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.

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

Author

Matthias Butterweck, Mar 16 2019

Keywords

• nonn
• easy
• mult

Examples

			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).
		

Links

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

Crossrefs

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

Programs

Mathematica

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

PARI

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

Python

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()))

Formula

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

Extensions

Keyword mult added by Rémy Sigrist, Mar 17 2019