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## Sum of products of digit and their position in integers

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• Hi, The sum of the digits times the position of the digits in n appears to produce the integers which include the prime numbers. Let n = d(1)d(2)...d(n) where
Message 1 of 2 , Feb 6, 2009
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Hi,
The sum of the digits times the position of the digits in n appears
to produce the integers which include the prime numbers.
Let n = d(1)d(2)...d(n) where d(1),d(2),...d(n) are the digits of n.

We can also write n = d(1)10^n+d(2)^(n-1)+...+d(1)10 + d(0) and

Sumdigpos(n) = d(1)*1+d(2)*2+...+d(n)*n.

I want to prove Sum(n) generates the integers and therefore all prime
numbers as n approaches infinity.

For the first 200 numbers we have Sumdigpos

0,1,2,3,4,5,6,7,8,9,1,3,5,7,9,11,13,15,17,19,2,4,6,8,10,12,14,16,18,
20,3,5,7,9,11,13,15,17,19,21,4,6,8,10,12,14,16,18,20,22,5,7,9,11,13,
15,17,19,21,23,6,8,10,12,14,16,18,20,22,24,7,9,11,13,15,17,19,21,23,
25,8,10,12,14,16,18,20,22,24,26,9,11,13,15,17,19,21,23,25,27,1,4,7,
10,13,16,19,22,25,28,3,6,9,12,15,18,21,24,27,30,5,8,11,14,17,20,23,
26,29,32,7,10,13,16,19,22,25,28,31,34,9,12,15,18,21,24,27,30,33,36,
11,14,17,20,23,26,29,32,35,38,13,16,19,22,25,28,31,34,37,40,15,18,21,
24,27,30,33,36,39,42,17,20,23,26,29,32,35,38,41,44,19,22,25,28,31,34,
37,40,43,46,2

This sequence includes the integers 1 to 44

Maybe someone here can muster a proof.

For example Sums of first n integers = 1,3,6,10,15,21,28,36 do not
readily produce the integers but if we consider factors of these
numbers, the sums probably do.

It is interesting Sumdigpos(n) = n for only n=1,2,3,4,5,6,7,8,9,19.
This is easy to prove. For 19 = 1*1+9*2 = 19. We write a 2 digit
number as 10a+b. Then 10a+b=a+2b or 9a = b and b=9 and a = 1. I used
induction to prove for numbers greater than 2 digits. Maybe there is
a proof without induction.

Enjoy,
Cino Hilliard
• ... That s unconventional. Normally indices will match the power, so d(n)*10^n + ... + d(0). ... Ug. That s not what I imagined. I imagined the above in my
Message 2 of 2 , Feb 7, 2009
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--- On Sat, 2/7/09, Cino Hilliard <hillcino368@...> wrote:
> The sum of the digits times the position of the digits in n
>
> appears to produce the integers which include the prime numbers.
> Let n = d(1)d(2)...d(n) where d(1),d(2),...d(n) are the
> digits of n.
>
> We can also write n = d(1)10^n+d(2)^(n-1)+...+d(1)10 + d(0)

That's unconventional. Normally indices will match the power, so d(n)*10^n + ... + d(0).

> and
>
> Sumdigpos(n) = d(1)*1+d(2)*2+...+d(n)*n.

Ug. That's not what I imagined. I imagined the above in my notation, or d(n)*1 + ... + d(1)*n in your notation, such that the units are always weighted 1.

> I want to prove Sum(n) generates the integers and therefore
> all prime numbers as n approaches infinity.

To yield m, use a m-1 digit number, with a 1 in positions m-1 (worth 1), and
a 1 in position 1 (worth m-1), and zeroes elsewhere. Sum = m.

Phil
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