OFFSET
1,1
COMMENTS
m=1 is excluded because m^2 - 1 would be 0.
For all m > 1, both m^2 - 1 and m^2 + 1 are nonsquares, so each has an even number of divisors.
For k=1, m^2 + 1 is a prime, so T(n,1) == 0 (mod 2) for all n.
For n=1, m^2 - 1 = (m-1)*(m+1) is a prime, which occurs only at m=2; 2^2 + 1 = 5 is also a prime, so T(1,1) = 2 and T(1,k) = -1 for k > 1.
For n=2, m^2 - 1 = (m-1)*(m+1) has 4 divisors, so (except for T(2,2) = 3) T(2,k) is the average of a twin prime pair (A014574).
Is T(n,k) > 0 for all n > 1?
FORMULA
Define f(m) = tau(m^2 - 1) and g(m) = tau(m^2 + 1), where tau is the number of divisors function (A000005). Then
T(n,k) = min_{ m : f(m) = 2n and g(m) = 2k },
or -1 if no such m exists.
EXAMPLE
T(5,1) is the smallest integer m > 1 such that m^2 - 1 and m^2 + 1 have 10 and 2 divisors, respectively; since m^2 - 1 cannot be the 9th power of a prime, this requires that p^4 * q + 1 = m^2 = r - 1, where p, q, and r are distinct primes. The smallest such m is 28560, which gives a solution with p = 13, q = 28559, r = 815673601.
T(5,5) is the smallest integer m > 1 such that m^2 - 1 and m^2 + 1 each have 10 divisors; since neither m^2 - 1 nor m^2 + 1 can be the 9th power of a prime, this is the smallest m such that p^4 * q + 1 = m^2 = r^4 * s - 1, where p, q, r, and s are distinct primes: 22335421^4 * 248872305817685706212070112079 + 1 = 248872305817685706212070112080^2 = 13^4 * 2168601400616633822685176617536070987718973054081571441 - 1.
The first eight antidiagonals of the table are shown below.
.
n\k| 1 2 3 4 5 6 7 8
---+------------------------------------------------------------------
1 | 2 -1 -1 -1 -1 -1 -1 -1
2 | 4 3 18 72 16068 1620 1407318
3 | 10 8 168 360 369465818568 744768
4 | 14 5 32 68 182
5 | 28560 9 7 28398240
6 | 26 15 332
7 | 25071688922457240 728
8 | 56
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Jon E. Schoenfield, Jun 16 2024
STATUS
approved