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 A065081 Alternating bit sum (A065359) for n-th prime p: replace 2^k with (-1)^k in binary expansion of p. 2
 -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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Only 3d = 11b has an alternating sum of 0. LINKS Harry J. Smith, Table of n, a(n) for n=1,...,1000 William Paulsen, wpaulsen(AT)csm.astate.edu, Partitioning the [prime] maze EXAMPLE The sixth prime is 13d = 1101b -> -(1)+(1)-(0)+(1) = 1 = a(6) MATHEMATICA 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} ] PROG (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) ) } - \\ Harry J. Smith, Oct 06 2009 (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 CROSSREFS Cf. A065359. Sequence in context: A054868 A352517 A347981 * A366643 A212185 A270928 Adjacent sequences: A065078 A065079 A065080 * A065082 A065083 A065084 KEYWORD base,easy,sign AUTHOR Robert G. Wilson v, Nov 09 2001 STATUS approved

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Last modified December 7 11:41 EST 2023. Contains 367656 sequences. (Running on oeis4.)