(no subject)

[David Wasserman]

Dear Amarnath Murthy,
    These sequences are already in the database.  (1) is all ones, and (2) is the partial sums of tau(n).

Proof:
The divisors of 1, 2, ..., m are
m ones,
floor(m/2) twos,
floor(m/3) threes,
...
and one m.

For any k > 1, the divisors of k, k+1, k+2, ..., k+m-1 include
m ones,
at least floor(m/2) twos,
at least floor(m/3) threes,
...
one m,
and some larger numbers, such as k+m-1.

 - David

>Dear Neil and seq fans,
>Does this idea make any sense?
>
>1.Seq(1); Start of the least set of n successive
>numbers whose sum of the number of divisors ( tau(r)
>)is the least.
>1,1,1,1,1,...
>a(1) = 1, tau(1) = 1,a(2) = 1, tau(1) +tau(2) = 3,
>a(3) = 1, tau(1) +tau(2) + tau(3) = 5, etc.
>What is the first term >1?
>
>2. Seq(2); The least possible sum of the number of
>divisors of n successive numbers.
>1,3,5,8,10,...
>
>Similar sequences can be thought of for other
>number-theoretic functions like sigma(n) etc.
>thanks
>regards
>amarnath murthy
>
>
>
>
>	
>		
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