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51st All-Russian Mathematical Olympiad for Schoolchildren. Problem. Let us call a natural number "lopsided" if it is greater than 1 and all its prime divisors end with the same digit. Is there an increasing arithmetic progression with a difference not exceeding 2025, consisting of 150 natural numbers, each of which is "lopsided"? (A. Chironov)

All powers of primes (A000961) are terms.

For a triangle with leg lengths x,y, the square has side length x*y*z/(x*y + z^2) and the area rounded down is a(n) = f(x,y,z) = floor((x*y*z/(x*y + z^2))^2) .

For a triangle with leg lengths x,y, the square has side length x*y/(x+y) and the area rounded down is a(n) = f(x,y) = floor( (x*y/(x+y))^2 ).

The adjacent sides of an octagon are not equal, the ratio of the larger side to the smaller one is sqrt(2), its area is 7 times the square of the shorter side. Using Pick's formula N = S - V/2 + 1 we obtain N = S - 3 = 7*A004613(n) - 3.

Values of x + 3*y in solutions of x^2 = 5*y^2 - 4*y in positive integers. In the solutions, the values of x and y are given by Fibonacci(4*n + 2) and Fibonacci(2*n + 1)^2 respectively.

The above Diophantine equation arises out of the following problem regarding the subdivision of a square into four triangles of integer area. For n >= 1, the sequence gives the areas of the squares in the solutions (see illustration in Links). Two lines are drawn from a corner of a square to points on the opposing sides. A third line is added between the two points so that the square is divided into four triangles. The area of each triangle is required to be an integer and those of the right triangles to form an arithmetic progression with difference 1. The smallest right triangle by area is the one formed by the third line. In the solutions, the area of the inner triangle is given by Fibonacci(4*n + 2) and the total area of the three right triangles is 3*Fibonacci(2*n + 1)^2. The area of the square is then equal to a(n).

The sequence consists of groups of identical digits. The numbers representing each group, starting from the second group, form a cycle (5, 8, 1, 4, 7, 0, 3, 6, 9, 2).

The first digit of 2^k is A008952(k) = floor(10^{k*log_10(2)}), the first digit of 5^k is A111395(k) = floor(10^{k*log_10(5)}), and the first digit of 8^k is A008952(3*k) = floor(10^{k*log_10(8)}), where "{x}" denotes the fractional part of x.

All numbers 2^k, 5^k, 8^k for k = a(n) > 0 start only with 3.

From Jianing Song, Dec 26 2022: (Start)

Write lg = log_10. Note that {k*lg(5)} = 1 - {k*lg(2)} and that {k*lg(8)} = {k*lg(2)} + 0, 1, or 2, so we have {k > 0 : 2^k, 5^k, 8^k all start with a} = {k: {k*lg(2)} is in I_a}, where I_a = (lg(a), lg(a+1)) intersect (1-lg(a+1), 1-lg(a)) intersect (((lg(a))/3, (lg(a+1))/3) U ((lg(a)+1)/3, (lg(a+1)+1)/3) U ((lg(a)+2)/3, (lg(a+1)+2)/3)). Note that I_3 = ((lg(3)+1)/3, 1-lg(3)) and I_a is empty otherwise (since (lg(a), lg(a+1)) intersect (1-lg(a+1), 1-lg(a)) is empty). As a result, k > 0 is a term if and only if (lg(3)+1)/3 < {k*lg(2)} < 1-lg(3).

Except for 0, also numbers k in A358196 such that 2^k starts with 3 (see the comment in A358196).

Claim: for n > 1, a(n+1) - a(n) = 10, 73, or 83. Proof: write a = (lg(3)+1)/3, b = 1-lg(3).

Step 1. If k > 0 is a term, then:

(a) {k*lg(2)} is in (a, b-(10*lg(2)-3)) => {(k+10)*lg(2)} is in (a+(10*lg(2)-3), b) => k+10 is a term;

(b) {k*lg(2)} is in (a+(25-83*lg(2)), b) => {(k+83)*lg(2)} is in (a, b-(25-83*lg(2))) => k+83 is a term.

Note that (a, b-(10*lg(2)-3)) U (a+(25-83*lg(2)), b) = (a, b), so at least one of k+10 and k+83 is a term.

Step 2. If k > 0 and k+m are both terms, 0 < m <= 83, then {k*lg(2)} and {(k+m)*lg(2)} are both in (a, b), so m*lg(2) is the range (N-(b-a), N+(b-a)) for some integer N, which implies that m = 10, 20, 73, or 83.

Note that if {k*lg(2)} < 1 - 2*(10*lg(2)-3), then {(k+10)*lg(2)} = {k*lg(2)} + (10*lg(2)-3), {(k+20)*lg(2)} = {k*lg(2)} + 2*(10*lg(2)-3), so {k*lg(2)} < {(k+10)*lg(2)} < {(k+20)*lg(2)}, which means that if k and k+20 are both terms, so is k+10. This shows that the next term after k is either 10, 73, or 83 larger.

We can show similarly that, for k > 0 being a term of this sequence:

(a) if k+73 is a term, then k+83 is a term;

(b) if k+83 is a term, then k+93 is a term;

(c) if k+166 is a term, then k+73 is a term;

(d) if k+10, k+73 are not terms (i.e., the next term is k+83), then k+176, k+196 are terms.

As a result, if we write out the sequence of the first differences, 73 is always followed by two 10's, and 83 is followed by one or two 10's; 83, 10, 10 is always followed by 73, 10, 10, and 83, 10 is always followed by 83, 10, 10. (End)

The length of the side of the hexagon is determined using a triangular grid depending on the number of links, which reduces to nontrivial solutions of the Pell equation x^2 - 3y^2 = 1 for even x.

A polyline qualifies if its n-th link has length n; the angle between adjacent links is Pi/3; and links of the same parity are parallel.

Finding the possible numbers of links reduces to finding nontrivial solutions to the Pell equation x^2 - 3y^2 = 1 for even x. See the formulas below.