[seqfan] d(a(n)) = d(a(n)-a(n-1))

[Hagen von EItzen]

> %I A160689
> %S A160689 1,2,2,2,8,2,2,8,2,2,8,2,2,8,2,21,5
> %N A160689 a(1)=1. a(n) = the smallest positive integer such that d(a(n)) = d(sum{k=1 to n} a(k)), where d(m) = the number of divisors of m. 
> %C A160689 sum{k=1 to n} a(k) = A160690(n). d(A160689(n)) = d(A160690(n)) = A160691(n). 
> %Y A160689 A160690,A160691 
> %K A160689 more,nonn
> %O A160689 1,2
>
> %I A160690
> %S A160690 1,3,5,7,15,17,19,27,29,31,39,41,43,51,53,74,79
> %N A160690 a(1)=1. a(n) = the smallest integer > a(n-1) such that d(a(n)) = d(a(n)-a(n-1)), where d(m) = the number of divisors of m. 
> %C A160690 A160690(n)-A160690(n-1) = A160689(n), for n >= 2. d(A160689(n)) = d(A160690(n)) = A160691(n). 
> %Y A160690 A160689,A160691 
> %K A160690 more,nonn
> %O A160690 1,2
>
> %I A160691
> %S A160691 1,2,2,2,4,2,2,4,2,2,4,2,2,4,2,4,2
> %N A160691 a(n) = the number of divisors of A160689(n) = the number of divisors of A160690(n). 
> %Y A160691 A160689,A160690 
> %K A160691 more,nonn
> %O A160691 1,2
>
> Does every positive integer occur in A160691?

Among the first 30000 entries, we find
A160691(n)=1  once
A160691(n)=2  4593 times
A160691(n)=4  24015 times
A160691(n)=6  712 times
A160691(n)=8  678 times
A160691(n)=12 once

It appears that a nontrivial odd entry is quite unlikely and big entries are rare.
For a 3 to occur, A160690(n)=p^2 and A160690(n)-A160690(n-1)=q^2 must be squares of primes,
hence A160690(n-1)=p^2-q^2=(p+q)(p-q), which "usually" has more divisors than observed 
in the first 30000 terms:
Apart from small (hence irrelevant) counterexamples, p^2-q^2 is a multiple of 24.
If p^2-q^2 = 2^a * 3^b has at most 15 divisors, i.e. if (a+1)*(b+1) <= 15 and a>=3 and 
b>=1, then p^2-q^2 must be smaller than possible for n>=25000.
Hence p^2-q^2 = 2^a*3^b*c with a>=3, b>=1, c>1 and has at least 4*2*2=16 divisors.

In other words: The not yet seen value 16 must occur before we can even think of seeing 3.

Hagen