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

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• ... 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
>
>
• 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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