

A006521


Numbers n such that n divides 2^n + 1.
(Formerly M2806)


43



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
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OFFSET

1,2


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 BaileySmyth 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 n2^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 BaileySmyth 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.  JuriStepan 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 n2^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.


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 *)


PROG

(Haskell)
a006521 n = a006521_list !! (n1)
a006521_list = filter (\x > a000051 x `mod` x == 0) [1..]
 Reinhard Zumkeller, Jul 17 2014
(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) == n1] # Chai Wah Wu, Jul 25 2017
(MAGMA) [n: n in [1..6*10^5]  (2^n+1) mod n eq 0 ]; // Vincenzo Librandi, Dec 14 2018


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 == k1 (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.
Sequence in context: A271351 A036143 A248960 * A289257 A014953 A274627
Adjacent sequences: A006518 A006519 A006520 * A006522 A006523 A006524


KEYWORD

nonn


AUTHOR

N. J. A. Sloane


EXTENSIONS

More terms from David W. Wilson, Jul 06 2009


STATUS

approved



