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