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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A002321 Mertens's function: Sum_{k=1..n} mu(k), where mu is the Moebius function A008683.

Original entry on oeis.org

1, 0, -1, -1, -2, -1, -2, -2, -2, -1, -2, -2, -3, -2, -1, -1, -2, -2, -3, -3, -2, -1, -2, -2, -2, -1, -1, -1, -2, -3, -4, -4, -3, -2, -1, -1, -2, -1, 0, 0, -1, -2, -3, -3, -3, -2, -3, -3, -3, -3, -2, -2, -3, -3, -2, -2, -1, 0, -1, -1, -2, -1, -1, -1, 0, -1, -2, -2, -1, -2, -3, -3, -4, -3, -3, -3, -2, -3, -4, -4, -4
Offset: 1

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Author

Keywords

Comments

Partial sums of the Moebius function A008683.
Also determinant of n X n (0,1) matrix defined by A(i,j)=1 if j=1 or i divides j.
The first positive value of Mertens's function for n > 1 is for n = 94. The graph seems to show a negative bias for the Mertens function which is eerily similar to the Chebyshev bias (described in A156749 and A156709). The purported bias seems to be empirically approximated to - (6 / Pi^2) * (sqrt(n) / 4) (by looking at the graph) (see MathOverflow link, May 28 2012) where 6 / Pi^2 = 1 / zeta(2) is the asymptotic density of squarefree numbers (the squareful numbers having Moebius mu of 0). This would be a growth pattern akin to the Chebyshev bias. - Daniel Forgues, Jan 23 2011
All integers appear infinitely often in this sequence. - Charles R Greathouse IV, Aug 06 2012
Soundararajan proves that, on the Riemann Hypothesis, a(n) << sqrt(n) exp(sqrt(log n)*(log log n)^14), sharpening the well-known equivalence. - Charles R Greathouse IV, Jul 17 2015
Balazard & De Roton improve this (on the Riemann Hypothesis) to a(n) << sqrt(n) exp(sqrt(log n)*(log log n)^k) for any k > 5/2, where the implied constant in the Vinogradov symbol depends on k. Saha & Sankaranarayanan reduce the exponent to 5/4 on additional hypotheses. - Charles R Greathouse IV, Feb 02 2023

Examples

			G.f. = x - x^3 - x^4 - 2*x^5 - x^6 - 2*x^7 - 2*x^8 - 2*x^9 - x^10 - 2*x^11 - 2*x^12 - ...
		

References

  • E. Landau, Vorlesungen über Zahlentheorie, Chelsea, NY, Vol. 2, p. 157.
  • D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, pp. 7-10.
  • F. Mertens, "Über eine zahlentheoretische Funktion", Akademie Wissenschaftlicher Wien Mathematik-Naturlich Kleine Sitzungsber, IIa 106, (1897), p. 761-830.
  • D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section VI.1.
  • Biswajyoti Saha and Ayyadurai Sankaranarayanan, On estimates of the Mertens function, International Journal of Number Theory, Vol. 15, No. 02 (2019), pp. 327-337.
  • 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).
  • J. von zur Gathen and J. Gerhard, Modern Computer Algebra, Cambridge, 1999, see p. 482.

Crossrefs

First column of A134541.
First column of A179287.

Programs

  • Haskell
    import Data.List (genericIndex)
    a002321 n = genericIndex a002321_list (n-1)
    a002321_list = scanl1 (+) a008683_list
    -- Reinhard Zumkeller, Jul 14 2014, Dec 26 2012
    
  • Magma
    [&+[MoebiusMu(k): k in [1..n]]: n in [1..81]]; // Bruno Berselli, Jul 12 2021
  • Maple
    with(numtheory); A002321 := n->add(mobius(k),k=1..n);
  • Mathematica
    Rest[ FoldList[ #1+#2&, 0, Array[ MoebiusMu, 100 ] ] ]
    Accumulate[Array[MoebiusMu,100]] (* Harvey P. Dale, May 11 2011 *)
  • PARI
    a(n) = sum( k=1, n, moebius(k))
    
  • PARI
    a(n) = if( n<1, 0, matdet( matrix(n, n, i, j, j==1 || 0==j%i)))
    
  • PARI
    a(n)=my(s); forsquarefree(k=1,n, s+=moebius(k)); s \\ Charles R Greathouse IV, Jan 08 2018
    
  • Python
    from sympy import mobius
    def M(n): return sum(mobius(k) for k in range(1,n + 1))
    print([M(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 18 2017
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A002321(n):
        if n == 0:
            return 0
        c, j = n, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*A002321(k1)
            j, k1 = j2, n//j2
        return j-c # Chai Wah Wu, Mar 30 2021
    

Formula

Assuming the Riemann hypothesis, a(n) = O(x^(1/2 + eps)) for every eps > 0 (Littlewood - see Landau p. 161).
Lambert series: Sum_{n >= 1} a(n)*(x^n/(1-x^n)-x^(n+1)/(1-x^(n+1))) = x and -1/x. - Mats Granvik, Sep 09 2010 and Sep 23 2010
a(n)+2 = A192763(n,1) for n>1, and A192763(1,k) for k>1 (conjecture). - Mats Granvik, Jul 10 2011
Sum_{k = 1..n} a(floor(n/k)) = 1. - David W. Wilson, Feb 27 2012
a(n) = Sum_{k = 1..n} tau_{-2}(k) * floor(n/k), where tau_{-2} is A007427. - Enrique Pérez Herrero, Jan 23 2013
a(n) = Sum_{k=1..A002088(n)} exp(2*Pi*i*A038566(k)/A038567(k-1)) where i is the imaginary unit. - Eric Desbiaux, Jul 31 2014
Schoenfeld proves that |a(n)| < 5.3*n/(log n)^(10/9) for n > 1. - Charles R Greathouse IV, Jan 17 2018
G.f. A(x) satisfies: A(x) = (1/(1 - x)) * (x - Sum_{k>=2} (1 - x^k) * A(x^k)). - Ilya Gutkovskiy, Aug 11 2021

A156749 For all numbers k(n) congruent to -1 or +1 (mod 4) starting with k(n) = {3,5,7,9,11,...}, a(k(n)) is incremented by the congruence (mod 4) if k(n) is prime and by 0 if k(n) is composite.

Original entry on oeis.org

-1, 0, -1, -1, -2, -1, -1, 0, -1, -1, -2, -2, -2, -1, -2, -2, -2, -1, -1, 0, -1, -1, -2, -2, -2, -1, -1, -1, -2, -1, -1, -1, -2, -2, -3, -2, -2, -2, -3, -3, -4, -4, -4, -3, -3, -3, -3, -2, -2, -1, -2, -2, -3, -2, -2, -1, -1, -1, -1, -1, -1, -1, -2, -2, -3, -3, -3, -2, -3, -3
Offset: 1

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Author

Daniel Forgues, Feb 14 2009

Keywords

Comments

The fact that a(k(n)) is predominantly negative exhibits the Chebyshev Bias (where the congruences that are not quadratic residues generally lead in the prime number races, at least for "small" integers, over the congruences that are quadratic residues).
This bias seems caused (among other causes?) by the presence of all those squares (even powers) coprime to 4 taking away opportunities for primes to appear in the quadratic residue class +1 (mod 4), while the non-quadratic residue class -1 (mod 4) is squarefree.
The density of squares congruent to +1 (mod 4) is 1/(4*sqrt(k(n))) since 1/2 of squares are congruent to +1 (mod 4), while the density of primes in either residue class -1 or +1 (mod 4) is 1/(phi(4)*log(k(n))), with phi(4) = 2.
Here 1 is quadratic residue mod 4, but 3 (or equivalently -1) is quadratic non-residue mod 4. All the even powers (included in the squares) map congruences {-1, +1} to {+1, +1} respectively and so contribute to the bias, whereas all the odd powers map {-1, +1} to {-1, +1} respectively and so do not contribute to the bias.
One would then expect the ratio of this bias, if caused exclusively by the even powers, relative to the number of primes in either congruences to asymptotically tend towards to 0 as k(n) increases (since 1/(4*sqrt(k(n))) is o(1/(phi(4)*log(k(n))))).
The persistence or not of such bias in absolute value then does not contradict The Prime Number Theorem for Arithmetic Progressions (Dirichlet) which states that the asymptotic (relative) ratio of the count of prime numbers in each congruence class coprime to m tends to 1 in the limit towards infinity. (Cf. 'Prime Number Races' link below.)
Also, even if this bias grows in absolute value, it is expected to be drowned out (albeit very slowly) by the increasing fluctuations in the number of primes in each congruence class coprime to 4 since, assuming the truth of the Riemann Hypothesis, their maximum amplitude would be, with x standing for k(n) in our case, h(x) = O(sqrt(x)*log(x)) <= C*sqrt(x)*log(x) in absolute value which gives relative fluctuations of order h(x)/x = O(log(x)/sqrt(x)) <= C*log(x)/sqrt(x) in the densities of primes pi(x, {4, 1})/x and pi(x, {4, 3})/x in either congruence class.
Since 1/(4*sqrt(x)) is o(log(x)/sqrt(x)) the bias will eventually be overwhelmed by the "pink noise or nearly 1/f noise" corresponding to the fluctuations in the prime densities in either congruence class. The falsehood of the Riemann Hypothesis would imply even greater fluctuations since the RH corresponds to the minimal h(x).
We get pink noise or nearly 1/f noise if we consider the prime density fluctuations of pi(x, {4, k})/x as an amplitude spectrum over x (with a power density spectrum of (C*log(x)/sqrt(x))^2 = ((C*log(x))^2)/x and see x as the frequency f. This power density spectrum is then nearly 1/x and would have nearly equal energy (although slowly increasing as (C*log(x))^2) for each octave of x. (Cf. 'Prime Numbers: A Computational Perspective' link below.)
Among the positive integers k(n) up to 100000 that are congruent to -1 or +1 (mod 4) [indexed from n = 1 to 49999, with k(n) = 4*ceiling(n/2) + (-1)^n], a tie is attained or maintained, with a(k(n)) = 0, for only 34 integers and that bias favoring the non-quadratic residue class -1 (mod 4) gets violated only once, i.e., a(k(n)) = +1, for index n = 13430 (corresponding to the prime k(n) = 26861 congruent to +1 (mod 4) since n is even) where the congruence +1 leads once!

References

  • Richard E. Crandall and Carl Pomerance, Prime Numbers: A Computational Perspective

Crossrefs

Programs

  • Mathematica
    Table[Which[!PrimeQ[2*n+1], 0, Mod[2*n+1, 4] == 1, 1, True, -1], {n, 1, 100}] // Accumulate (* Jean-François Alcover, Dec 09 2014 *)

Formula

a(n) = -A066520(2*n+1) = A066339(2*n+1) - A066490(2*n+1). - Jonathan Sondow, May 17 2013

Extensions

Edited by Daniel Forgues, Mar 01 2009, Mar 29 2009

A156706 For all numbers k(n) congruent to +1 or -1 (mod 6) starting with k(n) = {5,7,11,13,...}, a(k(n)) is the congruence (mod 6) if k(n) is prime and 0 if k(n) is composite.

Original entry on oeis.org

-1, 1, -1, 1, -1, 1, -1, 0, -1, 1, 0, 1, -1, 1, -1, 0, -1, 0, -1, 1, 0, 1, -1, 1, 0, 1, -1, 0, -1, 0, 0, 1, -1, 1, -1, 1, -1, 0, 0, 0, 0, 1, -1, 0, -1, 1, 0, 0, -1, 1, 0, 1, 0, 1, -1, 0, -1, 0, -1, 1, 0, 0, -1, 1, -1, 1, 0, 0, 0, 1, 0, 0, 0, 1, -1, 1, -1, 0, -1, 1, 0, 0, -1, 0, -1, 0, -1, 0, -1, 1, 0
Offset: 1

Views

Author

Daniel Forgues, Feb 13 2009, Feb 14 2009

Keywords

Comments

Expression for k(n): k(n) = 6*ceiling(n/2) + (-1)^n, so the parity of n gives us the congruence (mod 6) of k(n). - Daniel Forgues, Mar 01 2009

Crossrefs

Cf. A075743.
The absolute values of this sequence give A075743. The partial sums of this sequence give A156709.
Showing 1-3 of 3 results.