cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A270833 Numbers n > 1 where all prime factors are Wieferich primes, i.e., terms of A001220.

Original entry on oeis.org

1093, 3511, 1194649, 3837523, 12327121, 1305751357, 4194412639, 13473543253, 43280521831, 1427186233201, 4584493014427, 14726582775529, 47305610361283, 151957912148641, 5010850864768711, 16096154973653197, 51705032124882319, 166089997978464613
Offset: 1

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Author

Felix Fröhlich, Mar 23 2016

Keywords

Comments

The prime terms are Wieferich primes.
All "Wieferich pseudoprimes", if any exist, are in the sequence (see second comment in A240719).

Examples

			4194412639 = 1093^2 * 3511. All prime factors are Wieferich primes, so 4194412639 is a term of the sequence.
		

Crossrefs

Programs

  • Mathematica
    Take[#, 19] &@ Rest@ Sort@ Map[1093^First@ # 3511^Last@ # &, Tuples[Range[0, 6], 2]] (* Michael De Vlieger, Mar 24 2016 *)
  • PARI
    is(n) = if(n==1, return(0)); my(f=factor(n)[, 1]); for(k=1, #f, if(Mod(2, f[k]^2)^(f[k]-1)!=1, return(0))); return(1)
    
  • PARI
    /* The following program is significantly faster; valid up to (p^x * q^y) < b, where b is the upper search bound for Wieferich primes (approximately 5*10^17 as of Mar 23 2016, see PrimeGrid PRPNet server statistics) */
    my(p=1093, q=3511, v=vector(0), w=vector(1)); for(x=0, 4, for(y=0, 4, w[1]=p^x*q^y; v=concat(v, w))); vecextract(vecsort(v,,8), "2..25")

A305184 Multiplicative order of 2 (mod p^2), where p is the n-th Wieferich prime (A001220).

Original entry on oeis.org

364, 1755
Offset: 1

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Author

Felix Fröhlich, May 30 2018

Keywords

Comments

Meissner discovered the congruence 2^364 == 1 (mod 1093^2) and thus proved that 1093 is a Wieferich prime, i.e., a term of A001220 (cf. Meissner, 1913).
Later, Beeger discovered the congruence 2^1755 == 1 (mod 3511^2) and proved that 3511 is also a Wieferich prime (cf. Beeger, 1922).
Let b(n) = (A001220(n)-1)/a(n). Then b(1) = 3 and b(2) = 2.
From the fact that a(1) and a(2) are composite it follows that A001220(1) = 1093 and A001220(2) = 3511 do not divide any terms of A001348 (cf. Dobson).
Curiously, both 364 and 1755 are repdigits in some base. 364 = 444 in base 9 and 1755 = 3333 in base 8. Compare this with Dobson's observation that 1092 and 3510 are 444 in base 16 and 6666 in base 8, respectively (cf. Dobson).

Crossrefs

Programs

  • PARI
    forprime(p=1, , if(Mod(2, p^2)^(p-1)==1, print1(znorder(Mod(2, p^2)), ", ")))

Formula

a(n) = A014664(A000720(A001220(n))) = A243905(A000720(A001220(n))). [Corrected by Jianing Song, Sep 20 2019]

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387
Offset: 1

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Author

Keywords

Comments

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

Examples

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
  • H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
  • 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).

Crossrefs

Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

Programs

  • GAP
    List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018
  • Haskell
    import Data.Ratio ((%), numerator)
    a001008 = numerator . sum . map (1 %) . enumFromTo 1
    a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016
    
  • Maple
    A001008 := proc(n)
        add(1/k,k=1..n) ;
        numer(%) ;
    end proc:
    seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016
  • Mathematica
    Table[Numerator[HarmonicNumber[n]], {n, 30}]
    (* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
    Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
    Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)
  • PARI
    A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009
    
  • PARI
    H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019
    
  • Python
    from sympy import Integer
    [sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017
    
  • Sage
    def harmonic(a, b):  # See the F. Johansson link.
        if b - a == 1:
            return 1, a
        m = (a+b)//2
        p, q = harmonic(a,m)
        r, s = harmonic(m,b)
        return p*s+q*r, q*s
    def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
    [A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012
    

Formula

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Extensions

Edited by Max Alekseyev, Oct 21 2011
Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A019434 Fermat primes: primes of the form 2^(2^k) + 1, for some k >= 0.

Original entry on oeis.org

3, 5, 17, 257, 65537
Offset: 1

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Comments

It is conjectured that there are only 5 terms. Currently it has been shown that 2^(2^k) + 1 is composite for 5 <= k <= 32 (see Eric Weisstein's Fermat Primes link). - Dmitry Kamenetsky, Sep 28 2008
No Fermat prime is a Brazilian number. So Fermat primes belong to A220627. For proof see Proposition 3 page 36 in "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
This sequence and A001220 are disjoint (see "Other theorems about Fermat numbers" in Wikipedia link). - Felix Fröhlich, Sep 07 2014
Numbers n > 1 such that n * 2^(n-2) divides (n-1)! + 2^(n-1). - Thomas Ordowski, Jan 15 2015
From Jaroslav Krizek, Mar 17 2016: (Start)
Primes p such that phi(p) = 2*phi(p-1); primes from A171271.
Primes p such that sigma(p-1) = 2p - 3.
Primes p such that sigma(p-1) = 2*sigma(p) - 5.
For n > 1, a(n) = primes p such that p = 4 * phi((p-1) / 2) + 1.
Subsequence of A256444 and A256439.
Conjectures:
1) primes p such that phi(p) = 2*phi(p-2).
2) primes p such that phi(p) = 2*phi(p-1) = 2*phi(p-2).
3) primes p such that p = sigma(phi(p-2)) + 2.
4) primes p such that phi(p-1) + 1 divides p + 1.
5) numbers n such that sigma(n-1) = 2*sigma(n) - 5. (End)
Odd primes p such that ratio of the form (the number of nonnegative m < p such that m^q == m (mod p))/(the number of nonnegative m < p such that -m^q == m (mod p)) is a divisor of p for all nonnegative q. - Juri-Stepan Gerasimov, Oct 13 2020
Numbers n such that tau(n)*(number of distinct ratio (the number of nonnegative m < n such that m^q == m (mod n))/(the number of nonnegative m < n such that -m^q == m (mod n))) for nonnegative q is equal to 4. - Juri-Stepan Gerasimov, Oct 22 2020
The numbers of primitive roots for the five known terms are 1, 2, 8, 128, 32768. - Gary W. Adamson, Jan 13 2022
Prime numbers such that every residue is either a primitive root or a quadratic residue. - Keith Backman, Jul 11 2022
If there are only 5 Fermat primes, then there are only 31 odd order groups which have a 2-group automorphism group. See the Miles Englezou link for a proof. - Miles Englezou, Mar 10 2025

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 137-141, 197.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • C. F. Gauss, Disquisitiones Arithmeticae, Yale, 1965; see Table 1, p. 458.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • Richard K. Guy, Unsolved Problems in Number Theory, A3.
  • Hardy and Wright, An Introduction to the Theory of Numbers, bottom of page 18 in the sixth edition, gives an heuristic argument that this sequence is finite.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 7, 70.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Crossrefs

Subsequence of A147545 and of A334101. Cf. also A333788, A334092.
Cf. A045544.

Programs

Formula

a(n+1) = A180024(A049084(a(n))). - Reinhard Zumkeller, Aug 08 2010
a(n) = 1 + A001146(n-1), if 1 <= n <= 5. - Omar E. Pol, Jun 08 2018

A001567 Fermat pseudoprimes to base 2, also called Sarrus numbers or Poulet numbers.

Original entry on oeis.org

341, 561, 645, 1105, 1387, 1729, 1905, 2047, 2465, 2701, 2821, 3277, 4033, 4369, 4371, 4681, 5461, 6601, 7957, 8321, 8481, 8911, 10261, 10585, 11305, 12801, 13741, 13747, 13981, 14491, 15709, 15841, 16705, 18705, 18721, 19951, 23001, 23377, 25761, 29341
Offset: 1

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A composite number n is a Fermat pseudoprime to base b if and only if b^(n-1) == 1 (mod n). Fermat pseudoprimes to base 2 are often simply called pseudoprimes.
Theorem: If both numbers q and 2q - 1 are primes (q is in the sequence A005382) and n = q*(2q-1) then 2^(n-1) == 1 (mod n) (n is in the sequence) if and only if q is of the form 12k + 1. The sequence 2701, 18721, 49141, 104653, 226801, 665281, 721801, ... is related. This subsequence is also a subsequence of the sequences A005937 and A020137. - Farideh Firoozbakht, Sep 15 2006
Also, composite odd numbers n such that n divides 2^n - 2 (cf. A006935). It is known that all primes p divide 2^(p-1) - 1. There are only two known numbers n such that n^2 divides 2^(n-1) - 1, A001220(n) = {1093, 3511} Wieferich primes p: p^2 divides 2^(p-1) - 1. 1093^2 and 3511^2 are the terms of a(n). - Alexander Adamchuk, Nov 06 2006
An odd composite number 2n + 1 is in the sequence if and only if multiplicative order of 2 (mod 2n+1) divides 2n. - Ray Chandler, May 26 2008
The Carmichael numbers A002997 are a subset of this sequence. For the Sarrus numbers which are not Carmichael numbers, see A153508. - Artur Jasinski, Dec 28 2008
An odd number n greater than 1 is a Fermat pseudoprime to base b if and only if ((n-1)! - 1)*b^(n-1) == -1 (mod n). - Arkadiusz Wesolowski, Aug 20 2012
The name "Sarrus numbers" is after Frédéric Sarrus, who, in 1819, discovered that 341 is a counterexample to the "Chinese hypothesis" that n is prime if and only if 2^n is congruent to 2 (mod n). - Alonso del Arte, Apr 28 2013
The name "Poulet numbers" appears in Monografie Matematyczne 42 from 1932, apparently because Poulet in 1928 produced a list of these numbers (cf. Miller, 1975). - Felix Fröhlich, Aug 18 2014
Numbers n > 2 such that (n-1)! + 2^(n-1) == 1 (mod n). Composite numbers n such that (n-2)^(n-1) == 1 (mod n). - Thomas Ordowski, Feb 20 2016
The only twin pseudoprimes up to 10^13 are 4369, 4371. - Thomas Ordowski, Feb 12 2016
Theorem (A. Rotkiewicz, 1965): (2^p-1)*(2^q-1) is a pseudoprime if and only if p*q is a pseudoprime, where p,q are different primes. - Thomas Ordowski, Mar 31 2016
Theorem (W. Sierpiński, 1947): if n is a pseudoprime (odd or even), then 2^n-1 is a pseudoprime. - Thomas Ordowski, Apr 01 2016
If 2^n-1 is a pseudoprime, then n is a prime or a pseudoprime (odd or even). - Thomas Ordowski, Sep 05 2016
From Amiram Eldar, Jun 19 2021, Apr 21 2024: (Start)
Erdős (1950) called these numbers "almost primes".
According to Erdős (1949) and Piza (1954), the term "pseudoprime" was coined by D. H. Lehmer.
Named after the French mathematician Pierre de Fermat (1607-1665), or, alternatively, after the Belgian mathematician Paul Poulet (1887-1946), or, the French mathematician Pierre Frédéric Sarrus (1798-1861). (End)
If m is a term of this sequence, then (m-1)/ord(2,m) >= 5, where ord(a,m) is the multiplicative order of a modulo m; see my link below. Actually, it seems that we always have (m-1)/ord(2,m) >= 9. - Jianing Song, Nov 04 2024

References

  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 80.
  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section A12, pp. 44-50.
  • George P. Loweke, The Lore of Prime Numbers. New York: Vantage Press (1982), p. 22.
  • Øystein Ore, Number Theory and Its History, McGraw-Hill, 1948.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 88-92.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 145.

Crossrefs

Cf. A001220 = Wieferich primes p: p^2 divides 2^(p-1) - 1.
Cf. A005935, A005936, A005937, A005938, A005939, A020136-A020228 (pseudoprimes to bases 3 through 100).

Programs

  • Magma
    [n: n in [3..3*10^4 by 2] | IsOne(Modexp(2,n-1,n)) and not IsPrime(n)]; // Bruno Berselli, Jan 17 2013
  • Maple
    select(t -> not isprime(t) and 2 &^(t-1) mod t = 1, [seq(i,i=3..10^5,2)]); # Robert Israel, Feb 18 2016
  • Mathematica
    Select[Range[3,30000,2], ! PrimeQ[ # ] && PowerMod[2, (# - 1), # ] == 1 &] (* Farideh Firoozbakht, Sep 15 2006 *)
  • PARI
    q=1;vector(50,i,until( !isprime(q) & (1<<(q-1)-1)%q == 0, q+=2);q) \\ M. F. Hasler, May 04 2007
    
  • PARI
    is_A001567(n)={Mod(2,n)^(n-1)==1 && !isprime(n) && n>1}  \\ M. F. Hasler, Oct 07 2012, updated to current PARI syntax and to exclude even pseudoprimes on Mar 01 2019
    

Formula

Sum_{n>=1} 1/a(n) is in the interval (0.015260, 33) (Bayless and Kinlaw, 2017). The upper bound was reduced to 0.0911 by Kinlaw (2023). - Amiram Eldar, Oct 15 2020, Feb 24 2024

Extensions

More terms from David W. Wilson, Aug 15 1996
Replacement of broken geocities link by Jason G. Wurtzel, Sep 05 2010
"Poulet numbers" added to name by Joerg Arndt, Aug 18 2014

A049094 Numbers m such that 2^m - 1 is divisible by a square > 1.

Original entry on oeis.org

6, 12, 18, 20, 21, 24, 30, 36, 40, 42, 48, 54, 60, 63, 66, 72, 78, 80, 84, 90, 96, 100, 102, 105, 108, 110, 114, 120, 126, 132, 136, 138, 140, 144, 147, 150, 155, 156, 160, 162, 168, 174, 180, 186, 189, 192, 198, 200, 204, 210, 216, 220, 222, 228, 231, 234, 240
Offset: 1

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Conjecture: 2^n-1 is squarefree iff gcd(n,2^n-1)=1. If true, the conjecture would imply that Mersenne numbers (A001348) are squarefree. - Vladeta Jovovic, Apr 12 2002. But the conjecture is not true: counterexamples are n = 364 and n = 1755, i.e., gcd(364,2^364-1) = 1 and (2^364-1) mod 1093^2 = 0 and gcd(1755,2^1755-1) = 1 and (2^1755-1) mod 3511^2 = 0, cf. A001220. - Vladeta Jovovic, Nov 01 2005. The conjecture is true with assumption that n is not a multiple of A002326((q-1)/2), where q is a Wieferich prime A001220. - Thomas Ordowski, Nov 17 2015
If d|n and 2^d-1 is not squarefree, then 2^n-1 cannot be squarefree. For example, if 6 is in the sequence then 6*d is also. - Enrique Pérez Herrero, Feb 28 2009
If p(p-1)|n then p^2|2^n-1, where p is an odd prime. - Thomas Ordowski, Jan 22 2014
The primitive elements of this sequence are A237043. - Charles R Greathouse IV, Feb 05 2014
Dilcher & Ericksen prove that this sequence is exactly the set of numbers divisible by either t(p)p for a Wieferich prime p>2 or t(p) for a non-Wieferich prime p, where t(p) is the order of 2 modulo p (see Proposition 3.1). - Kellen Myers, Jun 09 2015
If d^2 divides 2^n-1 then d divides n, where n is not a multiple of 364, 1755, ...; i.e., A002326((q-1)/2) for Wieferich primes q, A001220. - Thomas Ordowski, Nov 15 2015
(1, 2^n-1, 2^n) is an abc triple iff 2^n-1 is not squarefree. - William Hu, Jul 04 2024

Examples

			a(2)=12 because 2^12 - 1 = 4095 = 5*(3^2)*7*13 is divisible by a square.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, A3.

Crossrefs

Programs

  • Magma
    [n: n in [1..250] | not IsSquarefree(2^n-1)]; // Vincenzo Librandi, Jul 14 2015
  • Maple
    N:= 250:
    B:= Vector(N):
    for n from 1 to N do
      if B[n] <> 1 then
        F:= ifactors(2^n-1,easy)[2];
        if max(seq(t[2],t=F)) > 1 or (hastype(F,symbol)
                and not numtheory:-issqrfree(2^n-1)) then
           B[[seq(n*k,k=1..floor(N/n))]]:= 1;
        fi
      fi;
    od:
    select(t -> B[t]=1, [$1..N]); # Robert Israel, Nov 17 2015
  • Mathematica
    Select[Range[240], !SquareFreeQ[2^#-1]&] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2011 *)
  • PARI
    default(factor_add_primes,1);
    is(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)) \\ Charles R Greathouse IV, Feb 02 2014
    
  • PARI
    is(n)=!issquarefree(2^n-1) \\ Charles R Greathouse IV, Feb 04 2014
    

Extensions

More terms from Vladeta Jovovic, Apr 12 2002
Definition corrected by Jonathan Sondow, Jun 29 2010

A014127 Mirimanoff primes: primes p such that p^2 divides 3^(p-1) - 1.

Original entry on oeis.org

11, 1006003
Offset: 1

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Comments

Dorais and Klyve proved that there are no further terms up to 9.7*10^14.
These primes are so named after the celebrated result of Mirimanoff in 1910 (see below) that for a failure of the first case of Fermat's Last Theorem, the exponent p must satisfy the criterion stated in the definition. Lerch (see below) showed that these primes also divide the numerator of the harmonic number H(floor(p/3)). This is analogous to the fact that Wieferich primes (A001220) divide the numerator of the harmonic number H((p-1)/2). - John Blythe Dobson, Mar 02 2014, Apr 09 2015
The prime 1006003 was apparently discovered by K. E. Kloss (cf. Kloss, 1965) according to various sources. - Felix Fröhlich, Dec 08 2020
If there is no term other than 11 and 1006003, then the only solution (a, w, x, y, z) to the diophantine equation a^w + a^x = 3^y + 3^z is (5, 1, 1, 2, 3) (cf. Scott, Styer, 2006, Lemma 12). - Felix Fröhlich, Dec 10 2020
Named after the Russian mathematician Dmitry Semionovitch Mirimanoff (1861-1945). - Amiram Eldar, Jun 10 2021

References

  • Paulo Ribenboim, 13 Lectures on Fermat's Last Theorem, Springer, 1979, pp. 23, 152-153.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 233.
  • Alf van der Poorten, Notes on Fermat's Last Theorem, Wiley, 1996, p. 21.

Crossrefs

Sequences "primes p such that p^2 divides X^(p-1)-1": A001220 (X=2), A123692 (X=5), A212583 (X=6), A123693 (X=7), A045616 (X=10), A111027 (X=12), A128667 (X=13), A234810 (X=14), A242741 (X=15), A128668 (X=17), A244260 (X=18), A090968 (X=19), A242982 (X=20), A298951 (X=22), A128669 (X=23), A306255 (X=26), A306256 (X=30).

Programs

  • Mathematica
    Select[Prime[Range[1000000]], PowerMod[3, # - 1, #^2] == 1 &] (* Robert Price, May 17 2019 *)
  • PARI
    N=10^9; default(primelimit,N);
    forprime(n=2,N,if(Mod(3,n^2)^(n-1)==1,print1(n,", ")));
    \\ Joerg Arndt, May 01 2013
    
  • Python
    from sympy import prime
    from gmpy2 import powmod
    A014127_list = [p for p in (prime(n) for n in range(1,10**7)) if powmod(3,p-1,p*p) == 1] # Chai Wah Wu, Dec 03 2014

Extensions

Edited by Max Alekseyev, Oct 20 2010
Updated by Max Alekseyev, Jan 29 2012

A039951 a(n) is the smallest prime p such that p^2 divides n^(p-1) - 1.

Original entry on oeis.org

2, 1093, 11, 1093, 2, 66161, 5, 3, 2, 3, 71, 2693, 2, 29, 29131, 1093, 2, 5, 3, 281, 2, 13, 13, 5, 2, 3, 11, 3, 2, 7, 7, 5, 2, 46145917691, 3, 66161, 2, 17, 8039, 11, 2, 23, 5, 3, 2, 3
Offset: 1

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Comments

a(n^k) <= a(n) for any n,k > 1.
a(n) is currently unknown for n in {47, 72, 186, 187, 200, 203, 222, 231, 304, 311, 335, 355, 435, 454, 546, 554, 610, 639, 662, 760, 772, 798, 808, 812, 858, 860, 871, 983, 986, ...}. - Richard Fischer, Jul 15 2021
a(47) > 1.4*10^14, a(72) > 1.4*10^14 (see Fischer's tables).
For all nonnegative integers n and k, a(n^(n^k)) = a(n) (see Puzzle 762 in the links). Also a(n) = 3 if and only if mod(n, 36) is in the set {8, 10, 19, 26, 28, 35}. - Farideh Firoozbakht and Jahangeer Kholdi, Nov 01 2014

Crossrefs

Programs

  • Mathematica
    Table[p = 2; While[! Divisible[n^(p - 1) - 1, p^2], p = NextPrime@ p]; p, {n, 33}] (* Michael De Vlieger, Nov 24 2016 *)
    f[n_] := Block[{p = 2}, While[ PowerMod[n, p - 1, p^2] != 1, p = NextPrime@ p]; p]; Array[f, 33] (* Robert G. Wilson v, Jul 18 2018 *)
  • PARI
    a(n)={forprime(p=2, oo, if(Mod(n, p^2)^(p-1)==1, return(p))); oo} \\ Felix Fröhlich, Jul 24 2014

Formula

a(4k+1) = 2.
a(n) = A096082(n) for all n > 1 that are not of the form 4k+1. Note that A096082 begins with n = 2. [Corrected and clarified by Jonathan Sondow, Jun 17-18 2010]

Extensions

a(34)-a(46) from Helmut Richter (richter(AT)lrz.de), May 17 2004
Entry revised by N. J. A. Sloane, Nov 30 2006
Edited by Max Alekseyev, Oct 06, Oct 09 2009
Edited and updated by Max Alekseyev, Jan 29 2012

A049093 Numbers n such that 2^n - 1 is squarefree.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 61, 62, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92
Offset: 1

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Keywords

Comments

Numbers n such that gcd(n, 2^n - 1) = 1 and n is not a multiple of A002326((q - 1)/2), where q is a Wieferich prime A001220. - Thomas Ordowski, Nov 21 2015
If n is in the sequence, then so are all divisors of n. - Robert Israel, Nov 23 2015

Examples

			a(7) = 8 because 2^8 - 1 = 255 = 3 * 5 * 17 is squarefree.
		

Crossrefs

Complement of A049094.

Programs

  • Magma
    [n: n in [1..100] | IsSquarefree(2^n-1)]; // Vincenzo Librandi, Nov 22 2015
  • Maple
    N:= 400: # to get all terms <= N
    # This relies on the fact that the first N+1 members of A000225 have all been factored
    # without any further Wieferich primes being found.
    V:= Vector(N,1):
    V[364 * [$1..N/364]]:= 0:
    V[1755 * [$1..N/1755]]:= 0:
    for n from 2 to N do
    if V[n] = 0 then next fi;
    if igcd(n, 2 &^n - 1 mod n) > 1 then
      V[n * [$1..N/n]]:= 0
    fi;
    od:
    select(t -> V[t] = 1, [$1..N]); # Robert Israel, Nov 23 2015
  • Mathematica
    Select[Range@ 92, SquareFreeQ[2^# - 1] &] (* Michael De Vlieger, Nov 21 2015 *)
  • PARI
    isok(n) = issquarefree(2^n - 1); \\ Michel Marcus, Dec 19 2013
    

Extensions

Terms a(73)-a(910) in b-file from Max Alekseyev, Nov 15 2014, Sep 28 2015

A007663 Fermat quotients: (2^(p-1)-1)/p, where p=prime(n).

Original entry on oeis.org

1, 3, 9, 93, 315, 3855, 13797, 182361, 9256395, 34636833, 1857283155, 26817356775, 102280151421, 1497207322929, 84973577874915, 4885260612740877, 18900352534538475, 1101298153654301589, 16628050996019877513, 64689951820132126215, 3825714619033636628817
Offset: 2

Views

Author

N. J. A. Sloane, Sep 19 1994

Keywords

Comments

The only terms that are squares are a(2) = 1 and a(4) = 9. - Nick Hobson, May 20 2007
From Jonathan Sondow, Jul 19 2010: (Start)
a(n) == 0 (mod 3) if n > 2, since p = prime(n) > 3
and 0 = (-1)^(p-1)-1 == 2^(p-1)-1 (mod 3). (End)
p is in A001220 if and only if p | (2^(p-1)-1)/p, i.e., a(n) is divisible by prime(n). - Felix Fröhlich, Jun 20 2014
In general, every prime p that is 1 mod q-1 will create a numerator that is 0 mod q via Fermat's Little Theorem, meaning every p with this property (except q) will have a Fermat quotient divisible by q. - Roderick MacPhee, May 12 2017

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See pp. 47, 308.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 105.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, NY, 1986, p. 70.

Crossrefs

Programs

Formula

From Alexander Adamchuk, Oct 01 2006: (Start)
a(n) = 3*A096060(n) for n > 2.
a(n) = 3*A001045(prime(n)-1)/prime(n) for n > 1. (End)
a(n) = Sum_{i=0..(p-3)/2} 2^i*(p-i-2)!/((i+1)!*(p-2*(i+1))!) where p = prime(n), for n >= 2. - Vladimir Pletser, Jan 26 2023
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