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

1,2

Links

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

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

Example

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).

Mathematica

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

Prog

(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()))
(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

Crossrefs

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

Sequence in context: A098121 A115656 A076705 * A387201 A385517 A355063

Adjacent sequences: A306913 A306914 A306915 * A306917 A306918 A306919

Keyword

nonn,easy,mult

Author

Matthias Butterweck, Mar 16 2019

Extensions

Keyword mult added by Rémy Sigrist, Mar 17 2019

Status

approved