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A352881 a(n) is the minimal number z having the largest number of solutions to the Diophantine equation 1/z = 1/x + 1/y such that 1 <= x <= y <= 10^n.

%I A352881 #174 Mar 11 2025 22:09:26
%S A352881 2,12,60,840,9240,55440,720720,6126120,116396280,232792560,5354228880,
%T A352881 26771144400,465817912560,4813451763120,24067258815600,
%U A352881 144403552893600,2671465728531600,36510031623265200,219060189739591200,4709794079401210800,18839176317604843200,221360321731856907600
%N A352881 a(n) is the minimal number z having the largest number of solutions to the Diophantine equation 1/z = 1/x + 1/y such that 1 <= x <= y <= 10^n.
%C A352881 Solving for z gives z = (x*y) / (x+y), so x*y == 0 (mod x+y).
%C A352881 All known terms are from A025487:
%C A352881   a(1) =    2 = 2;
%C A352881   a(2) =   12 = 2^2 * 3;
%C A352881   a(3) =   60 = 2^2 * 3 * 5;
%C A352881   a(4) =  840 = 2^3 * 3 * 5 * 7;
%C A352881   a(5) = 9240 = 2^3 * 3 * 5 * 7 * 11.
%C A352881 If a solution to the equation 1/z = 1/x + 1/y is found such that gcd(x,y,z) is a square, then x+y, x*y*z, and (x-y)^2 + (2*z)^2 are also squares.
%C A352881 For all solutions, x^2 + y^2 + z^2 is a square.
%C A352881 The sequence is indeed a subsequence of A025487, and likely of A126098 as well. - _Max Alekseyev_, Mar 01 2023
%C A352881 a(n) < 5*10^(n-1). - _Max Alekseyev_, Mar 01 2023
%H A352881 Max Alekseyev, <a href="/A352881/b352881.txt">Table of n, a(n) for n = 1..30</a>
%H A352881 Project Euler, <a href="https://projecteuler.net/problem=108">Diophantine reciprocals I, Problem 108</a>.
%e A352881 For n=1, we have the following, where r = (x*y) mod (x+y). (In the last four columns, each number marked by an asterisk is a square.)
%e A352881 .
%e A352881   r  z  x  y  x*y  x+y  x*y*z  x^2+y^2+z^2
%e A352881   -  -  -  -  ---  ---  -----  -----------
%e A352881   0  1  2  2    4*   4*     4*           9* (solution)
%e A352881   2  1  2  4    8    6      8           21
%e A352881   4  1  2  6   12    8     12           41
%e A352881   6  1  2  8   16*  10     16*          69
%e A352881   3  1  3  3    9    6      9*          19
%e A352881   0  2  3  6   18*   9*    36*          49* (solution)
%e A352881   3  2  3  9   27   12     54           94
%e A352881   0  2  4  4   16*   8     32           36* (solution)
%e A352881   8  2  4  8   32   12     64*          84
%e A352881   5  2  5  5   25*  10     50           54
%e A352881   0  3  6  6   36*  12    108           81* (solution)
%e A352881   7  3  7  7   49*  14    147          107
%e A352881   0  4  8  8   64*  16*   256*         144* (solution)
%e A352881   9  4  9  9   81*  18    324*         178
%e A352881 .
%e A352881 z = 2 has the largest number of solutions, so a(1) = 2.
%e A352881 The number of solutions for the resulting z cannot exceed A018892(z).
%o A352881 (Python)
%o A352881 def a(n):
%o A352881     # k=x*y and d=x+y
%o A352881     bc, bk, L = 0, None, 10**n
%o A352881     for k in range(1, L):
%o A352881         c, k2 = 0, k * k
%o A352881         for d in range(max(1, k2 // (L - k) + 1), k + 1):
%o A352881             if k2 % d == 0: c += 1
%o A352881         if c > bc:
%o A352881             bc, bk = c, k
%o A352881     return bk
%o A352881 (PARI)
%o A352881 a(n)=my(bc=0,bk=0,L=10^n);for(k=1,L-1,my(c=0,k2=k^2);for(d=max(1,k2\(L-k)+1),k,if(k2%d==0,c++););if(c>bc,bc=c;bk=k););return(bk); \\ _Darío Clavijo_, Mar 03 2025
%Y A352881 Cf. A025487, A099829, A126098, A067128, A018892, A126098, A225206.
%K A352881 nonn
%O A352881 1,1
%A A352881 _Darío Clavijo_, Apr 06 2022
%E A352881 a(6) from _Chai Wah Wu_, Apr 10 2022
%E A352881 a(7)-a(22) from _Max Alekseyev_, Mar 01 2023