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Theorem: a(n)=a(n-1)+1+Sum_{k=1..n-1} d(k)

proof (by induction):

(Base: a(1) = 1 trivially.)

suppose we know a(n-1). 

we add a new term n to our set {1..n-1}. 

a(n) = all (arith.) sequences without new term + sequences with new term 
     = a(n-1) + (single seq. consisting of just the new term) + (sequences of len > 1 which end with the new term) 
     = a(n-1)+ 1 + (len > 1,terms ending with new term) 

To get the last term, loop over difference between max and min element of the subsets of {1...n}. 

Every divisor of a given difference i (from 1 to n-1 inclusive) gives us an arith. seq. 

Thus, the last term is the sum over i of d(i) => Sum_{i=1..n-1} d(k) 

- Daniel Hoying, May 19, 2020

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