--- 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