348, 2469908330348, 64471909888247, 274267749806485, 888906849689897, 896501949422459

Offset: 1

Seiichi Manyama, Oct 09 2016

nonn

No other terms below 10^15. - Max Alekseyev, Oct 14 2016

492385451091805616444 is a term.

			3^25 + 7 = 847288609450 = 25 * 33891544378, so 25 is a term.

Solutions to 3^n == k (mod n): A277340 (k=-11), this sequence (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==-7; \\ Joerg Arndt, Oct 09 2016

A277289_list = [1,2,4,5]+[n for n in range(6,10**6) if pow(3,n,n)==n-7] # Chai Wah Wu, Oct 12 2016

a(17) from Joerg Arndt, Oct 09 2016
a(18)-a(20) from Chai Wah Wu, Oct 12 2016
a(21)-a(26) 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

A327943 Numbers m that divide 6^m + 5.

Original entry on oeis.org

1, 11, 341, 186787, 8607491, 9791567, 11703131, 14320387, 50168819, 952168003, 71654478989, 1328490399527

Offset: 1

Juri-Stepan Gerasimov, Sep 30 2019

Conjecture: For k > 1, k^m == 1 - k (mod m) has infinitely many positive solutions.
Also includes 11834972807906571233 = 31*381773316384082943. - Robert Israel, Oct 03 2019
a(13) > 10^15. - Max Alekseyev, Nov 10 2022

Solutions to k^m == 1-k (mod m): A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), this sequence (k = 6).

Cf. A245318, A277288, A015891, A253209.

[1] cat [n: n in [1..10^8] | Modexp(6, n, n) + 5 eq n];

Join[{1},Select[Range[98*10^5],PowerMod[6,#,#]==#-5&]] (* The program generates the first six terms of the sequence. To generate more, increase the Range constant but the program may take a long time to run. *) (* Harvey P. Dale, Feb 05 2022 *)

a(11) from Giovanni Resta, Oct 02 2019

a(12) from Max Alekseyev, Nov 10 2022

A327840 Numbers m that divide 4^m + 3.

Original entry on oeis.org

1, 7, 16387, 4509253, 24265177, 42673920001, 103949349763, 12939780075073

Offset: 1

Juri-Stepan Gerasimov, Sep 27 2019

Number of solutions < 10^9 to k^n == k-1 (mod n): 1 (if k = 1), 188 (if k = 2, see A006521), 5 (if k = 3, see A015973), 5 (if k = 4, see this sequence), 5 (if k = 5), 10 (if k = 6), 10 (if k = 7), 7 (if k = 8), 5 (if k = 9), 8 (if k = 10), 11 (if k = 11), 8 (if k = 12), 9 (if k = 13), 4 (if k = 14), 3 (if k = 15), 6 (if k = 16), 7 (if k = 17), 7 (if k = 18), ...

a(9) > 10^15. - Max Alekseyev, Nov 10 2022

Solutions to k^n == 1-k (mod n): A006521 (k = 2), A015973 (k = 3), this sequence (k = 4), A123047 (k = 5), A327943 (k = 6).
Solutions to 4^n == k (mod n): A000079 (k = 0), A015950 (k = -1), A014945 (k = 1), A130421 (k = 2), this sequence (k = -3), A130422 (k = 3).
Cf. A015940, A253208.

[1] cat [n: n in [1..10^8] | Modexp(4,n,n) + 3 eq n];

Select[Range[10^7], IntegerQ[(PowerMod[4, #, # ]+3)/# ]&] (* Metin Sariyar, Sep 28 2019 *)

is(n)=Mod(4,n)^n==-3 \\ Charles R Greathouse IV, Sep 29 2019

a(6)-a(7) from Giovanni Resta, Sep 29 2019

a(8) from Max Alekseyev, Nov 10 2022

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A006521 Numbers n such that n divides 2^n + 1.

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A276671 Positive integers k such that 3^k == 2 (mod k).

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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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A277289 Positive integers n such that n | (3^n + 7).

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

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