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.

A328013 Decimal expansion of the growth constant for the partial sums of powerful part of n (A057521).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Oct 01 2019

Keywords

Examples

			3.51955505841710664719752940369857817186039808225407...
		

References

  • D. Suryanarayana and P. Subrahmanyam, The maximal k-full divisor of an integer, Indian J. Pure Appl. Math., Vol. 12, No. 2 (1981), pp. 175-190.

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 500; m = 500; c = LinearRecurrence[{0, 0, -2, 0, 1}, {0, 0, 6, 0, -5}, m]; RealDigits[(1 + 2/2^(3/2) - 1/2^(5/2))*(1 + 2/3^(3/2) - 1/3^(5/2))* Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/2] - 1/2^(n/2) - 1/3^(n/2))/n, {n, 3, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 100][[1]]
  • PARI
    prodeulerrat(1 + 2/p^3 - 1/p^5, 1/2) \\ Amiram Eldar, Jun 29 2023

Formula

The constant d1 in the paper by Cloutier et al. such that Sum_{k=1..x} A057521(k) = (d1/3)*x^(3/2) + O(x^(4/3)).
Sum_{k=1..x} 1/A055231(k) = d1*x^(1/2) + O(x^(1/3)).
Equals Product_{primes p} (1 + 2/p^(3/2) - 1/p^(5/2)).
Equals (zeta(3/2)/zeta(3)) * Product_{p prime} (1 + (sqrt(p)-1)/(p*(p-sqrt(p)+1))). - Amiram Eldar, Dec 26 2024

Extensions

More terms from Vaclav Kotesovec, May 29 2020

A370900 Partial sums of the powerfree part function (A055231).

Original entry on oeis.org

1, 3, 6, 7, 12, 18, 25, 26, 27, 37, 48, 51, 64, 78, 93, 94, 111, 113, 132, 137, 158, 180, 203, 206, 207, 233, 234, 241, 270, 300, 331, 332, 365, 399, 434, 435, 472, 510, 549, 554, 595, 637, 680, 691, 696, 742, 789, 792, 793, 795, 846, 859, 912, 914, 969, 976, 1033
Offset: 1

Views

Author

Amiram Eldar, Mar 05 2024

Keywords

References

  • Steven R. Finch, Mathematical Constants II, Cambridge University Press, 2018, p. 52.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, p, 1]; pfp[n_] := Times @@ f @@@ FactorInteger[n]; pfp[1] = 1; Accumulate[Array[pfp[#] &, 100]]
  • PARI
    pfp(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, f[i, 1], 1));}
    lista(kmax) = {my(s = 0); for(k = 1, kmax, s += pfp(k); print1(s, ", "))};

Formula

a(n) = Sum_{k=1..n} A055231(k).
a(n) = c * n^2 / 2 + O(R(n)), where c = Product_{p prime} (1 - (p^2+p-1)/(p^3*(p+1))) = 0.649606699337... (A191622), R(n) = x^(3/2) * exp(-c_1 * log(n)^(3/5) / log(log(n))^(1/5)) unconditionally, or x^(7/5) * exp(c_2 * log(n) / log(log(n))) assuming the Riemann hypothesis, and c_1 and c_2 are positive constants (Tóth, 2017).

A370901 Partial alternating sums of the powerfree part function (A055231).

Original entry on oeis.org

1, -1, 2, 1, 6, 0, 7, 6, 7, -3, 8, 5, 18, 4, 19, 18, 35, 33, 52, 47, 68, 46, 69, 66, 67, 41, 42, 35, 64, 34, 65, 64, 97, 63, 98, 97, 134, 96, 135, 130, 171, 129, 172, 161, 166, 120, 167, 164, 165, 163, 214, 201, 254, 252, 307, 300, 357, 299, 358, 343, 404, 342
Offset: 1

Views

Author

Amiram Eldar, Mar 05 2024

Keywords

Crossrefs

Similar sequences: A068762, A068773, A307704, A357817, A362028.

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, p, 1]; pfp[n_] := Times @@ f @@@ FactorInteger[n]; pfp[1] = 1; Accumulate[Array[(-1)^(# + 1) * pfp[#] &, 100]]
  • PARI
    pfp(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, f[i, 1], 1));}
    lista(kmax) = {my(s = 0); for(k = 1, kmax, s += (-1)^(k+1) * pfp(k); print1(s, ", "))};

Formula

a(n) = Sum_{k=1..n} (-1)^(k+1) * A055231(k).
a(n) = (5/38) * c * n^2 + O(R(n)), where c = Product_{p prime} (1 - (p^2+p-1)/(p^3*(p+1))) = 0.649606... (A191622), R(n) = x^(3/2) * exp(-c_1 * log(n)^(3/5) / log(log(n))^(1/5)) unconditionally, or x^(7/5) * exp(c_2 * log(n) / log(log(n))) assuming the Riemann hypothesis, and c_1 and c_2 are positive constants (Tóth, 2017).

A379583 Numerators of the partial sums of the reciprocals of the powerful part function (A057521).

Original entry on oeis.org

1, 2, 3, 13, 17, 21, 25, 51, 467, 539, 611, 629, 701, 773, 845, 1699, 1843, 1859, 2003, 2039, 2183, 2327, 2471, 2489, 62369, 65969, 198307, 201007, 211807, 222607, 233407, 467489, 489089, 510689, 532289, 532889, 554489, 576089, 597689, 600389, 621989, 643589, 665189
Offset: 1

Views

Author

Amiram Eldar, Dec 26 2024

Keywords

Examples

			Fractions begin with 1, 2, 3, 13/4, 17/4, 21/4, 25/4, 51/8, 467/72, 539/72, 611/72, 629/72, ...
		

Crossrefs

Cf. A057521, A191622, A370902, A370903, A379584 (denominators), A379585.

Programs

  • Mathematica
    f[p_, e_] := If[e > 1, p^e, 1]; powful[n_] := Times @@ f @@@ FactorInteger[n]; Numerator[Accumulate[Table[1/powful[n], {n, 1, 50}]]]
  • PARI
    powerful(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] > 1, f[i, 1]^f[i, 2], 1)); }
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += 1 / powerful(k); print1(numerator(s), ", "))};

Formula

a(n) = numerator(Sum_{k=1..n} 1/A057521(k)).
a(n)/A379584(n) = c * n + O(n^(1/2)), where c = A191622 (Cloutier et al., 2014). The error term was improved by Tóth (2017) to O(n^(1/2) * exp(-c1 * log(n)^(3/5) / log(log(n))^(1/5))) unconditionally, and O(n^(2/5) * exp(c2 * log(n) / log(log(n)))) assuming the Riemann hypothesis, where c1 and c2 are positive constants.

A379585 Numerators of the partial alternating sums of the reciprocals of the powerful part function (A057521).

Original entry on oeis.org

1, 0, 1, 3, 7, 3, 7, 13, 125, 53, 125, 107, 179, 107, 179, 349, 493, 53, 69, 65, 81, 65, 81, 79, 1991, 1591, 43357, 40657, 51457, 40657, 51457, 102239, 123839, 102239, 123839, 123239, 144839, 123239, 144839, 142139, 163739, 142139, 163739, 158339, 160739, 139139
Offset: 1

Views

Author

Amiram Eldar, Dec 26 2024

Keywords

Examples

			Fractions begin with 1, 0, 1, 3/4, 7/4, 3/4, 7/4, 13/8, 125/72, 53/72, 125/72, 107/72, ...
		

Crossrefs

Cf. A057521, A191622, A370902, A370903, A379583, A379586 (denominators).

Programs

  • Mathematica
    f[p_, e_] := If[e > 1, p^e, 1]; powful[n_] := Times @@ f @@@ FactorInteger[n]; Numerator[Accumulate[Table[(-1)^(n+1)/powful[n], {n, 1, 50}]]]
  • PARI
    powerful(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] > 1, f[i, 1]^f[i, 2], 1)); }
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += (-1)^(k+1) / powerful(k); print1(numerator(s), ", "))};

Formula

a(n) = numerator(Sum_{k=1..n} (-1)^(k+1)/A057521(k)).
a(n)/A379586(n) = (5/19) * A191622 * n + O(n^(1/2) * exp(-c1 * log(n)^(3/5) / log(log(n))^(1/5))) unconditionally, and with an improved error term O(n^(2/5) * exp(c2 * log(n) / log(log(n)))) assuming the Riemann hypothesis, where c1 and c2 are positive constants.
Showing 1-5 of 5 results.