Quote:
Originally Posted by Siemelink
There are 2 squares remaining, 2500 and 13456, but that is just coincidence I think. There are many square k's that do have a prime.
Also, 24 = 6*2*2. Several of the conjectures remove k = 6*square, but I don't understand why. How can I check if it can be removed here also?
Willem.
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It's not always very easy. The way I do it is to look for patterns in the factors of the various n-values for specific k-values. If there are algebraic factors, it's most common for them to be in a pattern of f*(f+2), i.e.:
11*13
179*181
etc.
In other cases there may be a consistent steady increase in the differences of their factors, which is especially tricky to find but indicates the existence of algebraic factors. That's what I ran into on base 24.
For your 3 cases here, you have:
k=24:
n-value : factors
1 : 13*83
2 : 23*2113
3 : 17*103*1249
4 : 23*163*26251
9 : 2843*6387736694293
Analysis:
For n=3 & 4, multiplying the 2 lower prime factors together does not come close to the higher prime factor so little chance of algebraic factors.
For n=9, the large lowest prime factor that bears no relation to the other prime factor means that there is unlikely to be a pattern to the occurrences of large prime factors so there must be a prime at some point.
k=2500:
n-value : factors
1 : 19*31*191
2 : 13*173*2251
3 : 89*2559691
4 : 19*73^2*103*983
9 : 9439*4280051*46824991
For n=9 same explanation as k=24.
k=13456:
n-value : factors
1 : 269*2251
2 : 17*23*227*307
3 : 31*39554129
4 : 7*23*467*503*1459
7 : 3319*1514943103721
For n=7 same explanation as k=24.
The prime factors for n=9, n=9, and n=7 respectively make it clear to me that these k-values should all yield primes at some point so you are correct to include them as remaining.
The higher-math folks may be able to chime in and answer why there are an abnormally large # of k's that are perfect squares that end up remaining even though they don't have known algebraic factors for most bases. IMHO, it's because there ARE algebraic factors for a subset of the universe of n-values on them but not for all of the n-values. Hence they are frequently lower weight than the other k's but NOT zero weight and so should eventually yield a prime.
Gary