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.

A384054 The number of integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is an exponentially odd number.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 8, 8, 10, 11, 9, 13, 14, 15, 15, 17, 16, 19, 15, 21, 22, 23, 24, 24, 26, 27, 21, 29, 30, 31, 32, 33, 34, 35, 24, 37, 38, 39, 40, 41, 42, 43, 33, 40, 46, 47, 45, 48, 48, 51, 39, 53, 54, 55, 56, 57, 58, 59, 45, 61, 62, 56, 63, 65, 66, 67, 51
Offset: 1

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Author

Amiram Eldar, May 18 2025

Keywords

Links

Crossrefs

Unitary analog of A384041.
Cf. A013662, A268335, A350388, A350389, A384046, A384047.
The number of integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is: A047994 (1), A384048 (squarefree), A384049 (cubefree), A384050 (powerful), A384051 (cubefull), A384052 (square), A384053 (cube), this sequence (exponentially odd), A384055 (odd), A384056 (power of 2), A384057 (3-smooth), A384058 (5-rough).

Programs

  • Mathematica
    f[p_, e_] := p^e - If[OddQ[e], 0, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a,100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^f[i,2] - if(f[i,2]%2, 0, 1));}
  • Python
    from math import prod
from sympy import factorint
def A384054(n): return prod(p**e-(e&1^1) for p,e in factorint(n).items()) # Chai Wah Wu, May 21 2025

Formula

Multiplicative with a(p^e) = p^e if e is odd, and p^e-1 if e is even.
a(n) = n * A047994(n) / A384052(n).
a(n) = A047994(A350388(n)) * A350389(n).
Dirichlet g.f.: zeta(s-1) * zeta(2*s) * Product_{p prime} (1 - 2/p^(2*s) + 1/p^(3*s-1)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(4) * Product_{p prime} (1 - 2/p^4 + 1/p^5) = 0.95692470821076622881... .

A384039 The number of integers k from 1 to n such that gcd(n,k) is a powerful number.

Original entry on oeis.org

1, 1, 2, 3, 4, 2, 6, 6, 7, 4, 10, 6, 12, 6, 8, 12, 16, 7, 18, 12, 12, 10, 22, 12, 21, 12, 21, 18, 28, 8, 30, 24, 20, 16, 24, 21, 36, 18, 24, 24, 40, 12, 42, 30, 28, 22, 46, 24, 43, 21, 32, 36, 52, 21, 40, 36, 36, 28, 58, 24, 60, 30, 42, 48, 48, 20, 66, 48, 44
Offset: 1

Views

Author

Amiram Eldar, May 18 2025

Keywords

Comments

The number of integers k from 1 to n such that the powerfree part (A055231) of gcd(n,k) is 1.

Links

Crossrefs

Cf. A000010, A001694, A005117, A050873, A055231, A063658, A206369.
The number of integers k from 1 to n such that gcd(n,k) is: A026741 (odd), A062570 (power of 2), A063659 (squarefree), A078429 (cube), A116512 (power of a prime), A117494 (prime), A126246 (1 or 2), A206369 (square), A254926 (cubefree), A372671 (3-smooth), this sequence (powerful), A384040 (cubefull), A384041 (exponentially odd), A384042 (5-rough).

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, p-1, (p^2-p+1)*p^(e-2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] == 1, f[i,1]-1, (f[i,1]^2-f[i,1]+1)*f[i,1]^(f[i,2]-2)));}

Formula

Multiplicative with a(p^e) = (p^2-p+1)*p^(e-2) if e >= 2, and p-1 otherwise.
a(n) >= A000010(n), with equality if and only if n is squarefree (A005117).
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^s + 1/p^(2*s)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^4) = 0.66922021803510257394... .

A384040 The number of integers k from 1 to n such that gcd(n,k) is a cubefull number.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 5, 6, 4, 10, 4, 12, 6, 8, 10, 16, 6, 18, 8, 12, 10, 22, 10, 20, 12, 19, 12, 28, 8, 30, 20, 20, 16, 24, 12, 36, 18, 24, 20, 40, 12, 42, 20, 24, 22, 46, 20, 42, 20, 32, 24, 52, 19, 40, 30, 36, 28, 58, 16, 60, 30, 36, 40, 48, 20, 66, 32, 44, 24
Offset: 1

Views

Author

Amiram Eldar, May 18 2025

Keywords

Comments

The number of integers k from 1 to n such that the cubefree part (A360539) of gcd(n,k) is 1.

Links

Crossrefs

Cf. A005117, A036966, A078429, A360539, A254926.
The number of integers k from 1 to n such that gcd(n,k) is: A026741 (odd), A062570 (power of 2), A063659 (squarefree), A078429 (cube), A116512 (power of a prime), A117494 (prime), A126246 (1 or 2), A206369 (square), A254926 (cubefree), A372671 (3-smooth), A384039 (powerful), this sequence (cubefull), A384041 (exponentially odd), A384042 (5-rough).

Programs

  • Mathematica
    f[p_, e_] := Switch[e, 1, p-1, 2, p^2-p, , (p^3-p^2+1)*p^(e-3)]; a[1] = 1; a[n] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] == 1, f[i,1]-1, if(f[i,2] == 2, f[i,1]*(f[i,1]-1), (f[i,1]^3-f[i,1]^2+1)*f[i,1]^(f[i,2]-3))));}

Formula

Multiplicative with a(p^e) = (p^3-p^2+1)*p^(e-3) if e >= 3, p*(p-1) if e = 2, and p-1 otherwise.
a(n) >= A384039(n), with equality if and only if n is squarefree (A005117).
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^s + 1/p^(3*s)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^6) = 0.62159731307414305346... .

A384042 The number of integers k from 1 to n such that gcd(n,k) is a 5-rough number (A007310).

Original entry on oeis.org

1, 1, 2, 2, 5, 2, 7, 4, 6, 5, 11, 4, 13, 7, 10, 8, 17, 6, 19, 10, 14, 11, 23, 8, 25, 13, 18, 14, 29, 10, 31, 16, 22, 17, 35, 12, 37, 19, 26, 20, 41, 14, 43, 22, 30, 23, 47, 16, 49, 25, 34, 26, 53, 18, 55, 28, 38, 29, 59, 20, 61, 31, 42, 32, 65, 22, 67, 34, 46
Offset: 1

Views

Author

Amiram Eldar, May 18 2025

Keywords

Links

Crossrefs

Cf. A000010, A003586, A007310, A065330, A065331, A372671.
The number of integers k from 1 to n such that gcd(n,k) is: A026741 (odd), A062570 (power of 2), A063659 (squarefree), A078429 (cube), A116512 (power of a prime), A117494 (prime), A126246 (1 or 2), A206369 (square), A254926 (cubefree), A372671 (3-smooth), A384039 (powerful), A384040 (cubefull), A384041 (exponentially odd), this sequence (5-rough).

Programs

  • Mathematica
    f[p_, e_] := If[p < 5, (p-1)*p^(e-1), p^e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] < 5, (f[i,1]-1)*f[i,1]^(f[i,2]-1), f[i,1]^f[i,2]));}

Formula

Multiplicative with a(p^e) = (p-1)*p^(e-1) if p <= 3 and p^e if p >= 5.
a(n) >= A000010(n), with equality if and only if n is 3-smooth (A003586).
a(n) = A000010(A065331(n)) * A065330(n).
a(n) = 2 * n * phi(n)/phi(6*n) = n * A000010(n) / A372671(n).
Dirichlet g.f.: zeta(s-1) * (1-1/2^s) * (1-1/3^s).
Sum_{k=1..n} a(k) ~ n^2 / 3.
Showing 1-4 of 4 results.

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