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 1-10 of 11 results. Next

A356191 a(n) is the smallest exponentially odd number that is divisible by n.

Original entry on oeis.org

1, 2, 3, 8, 5, 6, 7, 8, 27, 10, 11, 24, 13, 14, 15, 32, 17, 54, 19, 40, 21, 22, 23, 24, 125, 26, 27, 56, 29, 30, 31, 32, 33, 34, 35, 216, 37, 38, 39, 40, 41, 42, 43, 88, 135, 46, 47, 96, 343, 250, 51, 104, 53, 54, 55, 56, 57, 58, 59, 120, 61, 62, 189, 128, 65
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2022

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], p^e, 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]%2, f[i,1]^f[i,2], f[i,1]^(f[i,2]+1)))};
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1 - p^2*X^2) * (1 + p*X + p^3*X^2 - p^2*X^2))[n], ", ")) \\ Vaclav Kotesovec, Sep 09 2023

Formula

Multiplicative with a(p^e) = p^e if e is odd and p^(e+1) otherwise.
a(n) = n iff n is in A268335.
a(n) = A064549(n)/A007913(n).
a(n) = n*A336643(n).
a(n) = n^2/A350390(n).
From Vaclav Kotesovec, Sep 09 2023: (Start)
Let f(s) = Product_{p prime} (1 - p^(6-5*s) + p^(7-5*s) + 2*p^(5-4*s) - p^(6-4*s) + p^(3-3*s) - p^(4-3*s) - 2*p^(2-2*s)).
Sum_{k=1..n} a(k) ~ Pi^2 * f(2) * n^2 / 24 * (log(n) + 3*gamma - 1/2 + 12*zeta'(2)/Pi^2 + f'(2)/f(2)), where
f(2) = Product_{p prime} (1 - 4/p^2 + 4/p^3 - 1/p^4) = A256392 = 0.2177787166195363783230075141194468131307977550013559376482764035236264911...,
f'(2) = f(2) * Sum_{p prime} (11*p - 5) * log(p) / (p^3 + p^2 - 3*p + 1) = f(1) * 4.7165968208567630786609552448708126340725121316268495170070986645608062483...
and gamma is the Euler-Mascheroni constant A001620. (End)

A356192 a(n) is the smallest cubefull exponentially odd number (A335988) that is divisible by n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 8, 27, 1000, 1331, 216, 2197, 2744, 3375, 32, 4913, 216, 6859, 1000, 9261, 10648, 12167, 216, 125, 17576, 27, 2744, 24389, 27000, 29791, 32, 35937, 39304, 42875, 216, 50653, 54872, 59319, 1000, 68921, 74088, 79507, 10648, 3375, 97336
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2022

Keywords

Comments

First differs from A053149 and A356193 at n=16.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], p^Max[e, 3], p^(e + 1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50]
  • PARI
    a(n) = {my(f=factor(n)); prod(i=1, #f~, if(f[i,2]%2, f[i,1]^max(f[i,2],3), f[i,1]^(f[i,2]+1)))};

Formula

Multiplicative with a(p^e) = p^max(e,3) if e is odd and p^(e+1) otherwise.
a(n) = n iff n is in A335988.
a(n) = A356191(n) iff n is a powerful number (A001694).
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + (3*p^2-1)/(p^3*(p^2-1))) = 1.69824776889117043774... .
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(6)/4) * Product_{p prime} (1 - 1/p^2 + 1/p^5 - 2/p^6 + 1/p^8 + 1/p^9 - 1/p^10) = 0.1559368144... . - Amiram Eldar, Nov 13 2022

A356193 a(n) is the smallest cubefull number (A036966) that is a multiple of n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 8, 27, 1000, 1331, 216, 2197, 2744, 3375, 16, 4913, 216, 6859, 1000, 9261, 10648, 12167, 216, 125, 17576, 27, 2744, 24389, 27000, 29791, 32, 35937, 39304, 42875, 216, 50653, 54872, 59319, 1000, 68921, 74088, 79507, 10648, 3375, 97336
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2022

Keywords

Comments

First differs from A053149 and A356192 at n=16.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^Max[e, 3]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50]
  • PARI
    a(n) = {my(f=factor(n)); prod(i=1, #f~, f[i,1]^max(f[i,2],3))};

Formula

Multiplicative with a(p^e) = p^max(e,3).
a(n) = n iff n is in A036966.
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + (3*p-2)/(p^3*(p-1))) = 1.76434793373691907811... . - Amiram Eldar, Jul 29 2022
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(3)/4) * Product_{p prime} (1 - 1/p^2 - 1/p^3 + 2/p^5 - 1/p^6 - 1/p^8 + 2/p^9 - 1/p^10) = 0.1559111567... . - Amiram Eldar, Nov 13 2022
a(n) = n * A360541(n). - Amiram Eldar, Sep 01 2023

A356194 a(n) is the smallest multiple of n whose prime factorization exponents are all powers of 2.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 16, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 48, 25, 26, 81, 28, 29, 30, 31, 256, 33, 34, 35, 36, 37, 38, 39, 80, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 162, 55, 112, 57, 58, 59, 60, 61, 62, 63, 256, 65, 66, 67
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2022

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(2^Ceiling[Log2[e]]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    s(n) = {my(e=logint(n,2)); if(n == 2^e, n, 2^(e+1))};
    a(n) = {my(f=factor(n)); prod(i=1, #f~, f[i,1]^s(f[i,2]))};

Formula

Multiplicative with a(p^e) = p^(2^ceiling(log_2(e))).
a(n) = n iff n is in A138302.

A056553 Smallest 4th-power divisible by n divided by largest 4th-power which divides n.

Original entry on oeis.org

1, 16, 81, 16, 625, 1296, 2401, 16, 81, 10000, 14641, 1296, 28561, 38416, 50625, 1, 83521, 1296, 130321, 10000, 194481, 234256, 279841, 1296, 625, 456976, 81, 38416, 707281, 810000, 923521, 16, 1185921, 1336336, 1500625, 1296, 1874161, 2085136
Offset: 1

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(64) = 16 because smallest 4th power divisible by 64 is 256 and largest 4th power which divides 64 is 16 and 256/16 = 16.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^If[Divisible[e, 4], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^4; Array[a, 100] (* Amiram Eldar, Aug 29 2019*)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%4, f[i,1], 1))^4; } \\ Amiram Eldar, Oct 27 2022

Formula

a(n) = A053167(n)/A008835(n) = A056556(n)*A053165(n) = A056554(n)^4.
From Amiram Eldar, Oct 27 2022: (Start)
Multiplicative with a(p^e) = 1 if e is divisible by 4, and a(p^e) = p^4 otherwise.
Sum_{k=1..n} a(k) ~ c * n^5, where c = (zeta(20)/(5*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.123026157003... . (End)

A056554 Powerfree kernel of 4th-powerfree part of n.

Original entry on oeis.org

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

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(64) = 2 because 4th-power-free part of 64 is 4 and power-free kernel of 4 is 2.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] :=  p^If[Divisible[e, 4], 0, 1]; a[n_] := Times @@ (f @@@ FactorInteger[ n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
  • PARI
    a(n) = my(f=factor(n)); for (k=1, #f~, if (frac(f[k,2]/4), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ Michel Marcus, Feb 28 2019

Formula

a(n) = A007947(A053165(n)) = A053166(A053165(n)) = n/(A053164(n)*A000190(n)) = A053166(n)/A053164(n) = A056553(n)^(1/4).
If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 4 and Fj = 1 otherwise.
Multiplicative with a(p^e) = p^(if 4|e, then 0, else 1). - Mitch Harris, Apr 19 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(8)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3513111135... . - Amiram Eldar, Oct 27 2022

A056555 Smallest number k (k>0) such that n*k is a perfect 4th power.

Original entry on oeis.org

1, 8, 27, 4, 125, 216, 343, 2, 9, 1000, 1331, 108, 2197, 2744, 3375, 1, 4913, 72, 6859, 500, 9261, 10648, 12167, 54, 25, 17576, 3, 1372, 24389, 27000, 29791, 8, 35937, 39304, 42875, 36, 50653, 54872, 59319, 250, 68921, 74088, 79507, 5324, 1125
Offset: 1

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(64) = 4 because the smallest 4th power divisible by 64 is 256 and 64*4 = 256.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^Mod[4 - e, 4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)
  • PARI
    a(n,f=factor(n))=f[,2]=-f[,2]%4; factorback(f) \\ Charles R Greathouse IV, Apr 24 2020

Formula

a(n) = A053167(n)/n = n^3/A000190(n)^4 = A056553(n)/A053165(n).
Multiplicative with a(p^e) = p^((4 - e) mod 4). - Amiram Eldar, Sep 08 2020
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(16)/(4*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.1537848996... . - Amiram Eldar, Oct 27 2022

A365491 The number of divisors of the smallest number whose 4th power is divisible by n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 3, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 3, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Sep 05 2023

Keywords

Comments

First differs from A365210 at n = 25 and from A034444 at n = 32.
The number of divisors of the smallest 4th divisible by n, A053167(n), is A365492(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Ceiling[e/4] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
    With[{c=Range[100]^4},Table[DivisorSigma[0,Surd[SelectFirst[c,Mod[#,n]==0&],4]],{n,90}]] (* Harvey P. Dale, Jul 09 2024 *)
  • PARI
    a(n) = vecprod(apply(x -> (x-1)\4 + 2, factor(n)[, 2]));

Formula

a(n) = A000005(A053166(n)).
Multiplicative with a(p^e) = ceiling(e/4) + 1.
a(n) <= A000005(n) with equality if and only if n is squarefree (A005117).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 1/p^s - 1/p^(4*s)).
From Vaclav Kotesovec, Sep 06 2023: (Start)
Dirichlet g.f.: zeta(s)^2 * zeta(4*s) * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(4*s) + 1/p^(5*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(4*s) + 1/p^(5*s)).
Sum_{k=1..n} a(k) ~ zeta(4) * f(1) * n * (log(n) + 2*gamma - 1 + 4*zeta'(4)/zeta(4) + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^4 + 1/p^5) = 0.57615273538566705952061107826411727540624711680289618854325028459572487...,
f'(1) = f(1) * Sum_{p prime} (-5 + 4*p + 2*p^3) * log(p) / (1 - p - p^3 + p^5) = f(1) * 1.3011434396559802378314782600747661399223385669839998680418996210...
and gamma is the Euler-Mascheroni constant A001620. (End)
a(n) = A322483(A019554(n)) (the number of exponentially odd divisors of the smallest number whose square is divisible by n). - Amiram Eldar, Sep 08 2023

A062379 n divided by largest 4th-power-free factor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Henry Bottomley, Jun 18 2001

Keywords

Crossrefs

Programs

Formula

a(n) = n/A058035(n).
Multiplicative with a(p^e) = p^max(e-3, 0). - Amiram Eldar, Sep 07 2020

A365492 The number of divisors of the smallest 4th power divisible by n.

Original entry on oeis.org

1, 5, 5, 5, 5, 25, 5, 5, 5, 25, 5, 25, 5, 25, 25, 5, 5, 25, 5, 25, 25, 25, 5, 25, 5, 25, 5, 25, 5, 125, 5, 9, 25, 25, 25, 25, 5, 25, 25, 25, 5, 125, 5, 25, 25, 25, 5, 25, 5, 25, 25, 25, 5, 25, 25, 25, 25, 25, 5, 125, 5, 25, 25, 9, 25, 125, 5, 25, 25, 125, 5, 25
Offset: 1

Views

Author

Amiram Eldar, Sep 05 2023

Keywords

Comments

First differs from A082476 at n = 32.
The number of divisors of the 4th root of the smallest 4th power divisible by n, A053166(n), is A365491(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := 4*Ceiling[e/4] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> 4*((x-1)\4) + 5, factor(n)[, 2]));

Formula

a(n) = A000005(A053167(n)).
Multiplicative with a(p^e) = 4*ceiling(e/4) + 1.
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 4/p^s - 1/p^(4*s)).
Showing 1-10 of 11 results. Next