[seqfan] Re: Question from Harvey Dale about A233552

[Don Reble]

> Adding a bound on k is not an acceptable solution, imho!

Right: but A233552 might might already have an implicit bound on something.

For f(n,k) = n * 4^k - 1

If n is                     then that modulus divides f(n,k) if k is
-------                     ----------------------------------------
1,4 mod 5                   0,1 mod 2
1,2,4 mod 7                 0,1,2 mod 3
1,7,4 mod 9                 0,1,2 mod 3
1,10,9,12,3,4 mod 13        0,1,2,3,4,5 mod 6

One can combine those moduli (plus the given n == 1 mod 6) to
obtain covering sets, for example
    n == 1 mod 5, 1 mod 9, 4 mod 7, and 10 mod 13: n == 361 mod 8190
So 361 is in the sequence.
There are`2x3 choices for mod-5 and mod-7: those leave only two
choices for mod-9, and then mod-13 is forced.
The 12 choices yield
    n = {  361,  919, 1681, 1849, 2419, 2629,
          3301, 5209, 5539, 5581, 6421, 7771 } mod 8190
That yields the published A233552 sequence, except for 15661.

To show 15661 is there, use moduli 17 and 241. One can then ignore
either mod-5 or mod-13.
That ability to ignore complicates things, but one can compute a finite
list of values, modulo 2*5*7*9*13*17*241, describing those values which
can be easily proven using covering sets.
The implicit bound is on the moduli within the covering sets.

    ---

> Note that if n=m^2 is a square, n*4^k-1 = (m*2^k-1)*(m*2^k+1), so
> (if m*2^k-1 > 3) (n*4^k-1)/3 must be composite. Thus 25, 49, 121, 169,
> 289, 373, 529, 625, 751, 841, 961 should certainly be in the sequence. 

    That is, all 1-mod-6 squares, except for 1. (Not 373, 751.)

-- 
Don Reble  djr at nk.ca