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.

A384058 The number of integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is a 5-rough number (A007310).

Original entry on oeis.org

1, 1, 2, 3, 5, 2, 7, 7, 8, 5, 11, 6, 13, 7, 10, 15, 17, 8, 19, 15, 14, 11, 23, 14, 25, 13, 26, 21, 29, 10, 31, 31, 22, 17, 35, 24, 37, 19, 26, 35, 41, 14, 43, 33, 40, 23, 47, 30, 49, 25, 34, 39, 53, 26, 55, 49, 38, 29, 59, 30, 61, 31, 56, 63, 65, 22, 67, 51, 46
Offset: 1

Views

Author

Amiram Eldar, May 18 2025

Keywords

Crossrefs

Unitary analog of A384042.
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), A384054 (exponentially odd), A384055 (odd), A384056 (power of 2), A384057 (3-smooth), this sequence (5-rough).

Programs

  • Mathematica
    f[p_, e_] := p^e - If[p < 5, 1, 0]; 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,1] < 5, 1, 0));}

Formula

Multiplicative with a(p^e) = p^e-1 if p <= 3, and p^e if p >= 5.
a(n) = n * A047994(n) / A384057(n).
a(n) = A047994(A065331(n)) * A065330(n).
Dirichlet g.f.: zeta(s-1) * ((1 - 1/2^(s-1) + 1/2^(2*s-1))/(1 - 1/2^s)) * ((1 - 2/3^s + 1/3^(2*s-1))/(1 - 1/3^s)).
Sum_{k=1..n} a(k) ~ (55/144) * n^2.
In general, the average order of the number of integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is a p-rough number (i.e., not divisible by any prime smaller than the prime p) is (1/2) * Product_{q prime <= p} (1 - 1/q + 1/(q+1)) * n^2.

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.

Crossrefs

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.

Crossrefs

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... .

A384041 The number of integers k from 1 to n such that gcd(n,k) is an exponentially odd number.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 7, 8, 10, 11, 9, 13, 14, 15, 13, 17, 16, 19, 15, 21, 22, 23, 21, 24, 26, 25, 21, 29, 30, 31, 27, 33, 34, 35, 24, 37, 38, 39, 35, 41, 42, 43, 33, 40, 46, 47, 39, 48, 48, 51, 39, 53, 50, 55, 49, 57, 58, 59, 45, 61, 62, 56, 53, 65, 66, 67, 51
Offset: 1

Views

Author

Amiram Eldar, May 18 2025

Keywords

Crossrefs

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), this sequence (exponentially odd), A384042 (5-rough).

Programs

  • Mathematica
    f[p_, e_] := ((p^2+p-1)*p^(e-1) - (-1)^e)/(p+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]^2+f[i,1]-1)*f[i,1]^(f[i,2]-1) - (-1)^f[i,2])/(f[i,1] + 1));}

Formula

Multiplicative with a(p^e) = ((p^2+p-1)*p^(e-1) - (-1)^e)/(p+1).
a(n) >= A000010(n), with equality if and only if n = 1.
Dirichlet g.f.: (zeta(s-1)*zeta(2*s)/zeta(s)) * 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^2+1)) = 0.93749428273130025078... .
Showing 1-4 of 4 results.