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Re: [PrimeNumbers] Product identities

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  • Sebastian Martin
    I believe that I committed a mistake on having compiled (I use MATHEMATICA) the functions of the second identity. I am afraid that this second identity is not
    Message 1 of 4 , Dec 31, 2006
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      I believe that I committed a mistake on having
      compiled (I use MATHEMATICA) the functions of the
      second identity.

      I am afraid that this second identity is not true.

      I wait for your commentaries on the first and third
      identity.

      Sincerely

      Sebastián Martin Ruiz

      --- David Cleaver <wraithx@...> escribió:

      >
      >
      > Note:
      > LHS stands for Left Hand Side
      > RHS stands for Right Hand Side
      >
      > Sebastian Martin wrote:
      > >
      > >
      > > Dear colleagues:
      > >
      > > I have found some identities:
      > >
      > > 1) Product[k^Floor[n/k],{k,1,n}]=
      > > Product[Floor[n/k]!,{k,1,n}]
      >
      > I was able to confirm this for n=5 and n=6.
      >
      > >
      > > 2) Product[Floor[n/k]^k,{k,1,n}]=
      > > Product[k!,{k,1,n}]
      >
      > I was unable to confirm this, ie:
      >
      > For n = 5
      > LHS = 20
      > RHS = 34560
      >
      > For n = 6
      > LHS = 432
      > RHS = 24883200
      >
      > Did you perhaps type the wrong formula in?
      >
      > >
      > >
      >
      3)(Product[(i^j)*(j^i),{i,1,n},{j,1,n}]/Product[i!*j!,{i,1,n},{j,1,n}])^(1/n)=
      > > Product[C(n,i),{i,0,n}]
      >
      > I confirmed this for n = 5.
      > (The products in the LHS grow very rapidly!)
      > (For n=5 the Numerator was 2.37*10^65 and the
      > Denominator was 2.43*10^45)
      >
      > >
      > > OEIS A001142
      > >
      > > This identities are known?
      > >
      > > Sincerely
      > >
      > > Sebastian Martin Ruiz
      >
      > Good job on finding those identities, I personally
      > do not know if they
      > have been found before or not, someone else will
      > have to speak to that.
      > I just wanted to let you know about the problem
      > with the 2nd Identity
      > that you found.
      >
      > -David C.
      >


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    • Sebastian Martin
      Sebastian Martin escribió: Dear colleagues: I have found two identities: 1) Product[k^Floor[n/k],{k,1,n}]=
      Message 2 of 4 , Jan 17, 2007
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        Sebastian Martin <sebi_sebi@...> escribió:

        Dear colleagues:

        I have found two identities:

        1) Product[k^Floor[n/k],{k,1,n}]=
        Product[Floor[n/k]!,{k,1,n}]


        2)(Product[(i^j)*(j^i),{i,1,n},{j,1,n}]/Product[i!*j!,{i,1,n},{j,1,n}])^(1/n)=
        Product[C(n,i),{i,0,n}]

        OEIS A001142

        This identities are known?

        Sincerely

        Sebastian Martin Ruiz






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