|
|
A268387
|
|
Bitwise-XOR of the exponents of primes in the prime factorization of n.
|
|
16
|
|
|
0, 1, 1, 2, 1, 0, 1, 3, 2, 0, 1, 3, 1, 0, 0, 4, 1, 3, 1, 3, 0, 0, 1, 2, 2, 0, 3, 3, 1, 1, 1, 5, 0, 0, 0, 0, 1, 0, 0, 2, 1, 1, 1, 3, 3, 0, 1, 5, 2, 3, 0, 3, 1, 2, 0, 2, 0, 0, 1, 2, 1, 0, 3, 6, 0, 1, 1, 3, 0, 1, 1, 1, 1, 0, 3, 3, 0, 1, 1, 5, 4, 0, 1, 2, 0, 0, 0, 2, 1, 2, 0, 3, 0, 0, 0, 4, 1, 3, 3, 0, 1, 1, 1, 2, 1, 0, 1, 1, 1, 1, 0, 5, 1, 1, 0, 3, 3, 0, 0, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
The sums of the first 10^k terms, for k = 1, 2, ..., are 11, 139, 1427, 14207, 141970, 1418563, 14183505, 141834204, 1418330298, 14183245181, ... . Apparently, the asymptotic mean of this sequence is limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1.4183... . - Amiram Eldar, Sep 10 2022
|
|
LINKS
|
|
|
FORMULA
|
a(1) = 0; for n > 1: a(n) = A067029(n) XOR a(A028234(n)). [Here XOR stands for bitwise exclusive-or, A003987.]
Other identities and observations. For all n >= 1:
From Peter Munn, Dec 02 2019 with XOR used as above: (Start)
Defined by: a(p^k) = k, for prime p; a(A059897(n,k)) = a(n) XOR a(k).
(End)
|
|
MATHEMATICA
|
Table[BitXor @@ Map[Last, FactorInteger@ n], {n, 120}] (* Michael De Vlieger, Feb 12 2016 *)
|
|
PROG
|
(Scheme, with memoization-macro definec)
(PARI) a(n) = {my(f = factor(n)); my(b = 0); for (k=1, #f~, b = bitxor(b, f[k, 2]); ); b; } \\ Michel Marcus, Feb 06 2016
(Python)
from functools import reduce
from operator import xor
from sympy import factorint
|
|
CROSSREFS
|
Differs from A136566 for the first time at n=24, where a(24) = 2, while A136566(24) = 4.
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|