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 1-4 of 4 results.

A382902 The largest cubefree divisor of the n-th biquadratefree number.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 4, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 12, 25, 26, 9, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 20, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 18, 55, 28, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 36
Offset: 1

Views

Author

Amiram Eldar, Apr 08 2025

Keywords

Links

Crossrefs

Cf. A046100, A007948, A013662.
Similar sequences: A382903, A382904, A382905, A382906.

Programs

  • Mathematica
    f[p_, e_] := p^Min[e, 2]; s[n_] := Module[{fct = FactorInteger[n]}, If[AllTrue[fct[[;; , 2]], # < 4 &], Times @@ f @@@ fct, Nothing]]; Array[s, 100]
  • PARI
    list(lim) = {my(f); print1(1, ", "); for(k = 2, lim, f = factor(k); if(vecmax(f[, 2]) < 4, print1(prod(i = 1, #f~, f[i, 1]^min(f[i, 2], 2)), ", ")));}

Formula

a(n) = A007948(A046100(n)).
a(n) = (A382903(n) * A046100(n)^2)^(1/3).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(4)^2 * Product_{p prime} (1 - 1/p^3 + 1/p^4 - 1/p^5) = 1.01974824991243823979... .

A382903 The largest cubefree unitary divisor of the n-th biquadratefree number.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 1, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 3, 25, 26, 1, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 5, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 2, 55, 7, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 9, 73
Offset: 1

Views

Author

Amiram Eldar, Apr 08 2025

Keywords

Links

Crossrefs

Cf. A046100, A013662, A360539.
Similar sequences: A382902, A382904, A382905, A382906.

Programs

  • Mathematica
    f[p_, e_] := p^If[e < 3, e, 0]; s[n_] := Module[{fct = FactorInteger[n]}, If[AllTrue[fct[[;; , 2]], # < 4 &], Times @@ f @@@ fct, Nothing]]; Array[s, 100]
  • PARI
    list(lim) = {my(f); print1(1, ", "); for(k = 2, lim, f = factor(k); if(vecmax(f[, 2]) < 4, print1(prod(i = 1, #f~, f[i, 1]^if(f[i, 2] < 3, f[i, 2], 0)), ", ")));}

Formula

a(n) = A360539(A046100(n)).
a(n) = A382902(n)^3 / A046100(n)^2.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(4)^2 * Product_{p prime} (1 - 1/p^3 + 1/p^6 - 1/p^7) = 0.98420942697128846925... .

A382904 The squarefree kernel of the n-th biquadratefree number.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74
Offset: 1

Views

Author

Amiram Eldar, Apr 08 2025

Keywords

Links

Crossrefs

Cf. A007947, A013662, A046100.
Similar sequences: A382902, A382903, A382905, A382906.

Programs

  • Mathematica
    s[n_] := Module[{fct = FactorInteger[n]}, If[AllTrue[fct[[;; , 2]], # < 4 &], Times @@ fct[[;;, 1]], Nothing]]; Array[s, 100]
  • PARI
    list(lim) = {my(f); print1(1, ", "); for(k = 2, lim, f = factor(k); if(vecmax(f[, 2]) < 4, print1(vecprod(f[, 1]), ", ")));}

Formula

a(n) = A007947(A046100(n)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(4)^2 * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.8197259315983831456004... .

A382906 The powerful part of the n-th biquadratefree number.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 8, 9, 1, 1, 4, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 8, 25, 1, 27, 4, 1, 1, 1, 1, 1, 1, 36, 1, 1, 1, 8, 1, 1, 1, 4, 9, 1, 1, 49, 25, 1, 4, 1, 27, 1, 8, 1, 1, 1, 4, 1, 1, 9, 1, 1, 1, 4, 1, 1, 1, 72, 1, 1, 25, 4, 1, 1, 1, 1, 1, 4, 1, 1, 1, 8, 1, 9, 1
Offset: 1

Views

Author

Amiram Eldar, Apr 08 2025

Keywords

Links

Crossrefs

Cf. A013662, A046100, A057521.
Similar sequences: A382902, A382903, A382904, A382905.

Programs

  • Mathematica
    f[p_, e_] := p^If[e > 1, e, 0]; s[n_] := Module[{fct = FactorInteger[n]}, If[AllTrue[fct[[;; , 2]], # < 4 &], Times @@ f @@@ fct, Nothing]]; Array[s, 100]
  • PARI
    list(lim) = {my(f); print1(1, ", "); for(k = 2, lim, f = factor(k); if(vecmax(f[, 2]) < 4, print1(prod(i = 1, #f~, f[i, 1]^if(f[i, 2] > 1, f[i, 2], 0)), ", ")));}

Formula

a(n) = A057521(A046100(n)).
a(n) = A046100(n)/A382905(n).
Sum_{k=1..n} a(k) ~ c * n^(3/2) / 3, where c = zeta(4)^(3/2) * Product_{p prime} (1 + 2/p^(3/2) - 1/p^2 - 2/p^(5/2)) = 2.21177275344948791706... .
Showing 1-4 of 4 results.

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