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Re: q | L(2^n * p^m)?

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  • David Broadhurst
    ... If you omit the requirement that the index be a natural number, then a much funnier thing than that happens: 2|L(0) is not a record because 3|L(-2) ?
    Message 1 of 12 , Oct 30, 2002
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      Shane:

      > q | L(0)
      > q=2

      If you omit the requirement that the index
      be a natural number, then a much funnier
      thing than that happens:

      2|L(0) is not a record because 3|L(-2)

      ? L(n)=if(n,fibonacci(2*n)/fibonacci(n),2);
      ? if(L(-2)%3==0,print(ok))
      ok

      3|L(-2) is not a record because 7|L(-4):

      ? if(L(-4)%7==0,print(ok))
      ok

      and so on, showing that *no* record exists :-)

      > lol
      loss of logic?

      David
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