Sum of products of digit and their position in integers
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
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.
- --- On Sat, 2/7/09, Cino Hilliard <hillcino368@...> wrote:
> The sum of the digits times the position of the digits in nThat's unconventional. Normally indices will match the power, so d(n)*10^n + ... + d(0).
> 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)
> andUg. 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.
> Sumdigpos(n) = d(1)*1+d(2)*2+...+d(n)*n.
> I want to prove Sum(n) generates the integers and thereforeTo yield m, use a m-1 digit number, with a 1 in positions m-1 (worth 1), and
> all prime numbers as n approaches infinity.
a 1 in position 1 (worth m-1), and zeroes elsewhere. Sum = m.