

A212496


a(n) = Sum_{k=1..n} (1)^(kOmega(k)) with Omega(k) the total number of prime factors of k (counted with multiplicity).


1



1, 2, 1, 0, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 5, 6, 5, 6, 7, 6, 7, 6, 7, 6, 5, 6, 5, 6, 7, 8, 7, 8, 9, 8, 9, 8, 9, 10, 11, 10, 9, 8, 7, 6, 7, 8, 7, 8, 7, 8, 9, 10
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OFFSET

1,2


COMMENTS

On May 16 2012, ZhiWei Sun conjectured that a(n) is positive for each n > 4. He has verified this for n up to 10^10, and shown that the conjecture implies the Riemann Hypothesis. Moreover, he guessed that a(n) > sqrt(n) for any n > 324 (and also a(n) < sqrt(n)*log(log(n)) for n > 5892); this implies that the sequence contains all natural numbers.
Sun also conjectured that b(n) = Sum_{k=1..n} (1)^(kOmega(k))/k < 0 for all n=1,2,3,..., and verified this for n up to 2*10^9. Moreover, he guessed that b(n) < 1/sqrt(n) for all n > 1, and b(n) > log(log(n))/sqrt(n) for n > 2008.


LINKS



EXAMPLE

We have a(4)=0 since (1)^(1Omega(1)) + (1)^(2Omega(2)) + (1)^(3Omega(3)) + (1)^(4Omega(4)) = 1  1 + 1 + 1 = 0.


MAPLE

ListTools:PartialSums([seq((1)^(knumtheory:bigomega(k)), k=1..60)]); # Robert Israel, Jan 03 2023


MATHEMATICA

PrimeDivisor[n_]:=Part[Transpose[FactorInteger[n]], 1]
Omega[n_]:=If[n==1, 0, Sum[IntegerExponent[n, Part[PrimeDivisor[n], i]], {i, 1, Length[PrimeDivisor[n]]}]]
s[0]=0
s[n_]:=s[n]=s[n1]+(1)^(nOmega[n])
Do[Print[n, " ", s[n]], {n, 1, 100000}]
Accumulate[Table[(1)^(nPrimeOmega[n]), {n, 1000}]] (* Harvey P. Dale, Oct 07 2013 *)


PROG

(Python)
from functools import reduce
from operator import ixor
from sympy import factorint
def A212496(n): return sum(1 if reduce(ixor, factorint(i).values(), i)&1 else 1 for i in range(1, n+1)) # Chai Wah Wu, Jan 03 2023


CROSSREFS



KEYWORD

sign,nice


AUTHOR



STATUS

approved



