A327840 -id:A327840 - 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

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

A328033 Numbers m that divide 7^m + 6.

Original entry on oeis.org

1, 13, 793, 1943, 150341, 183793, 2348789, 26052527, 27982637, 54789869, 1588344433, 3928538029, 8115802931, 16936276919, 17786709541, 47778790033, 973094452518029

Offset: 1

Juri-Stepan Gerasimov, Oct 02 2019

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

Also includes 2073273696480171732497. - Giovanni Resta, Oct 04 2019

Solutions to k^m == 1-k (mod m): A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), this sequence (k = 7), A327468 (k = 8).

Cf. A253210 (7^n + 6).

[1] cat [n: n in [1..10^8] | Modexp(7, n, n) + 6 eq n];

a(12)-a(16) from Giovanni Resta, Oct 04 2019

a(17) from Max Alekseyev, Feb 07 2024

A327468 Numbers m that divide 8^m + 7.

Original entry on oeis.org

1, 3, 5, 25, 519, 290502305, 821808425, 979288025, 982989263, 25783323897, 27771237541, 31045665345, 65130752425, 3708883906025, 15079242289703, 973336048301405

Offset: 1

Juri-Stepan Gerasimov, Oct 04 2019

Conjecture: For k > 1, k^m == 1-k (mod m) has an infinite number of positive solutions.
Integer m not divisible by 3 is a term if and only if 3m is a term of A240941. - Max Alekseyev, Feb 07 2024
Also terms 930486448009391617725 and 21036656390681764555645540794214294457925. - Giovanni Resta, Oct 04 2019
Other terms 71245661271703622047, 7093208961478946798805, 7807963392818324067361574236385. - Max Alekseyev, Feb 07 2024

Solutions to k^m == 1-k (mod m): 1 (k = 1), A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), A328033 (k = 7), this sequence (k = 8).

Cf. A240941, A253211.

[m: m in [1..7] | (8^m + 7) mod m eq 0] cat [m: m in [8..10^8] | Modexp(8, m, m) + 7 eq m]; // Jon E. Schoenfield, Oct 05 2019

isok(n) = Mod(8, n)^n==-7; \\ Michel Marcus, Oct 05 2019

a(10)-a(13) from Giovanni Resta, Oct 04 2019

a(14)-a(16) from Max Alekseyev, Feb 07 2024

A328138 Numbers m that divide 9^m + 8.

Original entry on oeis.org

1, 17, 803, 1241, 20264753, 28214180783393, 228454543831049

Offset: 1

Juri-Stepan Gerasimov, Oct 04 2019

Conjecture: For n > 1, k^n == 1-k (mod n) has an infinite number of positive solutions.
No term can be a multiple of 2, 3, 5, 7, or 13. Also 4879573990210017348077958628152400091281634488825721395187 is a term. - Giovanni Resta, Oct 07 2019
Also 6788776064693081883870036833 is a term. - Max Alekseyev, Dec 27 2024

Subsequence of A008364.
Solutions to k^m == k-1 (mod m): 1 (k = 1), A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), A328033 (k = 7), A327468 (k = 8), this sequence (k = 9).
Cf. A253212 (9^n + 8).

[1] cat [n: n in [1..10^8] | Modexp(9, n, n) + 8 eq n];

isok(n) = Mod(9, n)^n==-8; \\ Michel Marcus, Oct 05 2019

a(n) > 15n for large enough n. (Surely the sequence grows superlinearly, but I can't prove it.) - Charles R Greathouse IV, Dec 27 2024

a(7) from Giovanni Resta confirmed and a(6) added by Max Alekseyev, Dec 27 2024

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

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A327943 Numbers m that divide 6^m + 5.

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A328033 Numbers m that divide 7^m + 6.

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A327468 Numbers m that divide 8^m + 7.

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A328138 Numbers m that divide 9^m + 8.

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