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Search: id:A065081
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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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+0
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
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Only 3d = 11b has an alternating sum of 0.
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LINKS
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Harry J. Smith, Table of n, a(n) for n=1,...,1000
William Paulsen, wpaulsen(AT)csm.astate.edu, Partitioning the [prime] maze
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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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PROGRAM
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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) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Oct 06 2009]
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CROSSREFS
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Cf. A065359.
Sequence in context: A064693 A072085 A054868 this_sequence A025909 A025899 A025869
Adjacent sequences: A065078 A065079 A065080 this_sequence A065082 A065083 A065084
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KEYWORD
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base,easy,sign
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AUTHOR
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Robert G. Wilson v (rgwv(AT)rgwv.com), Nov 09 2001
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