A005420 -id:A005420 - cp's OEIS frontend

A002326 Multiplicative order of 2 mod 2n+1.

1, 2, 4, 3, 6, 10, 12, 4, 8, 18, 6, 11, 20, 18, 28, 5, 10, 12, 36, 12, 20, 14, 12, 23, 21, 8, 52, 20, 18, 58, 60, 6, 12, 66, 22, 35, 9, 20, 30, 39, 54, 82, 8, 28, 11, 12, 10, 36, 48, 30, 100, 51, 12, 106, 36, 36, 28, 44, 12, 24, 110, 20, 100, 7, 14, 130, 18, 36, 68, 138, 46, 60, 28

Offset: 0

N. J. A. Sloane

In other words, least m > 0 such that 2n+1 divides 2^m-1.
Number of riffle shuffles of 2n+2 cards required to return a deck to initial state. A riffle shuffle replaces a list s(1), s(2), ..., s(m) with s(1), s((i/2)+1), s(2), s((i/2)+2), ... a(1) = 2 because a riffle shuffle of [1, 2, 3, 4] requires 2 iterations [1, 2, 3, 4] -> [1, 3, 2, 4] -> [1, 2, 3, 4] to restore the original order.
Concerning the complexity of computing this sequence, see for example Bach and Shallit, p. 115, exercise 8.
It is not difficult to prove that if 2n+1 is a prime then 2n is a multiple of a(n). But the converse is not true. Indeed, one can prove that a(2^(2t-1))=4t. Thus if n=2^(2t-1), where, for any m > 0, t=2^(m-1) then 2n is a multiple of a(n) while 2n+1 is a Fermat number which, as is well known, is not always a prime. It is an interesting problem to describe all composite numbers for which 2n is divisible by a(n). - Vladimir Shevelev, May 09 2008
For an algorithm of calculation of a(n) see author's comment in A179680. - Vladimir Shevelev, Jul 21 2010
From V. Raman, Sep 18 2012, Dec 10 2012: (Start)
If 2n+1 is prime, then the polynomial (x^(2n+1)+1)/(x+1) factors into 2n/a(n) polynomials of the same degree a(n) over GF(2).
If (x^(2n+1)+1)/(x+1) is irreducible over GF(2), then 2n+1 is prime, and 2 is a primitive root (mod 2n+1) (cf. A001122).
For all n > 0, a(n) is the degree of the largest irreducible polynomial factor for the polynomial (x^(2n+1)+1)/(x+1) over GF(2). (End)
a(n) is a factor of phi(2n+1) (A000010(2n+1)). - Douglas Boffey, Oct 21 2013
Conjecture: if p is an odd prime then a((p^3-1)/2) = p * a((p^2-1)/2). Because otherwise a((p^3-1)/2) < p * a((p^2-1)/2) iff a((p^3-1)/2) = a((p-1)/2) for a prime p. Equivalently p^3 divides 2^(p-1)-1, but no such prime p is known. - Thomas Ordowski, Feb 10 2014
A generalization of the previous conjecture: For each k>=2, if p is an odd prime then a(((p^(k+1))-1)/2) = p * a((p^k-1)/2). Computer testing of this generalized conjecture shows that there is no counterexample for k and p both up to 1000. - Ahmad J. Masad, Oct 17 2020
a(n) = a((N-1)/2), with odd N = 2*n+1 >= 3 (n >= 1), is also the primitive period length of (1/N) in binary notation: (1/N)2 = 0.repeat(a[1]a[2]...a[P(N)]), and P(N) = a((N-1)/2). E.g., N = 11 (n = 5), (1/11)_2 = 0.repeat(0001011101), with P(11) = 10 = a(5). Proof: Use a cyclic shift operation sigma (1 step to the left) on the cycle: sigma((1/N)_2) = .repeat(a[2]...a[P(N)]a[1]). Then one can prove for the composition sigma^[k] (k=0 is the identity map) written back in decimal notation the result (sigma^[k]((1/N)_2))_10 = (1/N)*2^k (mod N). E.g. N = 11, sigma^[2]((1/11)_2) = .repeat(0101110100), written in base 10 as 4/11, etc. Hence P(N) and the order of 2 modulo N coincide. - _Gary W. Adamson and Wolfdieter Lang, Oct 14 2020

			From _Vladimir Shevelev_, Oct 03 2017: (Start)
Our algorithm for the calculation of a(n) in the author's comment in A179680 (see also the Sage program below) could be represented in the form of a "finite continued fraction". For example let n = 8, 2*n+1 = 17. We have
    1 + 17
    ------- + 17
       2
    ------------- + 17
           2
    ------------------- + 17
              2
    -------------------------- = 1
                 32
Here the denominators are the A006519 of the numerators: A006519(1+17) = 2, A006519(9+17) = 2, A006519(13+17) = 2, A006519(15+17) = 32. Summing the exponents of these powers of 2, we obtain the required result: a(8) = 1 + 1 + 1 + 5 = 8. Indeed, we have (((1*32 - 17)*2 - 17)*2 - 17)*2 - 17 = 1. So 32*2*2*2 - 1 == 0 (mod 17), 2^8 - 1 == 0 (mod 17). In the general case, note that all "partial fractions" (which indeed are integers) are odd residues modulo 2*n+1 in the interval [1, 2*n-1]. It is easy to prove that the first 1 appears not later than in the n-th step. (End)

E. Bach and Jeffrey Shallit, Algorithmic Number Theory, I.
T. Folger, "Shuffling Into Hyperspace," Discover, 1991 (vol 12, no 1), pages 66-67.
M. Gardner, "Card Shuffles," Mathematical Carnival chapter 10, pages 123-138. New York: Vintage Books, 1977.
L. Lunelli and M. Lunelli, Tavola di congruenza a^n == 1 mod K per a=2,5,10, Atti Sem. Mat. Fis. Univ. Modena 10 (1960/61), 219-236 (1961).
J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 146, Exer. 21.3
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..10000
Jean-Paul Allouche, Manon Stipulanti, and Jia-Yan Yao, Doubling modulo odd integers, generalizations, and unexpected occurrences, arXiv:2504.17564 [math.NT], 2025.
M. Baake, U. Grimm and J. Nilsson, Scaling of the Thue-Morse diffraction measure, arXiv preprint arXiv:1311.4371 [math-ph], 2013.
D. Bayer and P. Diaconis, Trailing the dovetail shuffle to its lair, Ann. Appl. Prob. 2 (2) (1992) 294-313.
Matthew Brand, Choosing 1 of N with and without lucky numbers, arXiv:1808.07994 [math.NT], 2018.
J. Brillhart, J. S. Lomont and P. Morton, Cyclotomic properties of the Rudin-Shapiro polynomials, J. Reine Angew. Math.288 (1976), 37--65. See Table 2. MR0498479 (58 #16589).
Steve Butler, Persi Diaconis and R. L. Graham, The mathematics of the flip and horseshoe shuffles, arXiv:1412.8533 [math.CO], 2014.
Steve Butler, Persi Diaconis and R. L. Graham, The mathematics of the flip and horseshoe shuffles, The American Mathematical Monthly 123.6 (2016): 542-556.
A. J. C. Cunningham, On Binal Fractions, Math. Gaz., 4 (71) (1908), circa p. 266.
P. Diaconis, R. L. Graham, and W. M. Kantor, The mathematics of perfect shuffles, Adv. Appl. Math. 4 (2) (1983) 175-196
M. J. Gardner and C. A. McMahan, Riffling casino checks, Math. Mag., 50 (1) (1977), 38-41.
S. W. Golomb, Permutations by cutting and shuffling, SIAM Rev., 3 (1961), 293-297.
Jonas Kaiser, On the relationship between the Collatz conjecture and Mersenne prime numbers, arXiv preprint arXiv:1608.00862 [math.GM], 2016.
Torleiv Klove, On covering sets for limited-magnitude errors, Cryptogr. Commun. 8 (3) (2016) 415-433
V. I. Levenshtein, Conflict-avoiding codes and cyclic triple systems [in Russian], Problemy Peredachi Informatsii, 43 (No. 3, 2007), 39-53.
V. I. Levenshtein, Conflict-avoiding codes and cyclic triple systems, Problems of Information Transmission, September 2007, Volume 43, Issue 3, pp 199-212 (translated from Russian)
Yuan-Hsun Lo, Kenneth W. Shum, Wing Shing Wong and Yijin Zhang, Multichannel Conflict-Avoiding Codes of Weights Three and Four, arXiv:2009.11754 [cs.IT], 2020.
Jarkko Peltomäki and Aleksi Saarela, Standard words and solutions of the word equation X_1^2 ... X_n^2 = (X_1 ... X_n)^2, Journal of Combinatorial Theory, Series A (2021) Vol. 178, 105340. See also arXiv:2004.14657 [cs.FL], 2020.
Vladimir Shevelev, Gilberto Garcia-Pulgarin, Juan Miguel Velasquez-Soto and John H. Castillo, Overpseudoprimes, and Mersenne and Fermat numbers as primover numbers, arXiv preprint arXiv:1206:0606 [math.NT], 2012.
Vladimir Shevelev, G. Garcia-Pulgarin, J. M. Velasquez and J. H. Castillo, Overpseudoprimes, and Mersenne and Fermat Numbers as Primover Numbers, J. Integer Seq. 15 (2012) Article 12.7.7.
Eric Weisstein's World of Mathematics, Riffle Shuffle
Eric Weisstein's World of Mathematics, In-Shuffle
Eric Weisstein's World of Mathematics, Out-Shuffle
Eric Weisstein's World of Mathematics, Multiplicative Order
Wikipedia, Riffle Shuffle

Cf. A003571, A003573, A217469, A070667-A070683, A053447, A053451, A292239, A292265.
Cf. A024222, A006694 (number of cyclotomic cosets).
Cf. A014664 (order of 2 mod n-th prime).
Cf. A001122 (primes for which 2 is a primitive root).
Cf. A216838 (primes for which 2 is not a primitive root).
Cf. A000010, A000225, A005420, A006519, A007733, A025192, A048675, A056239, A179680.
Bisections give A274298, A274299.
Partial sums: A359147.

List([0..100],n->OrderMod(2,2*n+1)); # Muniru A Asiru, Feb 01 2019

import Data.List (findIndex)
import Data.Maybe (fromJust)
a002326 n = (+ 1) $ fromJust $
            findIndex ((== 0) . (`mod` (2 * n + 1))) $ tail a000225_list
-- Reinhard Zumkeller, Apr 22 2013

[ 1 ] cat [ Modorder(2, 2*n+1): n in [1..72] ]; // Klaus Brockhaus, Dec 03 2008

a := n -> `if`(n=0, 1, numtheory:-order(2, 2*n+1)):
seq(a(n), n=0..72);

Table[MultiplicativeOrder[2, 2*n + 1], {n, 0, 100}] (* Robert G. Wilson v, Apr 05 2011 *)

a(n)=if(n<0,0,znorder(Mod(2,2*n+1))) /* Michael Somos, Mar 31 2005 */

from sympy import n_order
[n_order(2, 2*n+1) for n in range(73)] # Hermann Stamm-Wilbrandt, Jul 27 2021

# From Peter Luschny, Oct 06 2017: (Start)
[Mod(2,n).multiplicative_order() for n in (0..145) if gcd(n,2) == 1]
# Algorithm from Vladimir Shevelev as described in A179680 and presented in Example.
def A002326VS(n):
    s, m, N = 0, 1, 2*n + 1
    while True:
        k = N + m
        v = valuation(k, 2)
        s += v
        m = k >> v
        if m == 1: break
    return s
[A002326VS(n) for n in (0..72)] # (End)

a((3^n-1)/2) = A025192(n). - Vladimir Shevelev, May 09 2008
Bisection of A007733: a(n) = A007733(2*n+1). - Max Alekseyev, Jun 11 2009
a((b(n)-1)/2) = n for odd n and even n such that b(n/2) != b(n), where b(n) = A005420(n). - Thomas Ordowski, Jan 11 2014
Note that a(2^n-1) = n+1 and a(2^n) = 2*(n+1). - Thomas Ordowski, Jan 16 2014
a(n) = A056239(A292239(n)) = A048675(A292265(n)). - Antti Karttunen, Oct 04 2017

More terms from David W. Wilson, Jan 13 2000

More terms from Benoit Cloitre, Apr 11 2003

A002034 Kempner numbers: smallest positive integer m such that n divides m!.

Original entry on oeis.org

1, 2, 3, 4, 5, 3, 7, 4, 6, 5, 11, 4, 13, 7, 5, 6, 17, 6, 19, 5, 7, 11, 23, 4, 10, 13, 9, 7, 29, 5, 31, 8, 11, 17, 7, 6, 37, 19, 13, 5, 41, 7, 43, 11, 6, 23, 47, 6, 14, 10, 17, 13, 53, 9, 11, 7, 19, 29, 59, 5, 61, 31, 7, 8, 13, 11, 67, 17, 23, 7, 71, 6, 73, 37, 10, 19, 11, 13, 79, 6, 9, 41, 83, 7

Offset: 1

N. J. A. Sloane

Sometimes named after Florentin Smarandache, although studied 60 years earlier by Aubrey Kempner and 35 years before that by Lucas.
Kempner originally defined a(1) to be 0, and there are good reasons to prefer that (see Hungerbühler and Specker), but we shall stay with the by-now traditional value a(1) = 1. - N. J. A. Sloane, Jan 02 2021
Kempner gave an algorithm to compute a(n) from the prime factorization of n. Partial solutions were given earlier by Lucas in 1883 and Neuberg in 1887. - Jonathan Sondow, Dec 23 2004
a(n) is the degree of lowest degree monic polynomial over Z that vanishes identically on the integers mod n [Newman].
Smallest k such that n divides product of k consecutive integers starting with n + 1. - Amarnath Murthy, Oct 26 2002
If m and n are any integers with n > 1, then |e - m/n| > 1/(a(n) + 1)! (see Sondow 2006).
Degree of minimal linear recurrence satisfied by Bell numbers (A000110) read modulo n. [Lunnon et al.] - N. J. A. Sloane, Feb 07 2009

			1! = 1, but clearly 8 does not divide 1.
2! = 2, but 8 does not divide 2.
3! = 6, but 8 does not divide 6.
4! = 24, and 8 does divide 24. Hence a(8) = 4.
However, 9 does not divide 24.
5! = 120, but 9 does not divide 120.
6! = 720, and 9 does divide 720. Hence a(9) = 6.

E. Lucas, Question Nr. 288, Mathesis 3 (1883), 232.
R. Muller, Unsolved problems related to Smarandache Function, Number Theory Publishing Company, Phoenix, AZ, ISBN 1-879585-37-5, 1993.
J. Neuberg, Solutions des questions proposées, Question Nr. 288, Mathesis 7 (1887), 68-69.
Donald J. Newman, A Problem Seminar. Problem 17, Springer-Verlag, 1982.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Florentin Smarandache, A Function in the Number Theory, Analele Univ. Timisoara, Fascicle 1, Vol. XVIII, 1980, pp. 79-88; Smarandache Function J., Vol. 1, No. 1-3 (1990), pp. 3-17.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 2.5.18 on page 77.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Charles Ashbacher, An Introduction to the Smarandache Function, Erhus Univ. Press, Vail, 62 pages, 1995 [WaybackMachine cached version added by _Felix Fröhlich_, Sep 10 2018].
Nicholas John Bizzell-Browning, LIE scales: Composing with scales of linear intervallic expansion, Ph. D. Thesis, Brunel Univ. (UK, 2024). See p. 147.
Constantin Dumitrescu, A brief history of the "Smarandache function", Bull. Pure Appl. Sci. Sec. E, Math., Vol. 12, No. 1-2 (1993), pp. 77-82 [WaybackMachine cached version added by _Felix Fröhlich_, Sep 10 2018].
Constantin Dumitrescu and Vasile Seleacu, The Smarandache Function, Erhus Univ. Press, Vail, 137 pages, 1996 [WaybackMachine cached version added by _Felix Fröhlich_, Sep 10 2018].
Jason Earls, The Smarandache sum of composites between factors function, in Smarandache Notions Journal, Vol. 14, No. 1 (2004), p. 246.
Paul Erdős and Ilias Kastanas, Solution 6674: The smallest factorial that is a multiple of n, Amer. Math. Monthly, Vol. 101, No. 2 (1994) p. 179.
Steven R. Finch, The average value of the Smarandache function, Smarandache Notions Journal, Vol. 10, No. 1-3 (1999), pp. 95-96.
Mats Granvik, Mathematica program to check the conjectured formula, Feb 28 2021.
Norbert Hungerbühler and Ernst Specker, A generalization of the Smarandache Function to several variables, INTEGERS, Vol. 6 (2006), #A23
Aleksandar Ivić, On a problem of Erdős involving the largest prime factor of n, arXiv:math/0311056 [math.NT], 2003-2004.
Aubrey J. Kempner, Miscellanea, The American Mathematical Monthly, Vol. 25, No. 5 (May, 1918), pp. 201-210. (See Section II, Concerning the smallest integer m! divisible by a given integer n.)
G. Kreweras, Sur quelques problèmes relatifs au vote pondéré, [Some problems of weighted voting], Math. Sci. Humaines No. 84 (1983), pp. 45-63.
Xiumei Li and Min Sha, A proof of Sondow's conjecture on the Smarandache function, arXiv preprint arXiv:1907.00370 [math.NT], 2019-2020. Amer. Math. Monthly, 127:10 (2020), 939-943.
W. F. Lunnon et al., Arithmetic properties of Bell numbers to a composite modulus I, Acta Arith., Vol. 35 (1979), pp. 1-16. [From _N. J. A. Sloane_, Feb 07 2009]
Jon Perry, Calculating the Smarandache Numbers, Smarandache Notions Journal, Vol. 14, No. 1 (2004), pp. 124-127.
Project Euler, Problem 549: Divisibility of factorials.
Sebastián Martín Ruiz, A Congruence with Smarandache's Function, Smarandache Notions Journal, Vol. 10, No. 1-3 (1999), pp. 130-132.
József Sándor, The Smarandache minimum and maximum functions, Scientia Magna, Vol. 1, No. 2 (2005), pp. 162-166. See p. 164.
Jonathan Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly, Vol. 113 (2006), pp. 637-641 and Vol. 114 (2007), p. 659.
Jonathan Sondow and Kyle Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, Vol. 517, Amer. Math. Soc., Providence, RI, 2010.
Jonathan Sondow and Eric W. Weisstein, MathWorld: Smarandache Function.
J. Thompson, A propriety (sic) of smarandache function, Abstract 878-11-758, Notices Amer. Math. Soc., Vol. 14 (1993), p. 41. [Annotated scanned copy]
Wikipedia, Kempner function.
Index entries for sequences related to factorial numbers.

Cf. A000142, A001113, A006530, A007672, A046022, A057109, A064759, A084945, A094371, A094372, A094404, A122378, A122379, A122416, A122417, A248937 (Fermi-Dirac analog: use unique representation of n > 1 as a product of distinct terms of A050376).

See A339594-A339596 for higher-dimensional generalizations.

import Data.List (elemIndex)
import Data.Maybe (fromJust)
a002034 1 = 1
a002034 n = fromJust (a092495 n `elemIndex` a000142_list)
-- Reinhard Zumkeller, Aug 24 2011

a:=proc(n) local b: b:=proc(m) if type(m!/n, integer) then m else fi end: [seq(b(m),m=1..100)][1]: end: seq(a(n),n=1..84); # Emeric Deutsch, Aug 01 2005
g:= proc(p,u)
  local i,t;
  t:= 0;
  for i from 1 while t < u do
    t:= t + 1 + padic[ordp](i,p);
  od;
  p*(i-1)
end;
A002034:= x -> max(map(g@op, ifactors(x)[2])); # Robert Israel, Apr 20 2014

Do[m = 1; While[ !IntegerQ[m!/n], m++ ]; Print[m], {n, 85}] (* or for larger n's *)
Kempner[1] := 1; Kempner[n_] := Max[Kempner @@@ FactorInteger[n]]; Kempner[p_, 1] := p; Kempner[p_, alpha_] := Kempner[p, alpha] = Module[{a, k, r, i, nu, k0 = alpha(p - 1)}, i = nu = Floor[Log[p, 1 + k0]]; a[1] = 1; a[n_] := (p^n - 1)/(p - 1); k[nu] = Quotient[alpha, a[nu]]; r[nu] = alpha - k[nu]a[nu]; While[r[i] > 0, k[i - 1] = Quotient[r[i], a[i - 1]]; r[i - 1] = r[i] - k[i - 1]a[i - 1]; i-- ]; k0 + Plus @@ k /@ Range[i, nu]]; Table[ Kempner[n], {n, 85}] (* Eric W. Weisstein, based on a formula of Kempner's, May 17 2004 *)
With[{facts = Range[100]!}, Flatten[Table[Position[facts, ?(Divisible[#, n] &), {1}, 1], {n, 90}]]] (* _Harvey P. Dale, May 24 2013 *)

a(n)=if(n<0,0,s=1; while(s!%n>0,s++); s)

a(n)=my(s=factor(n)[,1],k=s[#s],f=Mod(k!,n));while(f, f*=k++); k \\ Charles R Greathouse IV, Feb 28 2012

valp(n,p)=my(s);while(n\=p,s+=n);s
K(p,e)=if(e<=p,return(e*p));my(t=e*(p-1)\p*p);while(valp(t+=p,p)Charles R Greathouse IV, Jul 30 2013

from sympy import factorial
def a(n):
    m=1
    while True:
        if factorial(m)%n==0: return m
        else: m+=1
[a(n) for n in range(1, 101)] # Indranil Ghosh, Apr 24 2017

from sympy import factorint
def valp(n, p):
    s = 0
    while n: n //= p; s += n
    return s
def K(p, e):
    if e <= p: return e*p
    t = e*(p-1)//p*p
    while valp(t, p) < e: t += p
    return t
def A002034(n):
    return 1 if n == 1 else max(K(p, e) for p, e in factorint(n).items())
print([A002034(n) for n in range(1, 85)]) # Michael S. Branicky, Jun 09 2022 after Charles R Greathouse IV

A000142(a(n)) = A092495(n). - Reinhard Zumkeller, Aug 24 2011
From Joerg Arndt, Jul 14 2012: (Start)
The following identities were given by Kempner (1918):
a(1) = 1.
a(n!) = n.
a(p) = p for p prime.
a(p1 * p2 * ... * pu) = pu if p1 < p2 < ... < pu are distinct primes.
a(p^k) = p * k for p prime and k <= p.
Let n = p1^e1 * p2^e2 * ... * pu^eu be the canonical factorization of n, then a(n) = max( a(p1^e1), a(p2^e2), ..., a(pu^eu) ).
(End)
Clearly a(n) >= P(n), the largest prime factor of n (= A006530). a(n) = P(n) for almost all n (Erdős and Kastanas 1994, Ivic 2004). The exceptions are A057109. a(n) = P(n) if and only if a(n) is prime because if a(n) > P(n) and a(n) were prime, then since n divides a(n)!, n would also divide (a(n)-1)!, contradicting minimality of a(n). - Jonathan Sondow, Jan 10 2005
If p is prime then a(p^k) = k*p for 0 <= k <= p. Hence it appears also that if n = 2^m * p(1)^e(1) * ... * p(r)^e(r) and if there exists b, 1 <= b <= r, such that Max(2 * m + 2, p(i) * e(i), 1 <= i <= r) = p(b) * e(b) with e(b) <= p(b) then a(n) = e(b) * p(b). E.g.: a(2145986896455317997802121296896) = a(2^10 * 3^3 * 7^9 * 11^9 * 13^8) = 13 * 8 = 104, since 8 * 13 = Max (2 * 10 + 2, 3 * 3, 7 * 9, 11 * 9, 13 * 8) and 8 <= 13. - Benoit Cloitre, Sep 01 2002
It appears that a(2^m - 1) is the largest prime factor of 2^m - 1 (A005420).
a(n!) = n for all n > 0 and a(p) = p if p is prime. - Jonathan Sondow, Dec 23 2004
Conjecture: a(n) = 1 + n - Sum_{k=1..n} Sum_{m=1..n} cos(-2*Pi*k/n*m!)/n. Formula verified for the first 500 terms. - Mats Granvik, Feb 26 2021
Limit_{n->oo} (1/n) * Sum_{k=2..n} log(a(k))/log(k) = A084945 (Finch, 1999). - Amiram Eldar, Jul 04 2021

Error in 45th term corrected by David W. Wilson, May 15 1997

A049479 Smallest prime dividing 2^n - 1.

Original entry on oeis.org

3, 7, 3, 31, 3, 127, 3, 7, 3, 23, 3, 8191, 3, 7, 3, 131071, 3, 524287, 3, 7, 3, 47, 3, 31, 3, 7, 3, 233, 3, 2147483647, 3, 7, 3, 31, 3, 223, 3, 7, 3, 13367, 3, 431, 3, 7, 3, 2351, 3, 127, 3, 7, 3, 6361, 3, 23, 3, 7, 3, 179951, 3, 2305843009213693951, 3, 7, 3, 31

Offset: 2

Olivier Gérard

nonn

If p is prime then a(p) == 1 (mod p). Are there composite numbers k such that a(k) == 1 (mod k)? - Thomas Ordowski, Jan 27 2014
Yes, up to 1200, the following composites have the desired property: 169, 221, 323, 611, 779, 793, 923, 1121, 1159. - Michel Marcus, Jan 28 2014
a(n) <= a(lpf(n)) for every n, where lpf(n) = A020639(n). For which n is a(n) < a(lpf(n))? See A236769. - Thomas Ordowski, Jan 30 2014

			a(6)=3 since 2^6 - 1 = 63 = 3^2*7.

Max Alekseyev, Table of n, a(n) for n = 2..1236
Eric Weisstein's World of Mathematics, Mersenne Number.

Cf. A005420.

a = {}; Do[w = 2^n - 1; c = FactorInteger[w]; b = c[[1]][[1]]; AppendTo[a, b], {n, 2, 65}]; a (* Artur Jasinski, Dec 11 2007 *)
FactorInteger[#][[1,1]]&/@(2^Range[2,70]-1) (* Harvey P. Dale, Nov 18 2019 *)

from sympy import factorint
def A049479(n):
    return min(factorint(2**n-1)) # Chai Wah Wu, Jun 03 2019

a(n) > lpf(n) while a(2k) = 3 and a(2k+1) > 2*lpf(2k+1), where lpf(m) = A020639(m). - Thomas Ordowski, Jan 29 2014

For k >= 1, a(2k) = 3, a(6k-3)=7. - Zak Seidov, Mar 21 2014

More terms from Michael Lugo (mlugo(AT)thelabelguy.com), Dec 22 1999
Terms to a(500) in b-file from  T. D. Noe, Dec 06 2006
a(501)-a(1060) in b-file from Michel Marcus, Sep 15 2017
a(1061)-a(1236) in b-file added at the suggestion of Eric Chen by Max Alekseyev, Apr 25 2022

A274906 Largest prime factor of 4^n - 1.

Original entry on oeis.org

3, 5, 7, 17, 31, 13, 127, 257, 73, 41, 683, 241, 8191, 127, 331, 65537, 131071, 109, 524287, 61681, 5419, 2113, 2796203, 673, 4051, 8191, 262657, 15790321, 3033169, 1321, 2147483647, 6700417, 599479, 131071, 122921, 38737, 616318177, 525313, 22366891

Offset: 1

Vincenzo Librandi, Jul 11 2016

nonn

			4^7 - 1 = 16383 = 3*43*127, so a(7) = 127

Max Alekseyev, Table of n, a(n) for n = 1..1128
J. Brillhart et al., Factorizations of b^n +- 1, Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002.

Second bisection of A005420. - Michel Marcus, Jul 13 2016
Cf. largest prime factor of k^n-1: A005420 (k=2), A074477 (k=3), this sequence (k=4), A074479 (k=5), A274907 (k=6), A074249 (k=7), A274908 (k=8), A274909 (k=9), A005422 (k=10), A274910 (k=11).
Cf. A006530, A024036.

[Maximum(PrimeDivisors(4^n-1)): n in [1..40]];

Table[FactorInteger[4^n - 1][[-1, 1]], {n, 40}]

a(n) = A006530(A024036(n)). - Michel Marcus, Jul 11 2016

a(n) = max(A002587(n),A005420(n)). - Max Alekseyev, Apr 25 2022

Terms to a(100) in b-file from Vincenzo Librandi, Jul 13 2016
a(101)-a(603) in b-file from Amiram Eldar, Feb 08 2020
a(604)-a(1128) in b-file from Max Alekseyev, Jul 25 2023, Mar 15 2025

A237043 Numbers n such that 2^n - 1 is not squarefree, but 2^d - 1 is squarefree for every proper divisor d of n.

Original entry on oeis.org

6, 20, 21, 110, 136, 155, 253, 364, 602, 657, 812, 889, 979, 1081

Offset: 1

Charles R Greathouse IV, Feb 02 2014

Primitive elements of A049094: the elements of A049094 are precisely the positive multiples of members of this sequence.
If p^2 divides 2^n - 1 for some odd prime p, then by definition the multiplicative order of 2 mod p^2 divides n. The multiplicative order of 2 mod p^2 is p times the multiplicative order of 2 mod p unless p is a Wieferich prime, in which case the two orders are identical. Hence either p is a Wieferich prime or p*log_2(p+1) <= n. This should allow finding larger members of this sequence. - Charles R Greathouse IV, Feb 04 2014
If n is in the sequence and m>1 then m*n is not in the sequence. Because n is a proper divisor of m*n and 2^n-1 is not squarefree. - Farideh Firoozbakht, Feb 11 2014
a(15) >= 1207. - Max Alekseyev, Sep 28 2015
From Daniel Suteu, Jul 03 2019: (Start)
The following numbers are also in the sequence: {1755, 2265, 2485, 2756, 3081, 3164, 4112, 6757, 8251, 13861, 18533}.
Probably, the following numbers are also terms: {3422, 5253, 6806, 8164, 9316, 11342, 12550, 15025, 15026, 17030, 17404, 17468, 18145, 19670, 19701, 22052}. (End)

Peice Hua, Finite 2-groups having a cyclic or dihedral maximal subgroup and arc-transitive maps, arXiv:2508.05981 [math.GR], 2025. See p. 3.

Cf. A049094, A005420, A065069, A282631, A282632.

Select[Range@ 160, And[AllTrue[2^#2 - 1, SquareFreeQ], ! SquareFreeQ[2^First@ #1 - 1]] & @@ TakeDrop[Divisors@ #, -1] &] (* Michael De Vlieger, Jul 07 2019 *)

default(factor_add_primes, 1);
isA049094(n)=my(f=factor(n>>valuation(n, 2))[, 1], N, o); for(i=1, #f, if(n%(f[i]-1) == 0, return(1))); N=2^n-1; fordiv(n, d, f=factor(2^d-1)[, 1]; for(i=1, #f, if(d==n, return(!issquarefree(N))); o=valuation(N, f[i]); if(o>1, return(1)); N/=f[i]^o))
is(n)=fordiv(n,d,if(isA049094(d),return(d==n))); 0

\\ Simpler but slow
is(n)=fordiv(n,d,if(!issquarefree(2^d-1),return(d==n))); 0

a(14) from Charles R Greathouse IV, Sep 21 2015, following Womack's factorization of 2^991-1.

cp's OEIS Frontend

A359063 Integers k such that A005420(k) = A005420(2*k) = A005420(4*k) where A005420(k) is the largest prime factor of 2^k-1.

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A048857 A005420 prefixed with a 0.

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A128997 indices k such that A063670(k) differs from A005420(k).

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A002326 Multiplicative order of 2 mod 2n+1.

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A002034 Kempner numbers: smallest positive integer m such that n divides m!.

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A049479 Smallest prime dividing 2^n - 1.

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A274906 Largest prime factor of 4^n - 1.

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A359063 Integers k such that A005420(k) = A005420(2k) = A005420(4k) where A005420(k) is the largest prime factor of 2^k-1.