March 2019

[Xyzzy]

http://www.research.ibm.com/haifa/po...March2019.html

[uau]

The "bonus question" part seems badly thought out. It seems to assume a particular value for n1. But the requirement is symmetrical for n1 and n2, and for the question to not be completely trivial, n1 and n2 must be different numbers (though that's not quite spelled out in the question). So even if there is a "simplest" answer, you can at least exchange n1 and n2.

And there actually are infinitely many possible answers. There only finitely many possible combinations of values for (BAT(x, N1), BAT(x, N2)). Thus infinitely many primes have the same combination, and thus there are infinitely many possible answers. I doubt all of those have some meaningful connection between BAT(sqrt(n1), 10000) and IBM.

[uau]

The question has been updated to clarify that it's the BAT values that must be primes, not the answer numbers. It's still true that there are infinitely many possible answers.

[Xyzzy]

http://www.research.ibm.com/haifa/po...March2019.html

[Dieter]

The solution is containing a little error: N2 = 1.76E10 and not 1.76E9

[Dieter]

Searching and checking only numbers ni with BAT((sqrt(ni),10000) = 9100 from the beginning I had found immediately 53 and 477, but my "quadruple-precision floating point format" codes were not exact enough to compute BAT((sqrt(ni),N1) or BAT((sqrt(ni),N2).
So first of all I had to write routines for the octuple-precision floating point format (4 registers for one floating point number) (236 bits for the mantissa). Then I was able to see that
BAT((sqrt(53),N1) = BAT((sqrt(477),N1) and so on.

[uau]

Originally Posted by Dieter

my "quadruple-precision floating point format" codes were not exact enough to compute BAT((sqrt(ni),N1) or BAT((sqrt(ni),N2

You don't need floating point. In the case of square roots, the usual rational approximation algorithm works out so that all the intermediate values can be represented by small integers (you only get numbers of the form (sqrt(n) + a)/b, where a and b are below 2*sqrt(n)).


The code below demonstrates this. Note that it prints all optimal lower and upper bounds, not just optimal approximations.

#!/usr/bin/python3

n = 53
limit = 1e19

k = int(n**.5)
assert k**2 != n
e = 0
a, d, p = 0, 1, n
lb, ub = [0, 1], [1, 0]
while lb[1] < limit:
    print(('lower' if e == 0 else 'upper') + ' bounds:')
    i = (a+k)//d
    for _ in range(i):
        lb[0] += ub[0]
        lb[1] += ub[1]
        if lb[1] >= limit:
            break
        print(lb[0], '/', lb[1])
    a, d, p = i*d - a, p - i*(i*d -2*a), d
    assert a < (k+1) and d < 2*(k+1)
    lb, ub, e = ub, lb, 1-e