

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 famous 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

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, On a pair of zeta functions, preprint, arxiv:1204.6689.
ZhiWei Sun, On the parities of Omega(n)n, a message to Number Theory List, May 18, 2012.
ZhiWei Sun, Table of n, a(n) for n = 1..10^7 (rarcompressed)


EXAMPLE

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


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

(PARI) a(n)=sum(k=1, n, (1)^(bigomega(k)+k)) \\ Charles R Greathouse IV, Jul 31 2016


CROSSREFS

Cf. A008836, A002819.
Sequence in context: A002819 A307672 A037834 * A179765 A004074 A245615
Adjacent sequences: A212493 A212494 A212495 * A212497 A212498 A212499


KEYWORD

sign,nice


AUTHOR

ZhiWei Sun, May 19 2012


STATUS

approved



