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