Proof? Re: Are A055999 and A074171 somehow the same?

[Paul C. Leopardi]

On Thursday 07 October 2004 19:44, Lßbos ElemÚr wrote:
> On 6 Oct 2004, at 10:32, Alonso Del Arte wrote:
> > I think that if we can prove that A055999 and A074171 are the same
> > (except for the two initial terms), then the two sequences should be
> > merged, with "a(n)=n*(n+7)/2" as the primary definition; and "Start
> > with 1, add the next number if one gets a prime then subtract the next
> > number else add the next" as a comment.
> >
> > But what holds me back from asserting this is that I don't know how to
> > prove they are in fact the same. I have calculated a couple dozen more
> > terms for both and they agree, but I could calculate a million terms
> > and still stop short of the term that proves the two sequences are in
> > fact different.
> ...
> Do not delete A074171, because its definition is dependent on
> sequence of primes...Thus the coincidence with simple polynomial
> A055999 is surprizing.
...

I think I have a proof that A055999 and A074171 coincide, except for the first 
two terms of A074171.

For A055999, a(n)=n*(n+7)/2 is always composite. 
If n is even, say n=2*k, then a(n)=a(2*k)=k*(2*k+7).
If n is odd, n=2*k+1, then a(n)=(2*k+1)*(k+4).
The forward difference is 
a(n+1)-a(n) = ((n+1)(n+8) - n*(n+7)) / 2
            = (n^2+9*n+8-n^2-7*n)/2
            = n+4.
Thus the rule, 
"Start with a(0)=0. For n>=0, if a(n) is composite, then a(n+1):=a(n)+(n+4), 
otherwise a(n+1):=a(n)-(n+4)" will produce A055999.

For clarity, call the sequence for A074171 b rather than a.
A074171 starts with n=1, and has b(3)=0. The rule for A074171 applied to b(3) 
says that if b(3) is composite, add the next number, which is 4.
In general, the rule,
"Start with b(3)=0. For n>=3, if b(n) is composite, then b(n+1):=b(n)+(n+1), 
otherwise b(n+1):=b(n)-(n+1)" will produce A074171 from b(3) onward.
This rule can also be written
"Start with b(3)=0. For n>=0, if b(n+3) is composite, then 
b(n+4):=b(n+3)+(n+4), otherwise b(n+4):=b(n+3)-(n+4)."

Therefore, by induction, b(n+3)=a(n).