A366412 Number of nontrivial solutions to the P^*_k problem in base n.

0, 0, 1, 0, 5, 0, 2, 2, 8, 0, 12, 0, 9, 13, 7, 0, 16, 0, 13, 19, 15, 0, 17, 6, 14, 6, 24, 0, 46, 0, 4, 18, 21, 15, 48, 0, 14, 18, 35, 0, 56, 0, 17, 36, 32, 0, 28, 10, 35, 33, 29, 0, 32, 31, 46, 33, 28, 0, 79, 0, 21, 31, 21, 25, 114, 0, 17, 30, 109, 0, 36, 0, 16, 40, 48, 28, 132

Offset: 2

Author

Sayan Dutta, Oct 09 2023

Keywords

• nonn
• base

Comments

For a given fixed base, the number [a_1 a_2 ... a_(2*k+1)] is said to satisfy the property P_k^* if [a_1 ... a_k]*[a_(k+1) ... a_(2*k+1)] = [a_1 ... a_(k+1)]*[a_(k+2) ... a_(2*k+1)], where [...] is to be interpreted as a block of digits.

The property P^*_k is a subcase of a mathematically inaccurate method where cancelling the common digits of the numerator and denominator correctly reduces it (usually known as anomalous cancellation). This sequence deals only with those anomalously cancellable fractions where there are equal numbers of digits in the numerator and denominator, and the last digit of the numerator is cancelled with the first digit of the denominator.

This is also equivalent to solving the Diophantine equation (a*B + b)*c = a*(b*B^k + c) with 0 < b < B and 0 < a,c < B^k.

All the solutions of a(p^n) where p^n is a prime power are three-digit solutions (proved in the paper by Saha et al.). For example, see Example section.

For a given base B, the number of solutions of P_k^* become constant beyond k=max{5, 2*log_2(B - 1) + 2} (proved in the paper by Saha et al.).

If [a_1 ... a_k b c_1 ... c_k] is a solution, then so is [a_1 ... a_k b b b c_1 ... c_k]. The latter is called an extension of the former, and is counted as a trivial solution. See Proposition 1 of Saha et al. link.

A solution is always of the form [a_1 ... a_k b...b c_k] (see Theorem 2 in the paper by Saha et al.).

It has been conjectured that for a given composite base B, if there are no new nontrivial solutions (except for extensions) in (2k + 1) digits, then there would be no new solutions in (2k + 3) digits (see Saha et al. link).

Examples

			For n = 10, the a(10) = 8 solutions correspond to 16/64 = 1/4, 26/65 = 2/5, 19/95 = 1/5, 49/98 = 4/8, 217/775 = 21/75, 249/996 = 24/96, 1249/9992 = 124/992 and 34027/77776 = 3402/7776.
For n = 9 = 3^2, the only two solutions are 14/43 and 28/86.
		

Links

• R. P. Boas, Anomalous Cancellation, The Two-Year College Mathematics Journal, Vol. 3, No. 2 (Autumn, 1972), pp. 21-24.
• Shalosh B. Ekhad, Automated Generation of Anomalous Cancellations, arXiv:1709.03379 [math.HO], 2017.
• Satvik Saha, Sohom Gupta, Sayan Dutta, and Sourin Chatterjee, Characterising Solutions of Anomalous Cancellation, arXiv:2302.00479 [math.HO], 2023.

Crossrefs

Cf. A159975/A159976, A291093/A291094, A291965/A291966.

Programs

Python

import math
LEN = 79
carr=[]
for base in range(2, LEN):
    k = int(2 * math.log(base - 1) / math.log(2) + 2) + 1
    k = max(k, 5)
    I = (base ** k - 1) // (base - 1) - 1
    count = 0
    for b in range(2, base):
        for c_k in range(1, b):
            c = b * I + c_k
            a = b * c // (b * base - (base - 1) * c_k)
            if a < base ** (k - 1):
                continue
            if (a * base + b) * c == a * (b * base**k + c):
                count += 1
    carr.append(count)
print(carr)

Formula

a(p) = 0 if and only if p is a prime (see Theorem 3 of Saha et al. link).

a(n) <= (n-2)*(n-3)/2 (see Proposition 4 of Saha et al. link).