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A245615
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Sum_{lpf(j)=11, 11<j<=11*prime(n)}(-1)^(A000120(j)+1), where lpf=least prime factor.
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1
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1, 2, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 5, 6, 7, 6, 7, 8, 9, 8, 7, 8, 9, 10, 10, 11, 12, 11, 13, 12, 13, 12, 13, 13, 14, 13, 15, 16, 17, 18, 16, 16, 17, 16, 15, 16, 17, 17, 17, 18, 17, 18, 19, 20, 21, 23, 23, 22, 21, 22, 23, 24, 22, 23, 22, 23, 23, 22, 22, 21
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OFFSET
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5,2
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COMMENTS
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Since lpf(j)=11, the smallest prime divisor of j is always 11, while every other prime divisor of j, if one exists, is >=11. Since j>11, then the least suitable j>=121. Offset 5 is natural since prime(5)=11.
This is a special Newman's sum.
Conjecture: All terms are nonnegative.
A similar conjecture arises by changing 11 in the definition of the sequence by some other primes: 19,41,67,107,173,...
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LINKS
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EXAMPLE
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For n=5, j=121. So we have a(5)=(-1)^(A000120(121)+1)=(-1)^(5+1)=1;
for n=6, there is a second summand with j=11*13.
So a(6)=1+(-1)^(A000120(143)+1)=1+(-1)^(5+1)=2, etc.
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MAPLE
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N:= 100: # to get terms up to a(N)
Ks:= select(t -> min(numtheory:-factorset(t))>=11, {$11..ithprime(N)}):
f:= proc(n) local j, Kn, p;
p:= ithprime(n);
Kn:= map(`*`, select(`<=`, Ks, p), 11);
add(2*(convert(convert(j, base, 2), `+`) mod 2)-1, j=Kn);
end proc:
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PROG
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(PARI) a(n) = sum(j=12, 11*prime(n), if (factor(j)[1, 1]==11, (-1)^(hammingweight(j)+1), 0)); \\ Michel Marcus, Aug 04 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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