A008477 -id:A008477 - cp's OEIS frontend

A381614 If n = Product (p_j^k_j) then a(n) = Product (max(p_j, k_j)), with a(1) = 1.

1, 2, 3, 2, 5, 6, 7, 3, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 9, 5, 26, 3, 14, 29, 30, 31, 5, 33, 34, 35, 6, 37, 38, 39, 15, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 6, 55, 21, 57, 58, 59, 30, 61, 62, 21, 6, 65, 66, 67, 34, 69, 70, 71, 9, 73

Offset: 1

Paolo Xausa, Mar 01 2025

			a(18) = 6 because 18 = 2^1*3^2, max(2,1) = 2, max(3,2) = 3 and 2*3 = 6.
a(300) = 30 because 300 = 2^2*3^1*5^2, max(2,2) = 2, max(3,1) = 3, max(5,2) = 5 and 2*3*5 = 30.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A008473, A008477, A035306, A065463, A369008, A381588, A381613.

A381614[n_] := Times @@ Max @@@ FactorInteger[n];
Array[A381614, 100]

a(n) = my(f=factor(n)); prod(i=1, #f~, max(f[i,1], f[i,2])); \\ Michel Marcus, Mar 02 2025

a(p) = p, for p prime.

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = A065463 * Product_{p prime} (1 + 1/((p^2-1)*(p^2+p-1)*p^(2*p-2))) = 0.71628338157754073004... . - Amiram Eldar, Mar 07 2025

A306916 a(1)=1; for n > 1 replace each p^k in the prime factorization of n with prime(k)^p where prime(k) denotes the k-th prime number.

Original entry on oeis.org

1, 4, 8, 9, 32, 32, 128, 25, 27, 128, 2048, 72, 8192, 512, 256, 49, 131072, 108, 524288, 288, 1024, 8192, 8388608, 200, 243, 32768, 125, 1152, 536870912, 1024, 2147483648, 121, 16384, 524288, 4096, 243, 137438953472, 2097152, 65536, 800, 2199023255552, 4096

Offset: 1

Matthias Butterweck, Mar 16 2019

			For n = 12288 = 2^12 * 3^1 one gets a(12288) = prime(12)^2 * prime(1)^3 = 37^2 * 2^3 = 10952 (12288 is the smallest n > a(n) that is not a power of 2).

Matthias Butterweck, Table of n, a(n) for n = 1..3322

Cf. A008477 (replace p^k with k^p).

f[p_, e_] := Prime[e]^p; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 42] (* Amiram Eldar, Sep 14 2023 *)

a(n) = my(f=factor(n)); for (k=1, #f~, my(p = f[k,1]); f[k,1] = prime(f[k,2]); f[k,2] = p); factorback(f); \\ Michel Marcus, Mar 16 2019

from functools import reduce
from operator import mul
from sympy import factorint, prime
def a(n):
    return 1 if n == 1 else reduce(mul, (prime(k)**p for p,k in factorint(n).items()))

Sum_{n>=1} 1/a(n) = Product_{m>=1} (1 + Sum_{k>=1} 1/prime(k)^prime(m)) = Product_{m>=1} (1 + P(prime(m))) = 1.78279963787539257806..., where P(s) is the prime zeta function. - Amiram Eldar, Sep 14 2023, Oct 24 2023

Keyword mult added by Rémy Sigrist, Mar 17 2019

A323785 Iteratively swap prime factors of n with their exponent (unless the exponent is 1), until a cycle is reached, and then record the largest term in the cycle.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 9, 10, 11, 12, 13, 14, 15, 16, 17, 16, 19, 20, 21, 22, 23, 27, 32, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 32, 37, 38, 39, 45, 41, 42, 43, 44, 45, 46, 47, 48, 128, 32, 51, 52, 53, 54, 55, 63, 57, 58, 59, 60, 61, 62, 63, 32, 65, 66, 67

Offset: 1

Gabriel Antonius, Aug 31 2019

nonn

			For n=196, the iterative procedure evolves as follows: 196 = 2^2 * 7^2 --> 2^2 * 2^7 = 2^9 --> 9^2 = 3^4 --> 4^3 = 2^6 --> 6^2 = 3^2 * 2^2 --> 2^3 * 2^2 = 2^5 --> 5^2 --> 2^5 = 32.
Since the iterative procedure stabilizes in the cycle 2^5=32 <--> 5^2=25, of length 2, the largest ,member of which is 32, we get a(196)=32.

Gabriel Antonius, Table of n, a(n) for n = 1..10000

Cf. A008477.

p[x_, y_] := If[y > 1, y^x, x^y]; f[n_] := Times @@ (p[First[#], Last[#]] & /@ FactorInteger[n]); a[n_] := Module[{k = n, s = {k}}, While[! MemberQ[s, (k = f[k])], AppendTo[s, k]]; ind = Position[s, ?(# == k &)][[1, 1]]; Max[s[[ind ;; -1]]]]; Array[a, 57] (* _Amiram Eldar, Sep 02 2019 *)

def a(n):
    np = swap_prime_factors_with_exponents(n)
    npp = swap_prime_factors_with_exponents(np)
    return max(n, np) if n == npp else a(npp)
def swap_prime_factors_with_exponents(n):
    np = 1
    for p in primes:
        q = 0
        while n % p == 0:
            q += 1
            n = n // p
        if q > 1: np *= q ** p
        if q == 1: np *= p
    if n > 1: np *= n
    return np
N = 200
primes = []
for n in range(2, int(N ** (1/2)) + 1):
    if all(n % p > 0 for p in primes):
        primes.append(n)
for n in range(1,N+1):
    print(n, a(n))

A039784 phi(n) is equal to the product of the dual prime-power components of n.

Original entry on oeis.org

1, 2, 12, 48, 5832

Offset: 1

Olivier Gérard

nonn

			phi(5832)=1944, 5832=2^3*3^6, (3^2)*(6^3)=1944.

Cf. A000010, A008477.

No others < 7*10^7. - Naohiro Nomoto, Jun 23 2001

cp's OEIS Frontend

A381614 If n = Product (p_j^k_j) then a(n) = Product (max(p_j, k_j)), with a(1) = 1.

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A306916 a(1)=1; for n > 1 replace each p^k in the prime factorization of n with prime(k)^p where prime(k) denotes the k-th prime number.

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A323785 Iteratively swap prime factors of n with their exponent (unless the exponent is 1), until a cycle is reached, and then record the largest term in the cycle.

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Programs

Mathematica

Python

A039784 phi(n) is equal to the product of the dual prime-power components of n.

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Author

Keywords

Examples

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Extensions