A353397 Replace prime(k) with prime(2^k) in the prime factorization of n.
1, 3, 7, 9, 19, 21, 53, 27, 49, 57, 131, 63, 311, 159, 133, 81, 719, 147, 1619, 171, 371, 393, 3671, 189, 361, 933, 343, 477, 8161, 399, 17863, 243, 917, 2157, 1007, 441, 38873, 4857, 2177, 513, 84017, 1113, 180503, 1179, 931, 11013, 386093, 567, 2809, 1083
Offset: 1
Examples
The terms together with their prime indices begin: 1: {} 3: {2} 7: {4} 9: {2,2} 19: {8} 21: {2,4} 53: {16} 27: {2,2,2} 49: {4,4} 57: {2,8} 131: {32} 63: {2,2,4}
Links
- Amiram Eldar, Table of n, a(n) for n = 1..397 (calculated using the b-file at A033844)
Crossrefs
Programs
-
Mathematica
primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; Table[Times@@Prime/@(2^primeMS[n]),{n,100}]
-
PARI
a(n) = my(f=factor(n)); for(k=1, #f~, f[k,1] = prime(2^primepi(f[k,1]))); factorback(f); \\ Michel Marcus, May 20 2022
-
Python
from math import prod from sympy import prime, primepi, factorint def A353397(n): return prod(prime(2**primepi(p))**e for p, e in factorint(n).items()) # Chai Wah Wu, May 20 2022
Formula
If n = prime(e_1)...prime(e_k), then a(n) = prime(2^(e_1))...prime(2^(e_k)).
Sum_{n>=1} 1/a(n) = 1/Product_{k>=1} (1 - 1/prime(2^k)) = 1.90812936178871496289... . - Amiram Eldar, Dec 09 2022