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.

Showing 1-5 of 5 results.

A383278 The number of integers k such that A034444(k) * k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24
Offset: 1

Views

Author

Amiram Eldar, Apr 21 2025

Keywords

Comments

The number of terms of A383276 not exceeding n.

References

  • József Sándor, Dragoslav S. Mitrinovic, and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, Chapter II, section 31, page 72.

Crossrefs

Partial sums of A383277.
The unitary analog of A356005.

Programs

  • Mathematica
    Accumulate[Table[DivisorSum[n, 1 &, # * 2^PrimeNu[#] == n &], {n, 1, 100}]]
    (* second program: *)
    f[n_] := Module[{e = IntegerExponent[n, 2], w}, w = PrimeNu[n/2^e]; If[e > w + 1 || e == w, 1, 0]]; Accumulate[Array[f, 100]]
  • PARI
    list(lim) = my(s = 0); for(n = 1, lim, s += sumdiv(n, d, (1 << omega(d)) * d == n); print1(s, ", "));
    
  • PARI
    f(n) = {my(e = valuation(n, 2), w = omega(n >> e)); e > w + 1 || e == w;}
    list(lim) = my(s = 0); for(n = 1, lim, s += f(n); print1(s, ", "));

Formula

a(n) = Sum_{k=1..n} A383277(k).
a(n) = (c + o(1)) * n / sqrt(log(n)), where c = (1/sqrt(Pi)) * Product_{p prime} (p-1/2)/sqrt(p*(p-1)) = A087197 * A345288 = 0.61890644913204789046... (Abbott and Subbarao, 1989).

A345231 Decimal expansion of 1/sqrt(Pi) * Product_{p primes} sqrt(p*(p-1)) * log(p/(p-1)).

Original entry on oeis.org

5, 4, 6, 8, 5, 5, 9, 5, 5, 2, 8, 0, 4, 7, 4, 4, 6, 6, 8, 4, 5, 5, 1, 7, 1, 0, 0, 9, 9, 0, 7, 6, 1, 7, 8, 9, 9, 1, 0, 2, 1, 0, 4, 8, 5, 9, 2, 9, 7, 4, 2, 9, 4, 7, 8, 2, 8, 6, 8, 9, 3, 7, 1, 4, 9, 9, 3, 5, 1, 4, 8, 6, 2, 7, 3, 9, 1, 5, 5, 1, 7, 1, 5, 2, 7, 6, 8, 7, 1, 6, 0, 0, 2, 3, 7, 8, 3, 1, 0, 3, 2, 8, 7, 9, 8, 8
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 11 2021

Keywords

Examples

			0.5468559552804744668455171009907617899102104859297429478286893714993514862739...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; Clear[f]; f[p_] := Sqrt[p*(p - 1)]*Log[p/(p - 1)]; Do[cc = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x, m + 1]]; Print[1/Sqrt[Pi] * f[2] * Exp[N[Sum[Indexed[cc, n] * (PrimeZetaP[n] - 1/2^n), {n, 2, m}], 110]]], {m, 100, 500, 100}]

Formula

Equals lim_{n->infinity} sqrt(log(n))/n * Sum_{k=1..n} 1/d(k), where d(n) = A000005(n).
Equals A083281/sqrt(Pi).

A383160 a(n) is the numerator of the mean of the maximum exponents in the prime factorizations of the unitary divisors of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 5, 1, 3, 3, 2, 1, 5, 1, 5, 3, 3, 1, 7, 1, 3, 3, 5, 1, 7, 1, 5, 3, 3, 3, 3, 1, 3, 3, 7, 1, 7, 1, 5, 5, 3, 1, 9, 1, 5, 3, 5, 1, 7, 3, 7, 3, 3, 1, 11, 1, 3, 5, 3, 3, 7, 1, 5, 3, 7, 1, 2, 1, 3, 5, 5, 3, 7, 1, 9, 2, 3, 1, 11, 3, 3, 3
Offset: 1

Views

Author

Amiram Eldar, Apr 18 2025

Keywords

Comments

First differs from A296082 at n = 27.
a(n) depends only on the prime signature of n (A118914).

Examples

			Fractions begin with 0, 1/2, 1/2, 1, 1/2, 3/4, 1/2, 3/2, 1, 3/4, 1/2, 5/4, ...
4 has 2 unitary divisors: 1 and 4 = 2^2. The maximum exponents in their prime factorizations are 0 and 2, respectively. Therefore, a(4) = numerator((0 + 2)/2) = numerator(1) = 1.
12 has 4 unitary divisors: 1, 3 = 3^1, 4 = 2^2 and 12 = 2^2 * 3. The maximum exponents in their prime factorizations are 0, 1, 2 and 2, respectively. Therefore, a(12) = numerator((0 + 1 + 2 + 2)/4) = numerator(5/4) = 5.
		

Crossrefs

Programs

  • Mathematica
    emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := Numerator[DivisorSum[n, emax[#] &, CoprimeQ[#, n/#] &] / 2^PrimeNu[n]]; Array[a, 100]
  • PARI
    emax(n) = if(n == 1, 0, vecmax(factor(n)[,2]));
    a(n) = my(f = factor(n)); numerator(sumdiv(n, d, emax(d) * (gcd(d, n/d) == 1)) / (1 << omega(f)));

Formula

a(n) = numerator(Sum_{d|n, gcd(d, n/d) = 1} A051903(d) / A034444(n)) = numerator(A383159(n) / A034444(n)).
a(n)/A383161(n) >= A383157(n)/A383158(n), with equality if and only if n is either squarefree (A005117) or a power of prime (A000961), i.e., n is not in A126706.
a(n)/A383161(n) < 1 if and only if n is squarefree.
a(n)/A383161(n) = 1 if and only if n is a square of a prime (A001248).
Sum_{k=1..n} a(k)/A383161(k) ~ c_1 * n - c_2 * n / sqrt(log(n)), where c = m(2) + Sum_{k>=3} (k-1) * (m(k) - m(k-1)) = 1.36914082067166047512... is the asymptotic mean of the fractions a(k)/A383161(k), m(k) = Product_{p prime} (1 - 1/(2*p^k)) is the asymptotic mean of the ratio between the number of k-free unitary divisors and the number of unitary divisors. e.g., m(2) = A383057 and m(3) = A383058, and c_2 = A345288/sqrt(Pi) = 0.6189064491320478904... .

A083281 Decimal expansion of h = Product_{p prime}(sqrt(p(p-1))*log(1/(1-1/p))).

Original entry on oeis.org

9, 6, 9, 2, 7, 6, 9, 4, 3, 8, 2, 7, 4, 9, 1, 6, 3, 0, 7, 1, 6, 9, 5, 3, 7, 1, 4, 7, 2, 0, 9, 0, 7, 3, 2, 2, 6, 6, 2, 1, 3, 6, 8, 8, 6, 3, 8, 4, 9, 1, 6, 2, 1, 8, 1, 6, 1, 7, 8, 5, 8, 8, 7, 5, 1, 9, 5, 0, 5, 7, 0, 0, 2, 8, 3, 8, 7, 4, 0, 1, 9, 7, 3, 4, 7, 7, 8, 6, 5, 0, 8, 3, 3, 7, 3, 4, 2, 7, 6, 6, 5, 0, 9, 4, 8, 9
Offset: 0

Views

Author

Benoit Cloitre, Jun 02 2003

Keywords

Comments

Arises in formulas like: Sum_{k<=x} 1/tau(kd) = hx/sqrt(Pi*log(x))*{ g(d)+O((3/4)^omega(d)/log(x)) } where g satisfies Sum_{d<=x} g(d))=x/h/sqrt(Pi*log(x))*{ 1+O(1/log(x)) }.
The logarithm of the value has an expansion -P(2)/24 -P(3)/24 -109*P(4)/2880 -49*P(5)/1440-... in terms of the prime zeta functions P(.). - R. J. Mathar, Jan 31 2009
The average order of 1/tau(k) (where tau(k) is the number of divisors of k, A000005), Sum_{k<=x} 1/tau(k) ~ h*x/sqrt(Pi*log(x)), was found by Ramanujan in 1916 and was proven by Wilson in 1923. - Amiram Eldar, Jun 19 2019

Examples

			0.96927694382749163071695371472090732266213688638491621816178588751950570028...
		

References

  • G. Tenenbaum, Introduction à la théorie analytique et probabiliste des nombres, collection SMF no. 1, 1995, p. 210.

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; f[p_] := Sqrt[p*(p - 1)]*Log[p/(p - 1)]; c = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; RealDigits[Exp[NSum[Indexed[c, k]*PrimeZetaP[k], {k, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 100][[1]] (* Amiram Eldar, Jun 19 2019 *)
  • PARI
    prod(k=1,40000,sqrt(prime(k)*(prime(k)-1))*log(1/(1-1/prime(k))))

Formula

Equals A345231 * sqrt(Pi). - Vaclav Kotesovec, Jun 13 2021

Extensions

10 more digits from R. J. Mathar, Jan 31 2009
More terms from Amiram Eldar, Jun 19 2019
More digits from Vaclav Kotesovec, Jun 13 2021

A383055 Numerators of the partial sums of the reciprocals of the number of unitary divisors function (A034444).

Original entry on oeis.org

1, 3, 2, 5, 3, 13, 15, 17, 19, 5, 11, 23, 25, 13, 27, 29, 31, 8, 17, 35, 9, 37, 39, 10, 21, 43, 45, 23, 12, 97, 101, 105, 107, 109, 111, 113, 117, 119, 121, 123, 127, 16, 33, 67, 17, 69, 71, 18, 37, 75, 19, 77, 79, 20, 81, 41, 83, 21, 43, 173, 177, 179, 181, 185
Offset: 1

Views

Author

Amiram Eldar, Apr 15 2025

Keywords

Examples

			Fractions begin with 1, 3/2, 2, 5/2, 3, 13/4, 15/4, 17/4, 19/4, 5, 11/2, 23/4, ...
		

References

  • Jean-Marie De Koninck and Aleksandar Ivić, Topics in Arithmetical Functions, North-Holland Publishing Company, Amsterdam, Netherlands, 1980. See pp. 42-43.
  • Steven R. Finch, Mathematical Constants II, Cambridge University Press, 2018, p. 50.

Crossrefs

The unitary analog of A104528.
Cf. A002161, A034444, A345288, A383056 (denominators).
Similar sequences: A064608, A370898, A379513.

Programs

  • Mathematica
    Numerator[Accumulate[1/Array[2^PrimeNu[#] &, 100]]]
  • PARI
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += 1 / 2^omega(k); print1(numerator(s), ", "))};

Formula

a(n) = numerator(Sum_{k=1..n} 1/A034444(k)).
a(n)/A383056(n) = (c/sqrt(Pi)) * n / sqrt(log(n)) + O(n / log(n)^(3/2)), where c = A345288 (De Koninck and Ivić, 1980).
Showing 1-5 of 5 results.