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.

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A377818 Powerful numbers that have a single even exponent in their prime factorization.

Original entry on oeis.org

4, 9, 16, 25, 49, 64, 72, 81, 108, 121, 169, 200, 256, 288, 289, 361, 392, 432, 500, 529, 625, 648, 675, 729, 800, 841, 961, 968, 972, 1024, 1125, 1152, 1323, 1352, 1369, 1372, 1568, 1681, 1728, 1849, 2000, 2209, 2312, 2401, 2592, 2809, 2888, 3087, 3200, 3267, 3481
Offset: 1

Views

Author

Amiram Eldar, Nov 09 2024

Keywords

Comments

Each term can be represented in a unique way as m * p^(2*k), k >= 1, where m is a cubefull exponentially odd number (A335988) and p is a prime that does not divide m.
Powerful numbers k such that A350388(k) is a prime power with an even positive exponent (A056798 \ {1}).

Crossrefs

Intersection of A001694 and A377816.
Subsequence of A377819.

Programs

  • Mathematica
    With[{max = 3500}, Select[Union@ Flatten@ Table[i^2 * j^3, {j, 1, max^(1/3)}, {i, 1, Sqrt[max/j^3]}], Count[FactorInteger[#][[;; , 2]], _?EvenQ] == 1 &]]
  • PARI
    is(k) = if(k == 1, 0, my(e = factor(k)[, 2]); vecmin(e) > 1 && #select(x -> !(x%2), e) == 1);

Formula

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/(p*(p^2-1))) * Sum_{p prime} (p/(p^3-p+1)) = 0.61399274770712398109... .

A382061 Numbers whose number of divisors is divisible by their number of unitary divisors.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 72, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 88, 89, 91, 93, 94, 95, 96, 97
Offset: 1

Views

Author

Amiram Eldar, Mar 14 2025

Keywords

Comments

Numbers k such that A034444(k) | A000005(k).
The criterion according to which a number belongs to this sequence depends only on the prime signature of this number: if {e_1, e_2, ... } are the exponents in the prime factorization of k then k is a term if and only if A000005(k)/A344444(k) = Product_{i} (e_i + 1)/2 is an integer.
The exponentially odd numbers (A268335) are all terms, since their prime factorization has only odd exponents e_i, so (e_i + 1)/2 is an integer. This sequence first differs from A268335 at n = 53: a(53) = 72 = 2^3 * 3^2 is not a term of A268335. The next terms that are not in A268335 are 108, 200, 360, 392, 432, 500, ... .
All the squarefree numbers (A005117, which is a subsequence of A268335) are terms. These are the numbers k such that A034444(k) = A000005(k).
A number k is a term if and only if the powerful part of k, A057521(k), is a term. Therefore, the primitive terms of this sequence are the powerful terms, A382062.
The asymptotic density of this sequence is Sum_{n>=1} f(A382062(n)) = 0.72201619..., where f(n) = (n/zeta(2)) * Product_{prime p|n} (p/(p+1)).
The asymptotic density of a few subsequences can be evaluated more easily. For example:
1) Powerful numbers that are exponentially odd (A335988): When summing only over these numbers, the formula for the asymptotic density gives the density of the exponentially odd numbers: Product_{p prime} (1 - 1/(p*(p+1))) = 0.704442... (A065463).
2) Numbers of the form p^(2*k) * q^(2*m+1), where k and m >= 1, and p != q are primes: When summing only over these numbers, the density of the numbers whose powerful part is of this form is ((Sum_{p prime} p/((p^2-1)*(p+1))) * (Sum_{p prime} p^2/((p^4-1)*(p+1))) - Sum_{p prime} p^3/((p^2-1)^2*(p^2+1)*(p+1)^2)) / zeta(2) = 0.017174455422470834821... .

Crossrefs

Programs

  • Mathematica
    q[k_] := Divisible[DivisorSigma[0, k], 2^PrimeNu[k]]; Select[Range[100], q]
  • PARI
    isok(k) = {my(f = factor(k)); !(numdiv(f) % (1<
    				

Formula

2 is a term since A000005(2) = A034444(2) = 2, so 2 | 2.
24 is a term since A000005(24) = 8, A034444(24) = 4, and 4 | 8.

A368169 The number of divisors of the largest unitary divisor of n that is a cubefull exponentially odd number (A368167).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 4, 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
Offset: 1

Views

Author

Amiram Eldar, Dec 14 2023

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1 || EvenQ[e], 1, 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] == 1 || !(f[i, 2]%2), 1, f[i, 2]+1));}

Formula

Multiplicative with a(p^e) = e+1 if e is odd that is larger than 1, and 1 otherwise.
a(n) = A000005(A368167(n)).
a(n) >= 1, with equality if and only if n is in A335275.
a(n) <= A000005(n), with equality if and only if n is in A335988.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(2)^2 * Product_{p prime} (1 - 2/p^2 + 3/p^3 - 2/p^4 - 1/p^5 + 1/p^6) = 1.47140789970892803631... .

A368170 The largest cubefull exponentially odd divisor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 27, 1, 1, 1, 1, 32, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 27, 1, 8, 1, 1, 1, 1, 1, 1, 1, 32, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 27, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Dec 14 2023

Keywords

Comments

First differs from A008834 at n = 32, and from A366906 at n = 64.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e <= 2, 1, If[EvenQ[e], 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,2] <= 2, 1, if(!(f[i,2]%2), f[i,1]^(f[i, 2]-1), f[i, 1]^f[i, 2])))};

Formula

Multiplicative with a(p^e) = 1 if e <= 2, a(p^e) = p^e if e is odd and e > 1, and p^(e-1) otherwise.
a(n) = n/A368171(n).
a(n) >= 1, with equality if and only if n is cubefree (A004709).
a(n) <= n, with equality if and only if n is in A335988.

A369757 The number of divisors of the smallest cubefull exponentially odd number that is divisible by n.

Original entry on oeis.org

1, 4, 4, 4, 4, 16, 4, 4, 4, 16, 4, 16, 4, 16, 16, 6, 4, 16, 4, 16, 16, 16, 4, 16, 4, 16, 4, 16, 4, 64, 4, 6, 16, 16, 16, 16, 4, 16, 16, 16, 4, 64, 4, 16, 16, 16, 4, 24, 4, 16, 16, 16, 4, 16, 16, 16, 16, 16, 4, 64, 4, 16, 16, 8, 16, 64, 4, 16, 16, 64, 4, 16, 4
Offset: 1

Views

Author

Amiram Eldar, Jan 31 2024

Keywords

Crossrefs

Programs

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

Formula

a(n) = A000005(A356192(n)).
Multiplicative with a(p^e) = max(e,3) + 1 if e is odd, and e+2 if e is even.
a(n) >= A000005(n), with equality if and only if n is cubefull exponentially odd number (A335988).
Dirichlet g.f.: zeta(s) * zeta(2*s) * Product_{p prime} (1 + 3/p^s - 1/p^(2*s) - 3/p^(3*s) + 2/p^(4*s)).
From Vaclav Kotesovec, Feb 02 2024: (Start)
Dirichlet g.f.: zeta(s)^4 * Product_{p prime} (1 + (7*p^(2*s) + 2*p^(3*s) - 6*p^(4*s) - 7*p^s + 2) / ((p^s+1)*p^(5*s))).
Sum_{k=1..n} a(k) = c * n*log(n)^3/6 + O(n*log(n)^2), where c = Product_{p prime} (1 - (6*p^4 - 2*p^3 - 7*p^2 + 7*p - 2) / ((p+1)*p^5)) = 0.124604542136592401049820049658828040278... (End)

A369758 The sum of divisors of the smallest cubefull exponentially odd number that is divisible by n.

Original entry on oeis.org

1, 15, 40, 15, 156, 600, 400, 15, 40, 2340, 1464, 600, 2380, 6000, 6240, 63, 5220, 600, 7240, 2340, 16000, 21960, 12720, 600, 156, 35700, 40, 6000, 25260, 93600, 30784, 63, 58560, 78300, 62400, 600, 52060, 108600, 95200, 2340, 70644, 240000, 81400, 21960, 6240
Offset: 1

Views

Author

Amiram Eldar, Jan 31 2024

Keywords

Comments

First differs from A369720 at n = 16.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^If[OddQ[e], Max[e, 3] + 1, e + 2] - 1)/(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]^if(f[i,2]%2, max(f[i,2], 3) + 1, f[i,2] + 2) - 1)/(f[i,1] - 1));}
    
  • Python
    from math import prod
    from sympy import factorint
    def A369758(n): return prod((p**((3 if e==1 else e)+1+(e&1^1))-1)//(p-1) for p,e in factorint(n).items()) # Chai Wah Wu, Feb 03 2024

Formula

a(n) = A000203(A356192(n)).
Multiplicative with a(p) = p^3 + p^2 + p + 1, a(p^e) = (p^(e+1)-1)/(p-1) for an odd e >= 3, and a(p^e) = (p^(e+2)-1)/(p-1) for an even e.
a(n) >= A000203(n), with equality if and only if n is cubefull exponentially odd number (A335988).
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (1 + 1/p^(s-3) + 1/p^(s-2) + 1/p^(s-1) - 1/p^(2*s-2) - 1/p^(3*s-5) - 1/p^(3*s-4) - 1/p^(3*s-3) + 1/p^(4*s-5) + 1/p^(4*s-4)).
Sum_{k=1..n} a(k) ~ c * n^4 / 4, where c = zeta(4) * zeta(6) * Product_{p prime} (1 - 1/p^4 - 1/p^6 + 1/p^10 + 1/p^11 - 1/p^13) = 1.00040193512214077945... .
Equivalently, c = Product_{p prime} (1 + 1/(p^3*(p^4 - 1)*(p^4 + p^2 + 1))). - Vaclav Kotesovec, Feb 02 2024

A369759 The sum of unitary divisors of the smallest cubefull exponentially odd number that is divisible by n.

Original entry on oeis.org

1, 9, 28, 9, 126, 252, 344, 9, 28, 1134, 1332, 252, 2198, 3096, 3528, 33, 4914, 252, 6860, 1134, 9632, 11988, 12168, 252, 126, 19782, 28, 3096, 24390, 31752, 29792, 33, 37296, 44226, 43344, 252, 50654, 61740, 61544, 1134, 68922, 86688, 79508, 11988, 3528, 109512
Offset: 1

Views

Author

Amiram Eldar, Jan 31 2024

Keywords

Crossrefs

Programs

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

Formula

a(n) = A034448(A356192(n)).
Multiplicative with a(p) = p^3 + 1, a(p^e) = p^e + 1 for an odd e >= 3, and a(p^e) = p^(e+1) + 1 for an even e.
a(n) >= A034448(n), with equality if and only if n is cubefull exponentially odd number (A335988).
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (1 + 1/p^(s-3) - 1/p^(2*s-2) - 1/p^(3*s-5) + 1/p^(4*s-5) - 1/p^(4*s-3)).
Sum_{k=1..n} a(k) ~ c * n^4 / 4, where c = (zeta(4)*zeta(6)/zeta(2)) * Product_{p prime} (1 - 1/p^6 + 1/p^11 - 1/p^12) = 0.65813930591740259189... .

A381825 Odd cubefull exponentially odd numbers: numbers whose prime factorization has only odd primes and odd exponents that are larger than 1 (except for 1 whose prime factorization is empty).

Original entry on oeis.org

1, 27, 125, 243, 343, 1331, 2187, 2197, 3125, 3375, 4913, 6859, 9261, 12167, 16807, 19683, 24389, 29791, 30375, 35937, 42875, 50653, 59319, 68921, 78125, 79507, 83349, 84375, 103823, 132651, 148877, 161051, 166375, 177147, 185193, 205379, 226981, 273375, 274625
Offset: 1

Views

Author

Amiram Eldar, Mar 08 2025

Keywords

Comments

Differs from its subsequence A369118 by having the terms 1, 19683 = 3^9, 1953125 = 5^9, 2460375 = 3^9 * 5^3, 6751269 = 3^9 * 7^3, 14348907 = 3^15, ... .

Crossrefs

Intersection of A005408 and A335988.
Intersection A036966 and A376218.
Subsequence of A381824.
A369118 is a subsequence.
Cf. A065487.

Programs

  • Mathematica
    Join[{1}, Select[Range[3, 300000, 2], AllTrue[FactorInteger[#][[;;, 2]],  #1 > 1 && OddQ[#1] &] &]]
  • PARI
    isok(k) = k == 1 || (k % 2 && #select(x -> (x == 1) || !(x % 2), factor(k)[, 2]) == 0);

Formula

Sum_{n>=1} 1/a(n) = Product_{prime p >= 3} (1 + 1/(p*(p^2-1))) = (6/7) * A065487 = 1.05539241333308876809... .

A368172 The number of divisors of the largest cubefull exponentially odd divisor of n (A368170).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Dec 14 2023

Keywords

Comments

First differs from A365487 at n = 32.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e <= 2, 1, If[EvenQ[e], e, 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, 1, if(!(f[i,2]%2), f[i, 2], f[i, 2]+1)))};

Formula

a(n) = A000005(A368170(n)).
Multiplicative with a(p^e) = 1 if e <= 2, a(p^e) = e+1 if e is odd and e > 1, and a(p^e) = e otherwise.
a(n) >= 1, with equality if and only if n is cubefree (A004709).
a(n) <= A000005(n), with equality if and only if n is in A335988.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(2) * Product_{p prime} (1 - 1/p^2 + 3/p^3 - 1/p^5) = 1.69824776889117043774... .

A374291 Squares of powerful numbers.

Original entry on oeis.org

1, 16, 64, 81, 256, 625, 729, 1024, 1296, 2401, 4096, 5184, 6561, 10000, 11664, 14641, 15625, 16384, 20736, 28561, 38416, 40000, 46656, 50625, 59049, 65536, 82944, 83521, 104976, 117649, 130321, 153664, 160000, 186624, 194481, 234256, 250000, 262144, 279841, 331776
Offset: 1

Views

Author

Amiram Eldar, Jul 02 2024

Keywords

Comments

First differs from A340588 at n = 12.
4-full (or 3-full) squares.
Numbers whose exponents in their prime factorization are all even numbers >= 4.
This sequence is closed under multiplication.
The sequence {A000290(n)*A078615(A000290(n)), n>=1} is a permutation of this sequence, and the sequence {a(n)/A078615(a(n)), n>=1} is a permutation of {A000290(n), n>=1}.
The sequence {A335988(n)*A007947(A335988(n)), n>=1} is a permutation of this sequence, and the sequence {a(n)/A007947(a(n)), n>=1} is a permutation of A335988.

Crossrefs

Intersection of A000290 and A036967 (or A036966).
Intersection of A000290 and A337050.
Subsequence of A322449.

Programs

  • Mathematica
    powQ[n_] := n==1 || AllTrue[FactorInteger[n][[;; , 2]], # > 1 &]; Select[Range[600], powQ]^2
  • PARI
    is(k) = issquare(k) && ispowerful(sqrtint(k));
    
  • Python
    from math import isqrt
    from sympy import mobius, integer_nthroot
    def A374291(n):
        def squarefreepi(n):
            return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
        def bisection(f, kmin=0, kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c, l = n+x, 0
            j = isqrt(x)
            while j>1:
                k2 = integer_nthroot(x//j**2, 3)[0]+1
                w = squarefreepi(k2-1)
                c -= j*(w-l)
                l, j = w, isqrt(x//k2**3)
            c -= squarefreepi(integer_nthroot(x, 3)[0])-l
            return c
        return bisection(f,n,n)**2 # Chai Wah Wu, Sep 10 2024

Formula

a(n) = A000290(A001694(n)) = A001694(n)^2.
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/(p^2*(p^2-1))) = zeta(4)*zeta(6)/zeta(12) = 15015/(1382*Pi^2) = 1.10082313486953808844... .
Sum_{n>=1} 1/a(n)^s = Product_{p prime} (1 + 1/(p^(2*s)*(p^(2*s)-1))) = zeta(4*s)*zeta(6*s)/zeta(12*s), for s > 1/4.
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