[seqfan] https://oeis.org/A254211 Finite?

[Robert Gerbicz]

2015-01-27 3:12 GMT+01:00 Ron Hardin <rhhardin at att.net>:
>
> The numbers for https://oeis.org/A254211 suggest that it might be finite
but I have no insight on the matter.
>
>
> Number of length n 1..(1+2) arrays with no leading or trailing partial
sum equal to a prime and no consecutive values equal
>
> rhhardin at mindspring.com
> rhhardin at att.net (either)
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/

In fact a(n)>0 is true for large n. Firstly a proof for infinite many n
values: let
z=[1,3,2,3,1,2,3,1,2,3,1,2,1,3,2,3,2,3,1,3,2,1,3,2,1,3,1,3,2,3,2,3,1,3,2,1,3,2,1,3,1,3,2,3,2,3,1,3,2,1,3,2,1,3,2,3,1];
w=[2,3,1,3,1,2,3,1,2,3,1,3,2,3];
and u=concat(z,w^k,z), where w^k is w repeated k times.

For k>0 it it easy to see that in s there is no consecutive equal term.
The sum of the terms in z is 120 and for w this is 30, hence in each block
the sum of the terms is divisible by 30. In w all leading/partial sum is
divisible by 2,3 or 5, this gives that all "middle" sums that ends/starts
in w is composite (because the sum is greater than 5 it can't be 2,3 or 5).
The same is almost true for z, only the first and the last 1 gives a
problem, but if the only two sums are composite then u is a good sequence
(note that both of these sums appear as leading and trailing sums)
s1=sum(i=1,n-1,u[i])=30*k+239
s2=sum(i=1,n-56,u[i])=30*k+121
(Length of z is 57)

It is trivial that there are infinite many k values for that s1 and s2 is
composite. With CRT we can easily give that if k==66 mod 77 then s1 is
divisible by 7 and s2 is divisible by 11 (and bigger than 11), so these are
composites.
As length of u is n=14*k+2*57=14*(77*g+66)+2*57=1078*g+1038 so
a(1078*g+1038)>0 , where g is a nonnegative integer.

For the general case let
w1=w=[2,3,1,3,1,2,3,1,2,3,1,3,2,3];
w2=[2,1,2,1,3,1,2,3,1,2,3,1,3,2,3,2,3,1,3,1,2,3,1,2,3,1,3,2,3];
w3=[2,1,2,1,3,1,2,3,1,2,3,1,3,2,3,2,3,1,3,1,2,3,1,2,3,1,3,2,3,2,3,1,3,1,2,3,1,2,3,1,3,2,3];

Length of w1,w2,w3 is 14,29,43 and their sum is 30,60,90. Furthermore all
leading/trailing sum is divisible by 2,3 or 5. The first term is 2, the
last term is 3.
Observe that for
r1=concat(w2^13,w3)
r2=concat(w1^30)
their length is 420, the sum of r1 is 870, while for r2 the sum is 900.
r3=concat(w2^13,w3^2)
r4=concat(w1^30,w3)
their length is 463, the sum of r3 is 960, and the sum of r4 is 990.

420 and 463 are relative primes this gives that if n is large (actually
n>=193692) then
C1*420+C2*463=n-2*57 is solvable in nonnegative integers, in that case
let u=concat(z,(r1 or r2 C1 times),(r3 or r4 C2 times),z)
length of u is n.
As above we have to consider only the
s1=sum(i=1,n-1,u[i])
s2=sum(i=1,n-56,u[i])
values, as all other conditions are still true for our sequence. C1 or
C2>n/1000 (if n is large), say C1 and fix the r3^C2 sequence, and vary only
the (r1 or r2 C1 times) part of the sequence. We have C1+1>n/1000 choices,
if all of them is wrong then we have n/1000 different(!) primes less than
3*n, what is impossible. (s1 and s2 values are different for all sequences
because s1%30=29 and s2%30=1).