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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.

%I A306916 #31 May 22 2025 10:21:48
%S A306916 1,4,8,9,32,32,128,25,27,128,2048,72,8192,512,256,49,131072,108,
%T A306916 524288,288,1024,8192,8388608,200,243,32768,125,1152,536870912,1024,
%U A306916 2147483648,121,16384,524288,4096,243,137438953472,2097152,65536,800,2199023255552,4096
%N 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.
%H A306916 Matthias Butterweck, <a href="/A306916/b306916.txt">Table of n, a(n) for n = 1..3322</a>
%F A306916 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
%e A306916 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).
%t A306916 f[p_, e_] := Prime[e]^p; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 42] (* _Amiram Eldar_, Sep 14 2023 *)
%o A306916 (Python)
%o A306916 from functools import reduce
%o A306916 from operator import mul
%o A306916 from sympy import factorint, prime
%o A306916 def a(n):
%o A306916     return 1 if n == 1 else reduce(mul, (prime(k)**p for p,k in factorint(n).items()))
%o A306916 (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
%Y A306916 Cf. A008477 (replace p^k with k^p).
%K A306916 nonn,easy,mult
%O A306916 1,2
%A A306916 _Matthias Butterweck_, Mar 16 2019
%E A306916 Keyword mult added by _Rémy Sigrist_, Mar 17 2019