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

Author

N. J. A. Sloane

Keywords

• nonn

Comments

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

References

• 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).

Links

• 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.

Crossrefs

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.

Programs

Haskell

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

Magma

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

Maple

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

Mathematica

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 *)

PARI

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

Python

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

Extensions

More terms from David W. Wilson, Jul 06 2009