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 1-6 of 6 results.

A365498 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)).

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 1, 2, 4, 2, 4, 2, 4, 4, 1, 2, 4, 2, 4, 4, 4, 2, 2, 2, 4, 1, 4, 2, 8, 2, 1, 4, 4, 4, 4, 2, 4, 4, 2, 2, 8, 2, 4, 4, 4, 2, 2, 2, 4, 4, 4, 2, 2, 4, 2, 4, 4, 2, 8, 2, 4, 4, 1, 4, 8, 2, 4, 4, 8, 2, 2, 2, 4, 4, 4, 4, 8, 2, 2, 1, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Vaclav Kotesovec, Sep 06 2023

Keywords

Comments

The number of unitary divisors of n that are cubefree numbers (A004709). - Amiram Eldar, Sep 06 2023

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e <= 2, 2, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 06 2023 *)
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1-X) * (1 + X - X^3))[n], ", "))

Formula

Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085026313185459506482223745141452711510108346133288...,
f'(1) = f(1) * Sum_{p prime} (-4 + 3*p + 2*p^2) * log(p) / (1 - p - p^2 + p^4) = f(1) * 1.4525924794451595590371439593828547341482465114411929136723476679...
and gamma is the Euler-Mascheroni constant A001620.
Multiplicative with a(p^e) = 2 if e <= 2, and 1 otherwise. - Amiram Eldar, Sep 06 2023
From Vaclav Kotesovec, Jan 27 2025: (Start)
Following formulas have been conjectured for this sequence by Sequence Machine, with each one giving the first 1000000 terms correctly:
a(n) = A056671(n) * A368885(n).
a(n) = A034444(n) / A368248(n).
a(n) = A158522(n) / A307428(n).
a(n) = A369310(n) / A190867(n).
a(n) = A286324(n) / A368172(n). (End)

A385048 The sum of the unitary divisors of n that are cubefull numbers (A036966).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 28, 1, 1, 1, 1, 33, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 28, 1, 9, 1, 1, 1, 1, 1, 1, 1, 65, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 17, 82, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 16 2025

Keywords

Comments

The number of these divisors is A368248(n), and the largest of them is A360540(n).

Crossrefs

The unitary analog of A385005.
The sum of unitary divisors of n that are: A092261 (squarefree), A192066 (odd), A358346 (exponentially odd), A358347 (square), A360720 (powerful), A371242 (cubefree), A380396 (cube), A383763 (exponentially squarefree), A385043 (exponentially 2^n), A385045 (5-rough), A385046 (3-smooth), A385047 (power of 2), this sequence (cubefull), A385049 (biquadratefree).

Programs

  • Mathematica
    f[p_, e_] := If[e <= 2, 1, p^e + 1]; 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] < 3, 1, f[i, 1]^f[i, 2] + 1));}

Formula

Multiplicative with a(p^e) = 1 if e <= 2, and a(p^e) = p^e + 1 if e >= 3.
a(n) = A034448(n) / A371242(n).
a(n) <= A034448(n), with equality if and only if n is cubefull (A036966).
a(n) <= A385005(n), with equality if and only if n is biquadratefree (A046100).
Dirichlet g.f.: zeta(s)*zeta(s-1)*Product_{p prime} (1 - 1/p^(s-1) + 1/p^(3*s-3) - 1/p^(4*s-3)).

A368251 The number of nonsquarefree divisors of n that are powers of squarefree numbers (A072777).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 0, 0, 3, 0, 1, 0, 1, 0, 0, 0, 2, 1, 0, 2, 1, 0, 0, 0, 4, 0, 0, 0, 3, 0, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 3, 1, 1, 0, 1, 0, 2, 0, 2, 0, 0, 0, 1, 0, 0, 1, 5, 0, 0, 0, 1, 0, 0, 0, 4, 0, 0, 1, 1, 0, 0, 0, 3, 3, 0, 0, 1, 0, 0, 0
Offset: 1

Views

Author

Amiram Eldar, Dec 19 2023

Keywords

Comments

First differs from A046660 and A066301 at n = 36, and from A183094 at n = 72.
Let b(n, k) be the sequence that counts the divisors of n that are k-th powers of squarefree numbers. Then, b(n, 1) = A034444(n), b(n, 2) = A323308(n), b(n, 3) = A368248(n). b(n, k) is multiplicative with b(p^e, k) = 2 if e >= k, and 1 otherwise. The asymptotic mean of b(n, k) for k >= 2 is lim_{m->oo} (1/m) * Sum_{n=1..m} b(n, k) = zeta(k)/zeta(2*k). Since a(n) = Sum_{k>=2} (b(n, k) - 1), the formula for the asymptotic mean of this sequence follows (see the Formula section).

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, 1 + Total[2^Accumulate[Count[e, #] & /@ Range[Max[e], 1, -1]] - 1] - 2^Length[e]]; a[1] = 0; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), e, m, h, c); if(n == 1, 0, e = f[,2]; m = vecmax(e); h = vector(m); for(i = 1,m, c = 0; for(j = 1, #e, if(e[j] == (m+1-i), c++)); h[i] = c); for(i = 2, m, h[i] += h[i-1]); for(i = 1, m, h[i] = 2^h[i]-1); 1 + vecsum(h) - 1<<#e);}

Formula

a(n) = A327527(n) - A034444(n).
a(n) = 0 if and only if n is squarefree (A005117).
Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=2} (zeta(k)/zeta(2*k) - 1) = 0.848633... (A368250).

A385042 The number of unitary divisors of n whose exponents in their prime factorizations are all powers of 2 (A138302).

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 1, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 2, 2, 4, 1, 4, 2, 8, 2, 1, 4, 4, 4, 4, 2, 4, 4, 2, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 2, 4, 2, 4, 4, 2, 8, 2, 4, 4, 1, 4, 8, 2, 4, 4, 8, 2, 2, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Jun 16 2025

Keywords

Comments

First differs from A367515 at n = 128.
The sum of these divisors is A385043(n), and the largest of them is A367168(n).

Crossrefs

The unitary analog of A353898.
The number of unitary divisors of n that are: A000034 (power of 2), A055076 (exponentially odd), A056624 (square), A056671 (squarefree), A068068 (odd), A323308 (powerful), A365498 (cubefree), A365499 (biquadratefree), A368248 (cubefull), A380395 (cube), A382488 (3-smooth), this sequence (exponentially 2^n), A385044 (5-rough).

Programs

  • Mathematica
    f[p_, e_] := Boole[e == 2^IntegerExponent[e, 2]] + 1; a[ 1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> (x == 1<
    				

Formula

Multiplicative with a(p^e) = A209229(e) + 1.
a(n) <= A034444(n), with equality if and only if n is in A138302.
a(n) <= A353898(n), with equality if and only if n is squarefree (A005117).

A385044 The number of unitary divisors of n that are 5-rough numbers (A007310).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 4, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 4, 2, 2, 2, 2, 2, 2, 2, 2, 1, 4, 2, 2, 2, 2, 4, 2, 1, 2, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 4, 2, 2
Offset: 1

Views

Author

Amiram Eldar, Jun 16 2025

Keywords

Comments

The sum of these divisors is A385045(n), and the largest of them is A065330(n).

Crossrefs

The unitary analog of A035218.
The number of unitary divisors of n that are: A000034 (power of 2), A055076 (exponentially odd), A056624 (square), A056671 (squarefree), A068068 (odd), A323308 (powerful), A365498 (cubefree), A365499 (biquadratefree), A368248 (cubefull), A380395 (cube), A382488 (3-smooth), A385042 (exponentially 2^n), this sequence (5-rough).

Programs

  • Mathematica
    f[p_, e_] := If[p <= 3, 1, 2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> if(x <= 3, 1, 2), factor(n)[, 1]));

Formula

Multiplicative with a(p^e) = 1 if p <= 3, and 2 if p >= 5.
a(n) = A034444(n)/A382488(n).
a(n) <= A034444(n), with equality if and only if n is 5-rough.
a(n) <= A035218(n).
Dirichlet g.f.: (zeta(s)^2/zeta(2*s)) * (1/((1+1/2^s)*(1+1/3^s))).
Sum_{k=1..n} a(k) ~ (n / (2 * zeta(2))) *(log(n) + 2*gamma - 1 + log(2)/3 + log(3)/4 - 2*zeta'(2)/zeta(2)), where gamma is Euler's constant (A001620).

A385418 The number of unordered factorizations of n into powers of primes of the form p^(2^k-1) where p is prime and k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 28 2025

Keywords

Comments

First differs from A304327 and A368248 at n = 64.
First differs from A061704 and A362852 at n = 128.
The number of unordered factorizations of n into powers of primes in A036537.

Examples

			  n | a(n) | factorizations
  --+------+-------------------------------------------------------------------
  2 |    8 | 2 * 2 * 2, 2^3
  3 |   64 | 2 * 2 * 2 * 2 * 2 * 2, 2 * 2 * 2 * 2^3, 2^3 * 2^3
  4 |  128 | 2 * 2 * 2 * 2 * 2 * 2 * 2, 2 * 2 * 2 * 2 * 2^3, 2 * 2^3 * 2^3, 2^7
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := T[n, k] = If[k <= n, T[n - k, k] + T[n, 2*k + 1], Boole[n == 0]]; f[p_, e_] := T[e, 1];
    a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    t(n, k) = if(k <= n, t(n-k, k) + t(n, 2*k+1), n == 0);
    a(n) = vecprod(apply(x -> t(x, 1), factor(n)[,2]));

Formula

Multiplicative with a(p^e) = A000929(e).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{k>=2} zeta(2^k-1) = 1.21213028603089660618... .
Showing 1-6 of 6 results.