Optimize $a$ and $b$ for $2^a\times3^b\approx143$

Question

I am interested in approximating positive integers with numbers in the form $N=2^a3^b$ where $a,b\in\mathbb{Z}$. For example, let's say that we want to approximate $143=11\times13$. We have,

$$2^a3^b\approx11\times13\Rightarrow a\log(2)+b\log(3)=\log(11)+\log(13)$$

$$a\log(2)+b\log(3)-\log(11)-\log(13)=0$$

The $\log p_1,\dots,\log p_t$ is linearly independent over $\mathbb{Q}$ (see), so obviously we cannot determine $a,b$ to have the above equation. I'm wondering if it is possible to optimize $a$ and $b$ to minimize the absolute error between $2^a3^b$ and $143$. Furthermore, I would be glad to solve this optimization problem in general case. In general, the problem can be stated in the form related to linear combination of the logarithm functions:

$$a \log (2) + b \log (3) + n_1\log(p_1)+n_2\log(p_2)+\cdots + n_k \log (p_k) = 0 $$

where the integers $n_1,\ldots,n_k$ and the prime numbers $p_1,\ldots,p_k$ are given and we want to optimize the integers $a$ and $b$.

---

I used a python script for $2^a\times 3^b\approx 143$ and the result was not so good. I looped over $a,b\in\{-10^6,\ldots,10^6\}$ and after about $35$ minutes of running, I got only two results for $|2^a3^b-143|<0.1$:

\begin{array}{|c|c|c|} \hline a & b & \text{Error} \\ \hline 825 & -516 & 0.05170246118419186 \\ -229 & 149 & 0.04546204418605271 \\ \hline \end{array}

The python script:

from numba import njit, prange
@njit(parallel=True)
def fn():
    for a in prange(-1000000, 1000001):
        i_a = 2.0a
        for b in prange(-1000000, 1000001):
            i_b = i_a * 3.0b
            v = i_b - 143.0
            if abs(v) <= 1e-01:
                print(
                    "a=",
                    a,
                    ", b=",
                    b,
                    ", error=",
                    abs(v),
                )
fn()

Answer 1

Convert the equation to $a+br = t$, $a,b \in \mathbb{Z}$ and $r,t \in \mathbb{R}$ and $|r|<1$.

Make good enough rational approximation to $r,t \implies r \approx \frac{p_k}{q_k}, t \approx \frac{r_k}{s_k}$.

$a+b\frac{p_k}{q_k} = \frac{r_k}{s_k} \implies q_ks_k a +s_kp_k b = q_k r_k \implies s_k | q_k, \ell_k s_k = q_k \implies q_k a +p_k b = \ell_k r_k$

$\gcd(p_k,q_k) = 1 \implies$ there is a solution $a,b$ to above equation by Euclidean algorithm to find gcd.

So find rational approximations $r \approx \frac{p_k}{\ell_k s_k}$ and $t \approx \frac{r_k}{s_k}$ such that $\gcd(p_k,\ell_k s_k) = 1$ and find solution $a,b$ by Euclidean algorithm.

Rational approximations can be found by decimal expansion with +1 added to numerator if needed to make gcd = 1 condition work.

Answer 2

I used @DanielV's and @DanielMathias's suggestions in the comments:

$$2^a3^b=143\implies a+b\log_23=\log_2143$$ $$b\approx\dfrac{(\log_2143)-a}{\log_23}$$ And iterated over $a$ to minimize the absolute value of the difference between $b$ and $\dfrac{(\log_2143)-a}{\log_23}$.

Here is the optimized code:

from numba import njit
from numpy import log2
from mpmath import mp
mp.dps = 50
@njit(parallel=False)
def fn():
    results = []
    for a in range(-1011, 1011+1):
        v = (log2(143) - a) / log2(3)
        s = abs(v - round(v))
        if s < 10**(-5):
            results.append((a, v))
    return results
results = fn()
for a, v in results:
    b = round(v)
    error = abs(143 - (mp.exp(mp.log(2) * a + mp.log(3) * b)))
    if error < mp.mpf('1e-8'):
        print("a: ", a, " b: ", b, " error: ", error)

It took $11$ minute and $51$ seconds to run this code using Google Colab's processor. Here is the output with the error term, $|2^a3^b-143|$, less than $10^{-8}$:

\begin{array}{|c|c|c|} \hline a & b & \text{Error} \\ \hline -97766741041 & 61683945837 & 0.0000000031507807699242165 \\ -87326880450 & 55097127167 & 0.0000000045983876908203010 \\ -77517158756 & 48907881876 & 0.0000000090896242338383130 \\ -76887019859 & 48510308497 & 0.0000000060459946117310390 \\ -67077298165 & 42321063206 & 0.0000000076420173130661400 \\ -66447159268 & 41923489827 & 0.0000000074936015326564330 \\ -56637437574 & 35734244536 & 0.0000000061944103922793130 \\ -56007298677 & 35336671157 & 0.0000000089412084535964800 \\ -46197576983 & 29147425866 & 0.0000000047468034714778310 \\ -35757716392 & 22560607196 & 0.0000000032991965506616947 \\ -25317855801 & 15973788526 & 0.0000000018515896298309042 \\ -14877995210 & 9386969856 & \color{blue}{0.0000000004039827089854595} \\ -4438134619 & 2800151186 & 0.0000000010436242118746395 \\ 6001725972 & -3786667484 & 0.0000000024912311327493929 \\ 16441586563 & -10373486154 & 0.0000000039388380536388005 \\ 26251308257 & -16562731445 & 0.0000000097491738709566810 \\ 26881447154 & -16960304824 & 0.0000000053864449745428625 \\ 36691168848 & -23149550115 & 0.0000000083015669501911850 \\ 37321307745 & -23547123494 & 0.0000000068340518954615790 \\ 47131029439 & -29736368785 & 0.0000000068539600294110340 \\ 47761168336 & -30133942164 & 0.0000000082816588163949490 \\ 57570890030 & -36323187455 & 0.0000000054063531086162290 \\ 58201028927 & -36720760834 & 0.0000000097292657373429740 \\ 68010750621 & -42910006125 & 0.0000000039587461878067696 \\ 78450611212 & -49496824795 & 0.0000000025111392669826560 \\ 88890471803 & -56083643465 & 0.0000000010635323461438878 \\ 99330332394 & -62670462135 & \color{blue}{0.0000000003840745747095345 }\\ \hline \end{array}

The colored numbers show the error lower than $10^{-9}$.

Answer 3

There is a generalisation of Euclid's algorithm to find (near) solutions to an affine linear equation $f(x, y) = a \cdot x + b \cdot y + c$ for some real coefficients. It goes as follows:

1. For any two vectors $v, w$ such that $f(w) \neq f(0)$ let $k(v, w) = \lfloor f(v)/(f(0) - f(w)) \rfloor$.
2. Initialize three vectors: $p \leftarrow (0,0)$, $v_1 \leftarrow (1,0)$, $v_2 \leftarrow (0, 1)$.
3. Let $k_0 = k(p, v_1)$ and $k_1 = k(p + v_2, v_1)$.
4. Substitute $(p, v_1, v_2) \leftarrow (p + k_0 v_1, v_2 + (k_1 - k_0) v1, v1).$
5. Now the next near solution is the best one among the four vectors $p$, $p+v_1$, $p+v_2$, $p + v_1 + v_2$.
6. Go to step 3. and repeat the procedure.

Note that the near solution may be stable for a few iterations, but will steadily improve overall. Here are the first subsequent different near solutions that are found for $f(x, y) = \log(2) \cdot x + \log(3) \cdot y - \log(143)$:

iteration	near solution	$\lvert f(x, y) \rvert$	$2^x 3^y$
1	$(7, 0)$	$1.1081$e-$1$	$128$
2 to 3	$(4, 2)$	$6.9687$e-$3$	$144$
4 to 6	$(23, -10)$	$6.5582$e-$3$	$142.0618$
7	$(-61, 43)$	$4.4942$e-$3$	$142.3588$
8 to 9	$(-229, 149)$	$3.1797$e-$4$	$142.9545$
10	$(17120, -10797)$	$5.8064$e-$6$	$143.0008$