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Fwd: [PrimeNumbers] Sum of products of digit and their position in integers

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  • Maximilian Hasler
    ... From: Maximilian Hasler Date: Sat, Feb 7, 2009 at 7:25 AM Subject: Re: [PrimeNumbers] Sum of products of digit and their
    Message 1 of 2 , Feb 7, 2009
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      ---------- Forwarded message ----------
      From: Maximilian Hasler <maximilian.hasler@...>
      Date: Sat, Feb 7, 2009 at 7:25 AM
      Subject: Re: [PrimeNumbers] Sum of products of digit and their
      position in integers
      To: cino hilliard <hillcino368@...>

      lets call a(n) the n-th term of your sequence

      a(1)=1, a(2)=2, a(11)=3, a(101)=4, a(1001)=5 etc.

      you get each number n>1 as a(10^(n-2)+1)-th term of the sequence.

      Maximilian

      On Sat, Feb 7, 2009 at 2:05 AM, cino hilliard <hillcino368@...> wrote:
      >
      >
      >> Date: Sat, 7 Feb 2009 01:58:13 -0400
      >> Subject: Re: [PrimeNumbers] Sum of products of digit and their position in
      >> integers
      >> From: maximilian.hasler@...
      >> To: hillcino368@...
      >>
      >> >> I think every integer n>2 appears as the n-1 digit number 10....01,
      >> >> doesn't it ?
      >> >
      >> > Sorry I do not understand this.
      >>
      >> The n-1 digit number 10...01 becomes
      >> d(1)*1 + ... + d(n-1)*(n-1) = 1*1 + 1*(n-1) = n
      >
      > Sorry, the old man is slipping. My org formula confused the integers
      > n = 1,2,3,4,5,...,1234567, etc to be multiplied and summed
      > with the the number of digits of n. They ain't the same.
      >
      > It should be
      > Let n = d(1)d(2)...d(m) where d(1),d(2),...d(m) are the digits of n and m is
      > the
      > number of digits n has.
      >
      > We can also write n = d(1)10^m+d(2)10^(m-1)+...+d(1)10 + d(0) and
      >
      > Sumdigpos(n) = d(1)*1+d(2)*2+...+d(m)*m.
      >
      > For n=1234567,m=7
      > For n=9999999,m=7
      >
      >
      > my formula is
      > sumdigpos = 1*1 + 2*2 + 3*3 + 4*4 + 5*5 + 6*6 + 7*7 = 140
      > 140 is now in the set.
      >
      > sumdigpos = 9*1 + 9*2 + 9*3 + 9*4 + 9*5 + 9*6 + 9*7 = 252
      >
      > So 252 is in the set
      >
      > I want to prove that after eternity and a day trying greater and greater n,
      > The set of sums will include the set of positive integers.
      >
      > Sum of products of digits and their position in m in n
      >
      > Cino
      >
      >
    • cino hilliard
      Hi Maximillian and others, Sorry for the sloppy first post. I confused n with the number of digits of n.The iteral n is so often used as an index or general
      Message 2 of 2 , Feb 7, 2009
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        Hi Maximillian and others,

        Sorry for the sloppy first post. I confused n with the number of digits of n.The iteral n is so often used as an index or general exponent, it is burned in the brain. Perhaps I should have used x as the number and n as the digit count thus avoiding possible puns. see how to construct the integers from the sum a(x) now.Thanks for the guidance. Actually, there are many ways to do this. for x = 1,2,11,21,201,2001,20001,...a(x) = 1,2,3,4,5,6,7,... Constructing the integers from a set with addition only is easy.All you need is the member 1. But there are sets such as the possible
        scores in American football which have one problem:-) a(x), x prime, does not generate all integers either. However, if we allow addition in the set with members a(x prime), we can generate the integers > 1.
        Enjoy,
        Cino hilliard
        > a(1)=1, a(2)=2, a(11)=3, a(101)=4, a(1001)=5 etc.> you get each number n>1 as a(10^(n-2)+1)-th term of the sequence.> Maximilian

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