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A065081
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Alternating bit sum (A065359) for n-th prime p: replace 2^k with (-1)^k in binary expansion of p.
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2
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-1, 0, 2, 1, -1, 1, 2, 1, 2, 2, 1, 1, -1, -2, -1, 2, -1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, -1, 1, 2, 1, -1, -1, -2, 2, 1, 1, -2, -1, -1, -1, 1, -1, 1, 2, 1, 1, 1, -1, 1, -1, -1, 1, -1, 2, 2, 2, 1, 4, 2, 1, 2, 1, 2, 1, 2, 1, 4, 2, 4, 2, 2, 1, 4, 1, 2, 2, 1, 2, 1, -1, 1, -1, 1, 1, -1, 2, 1, 2, 1, 2, 2, 1, -1, 1, 2, 2, -1, -2, 1
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OFFSET
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1,3
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COMMENTS
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Only 3d = 11b has an alternating sum of 0.
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LINKS
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EXAMPLE
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The sixth prime is 13d = 1101b -> -(1)+(1)-(0)+(1) = 1 = a(6)
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MATHEMATICA
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f[n_] := (d = Reverse[ IntegerDigits[n, 2]]; l = Length[d]; s = 0; k = 1; While[k < l + 1, s = s - (-1)^k*d[[k]]; k++ ]; s); Table[ Prime[ f[n]], {n, 1, 100} ]
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PROG
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(PARI)
baseE(x, b)=
{
local(d, e=0, f=1);
while (x>0, d=x-b*(x\b); x\=b; e+=d*f; f*=10);
return(e)
}
SumAD(x)=
{
local(a=1, s=0);
while (x>9, s+=a*(x-10*(x\10)); x\=10; a=-a);
return(s + a*x)
}
{ for (n=1, 1000, p=prime(n);
s=SumAD(baseE(p, 2)); write("b065081.txt", n, " ", s) )
(PARI)
f(p)=
{
v=binary(p);
L=#v; u=1; s=0;
forstep(k=L, 1, -1, if(v[k]==1, s+=u); u=-u; );
return(s)
};
for(n=1, 100, p=prime(n); an=f(p); print1(an, ", ")) \\ Washington Bomfim, Jan 16 2011
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CROSSREFS
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KEYWORD
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base,easy,sign
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AUTHOR
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STATUS
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approved
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