FW: [PrimeNumbers] Sum of products of digit and their position in integers
- 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.
> 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[Non-text portions of this message have been removed]