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For any relatively prime pair A and C, there are infinitely many solutions A*B^A=C*D^C, with B and D of the form B=C^W*A^X, D=C^Y*A^Z, and W,Y constrained by AW=CY+1 and X,Z constrained by CZ=AX+1. Details and examples on Munafo web page.

From Chai Wah Wu, May 17 2021: (Start)

Sequence is infinite.

If a, b > 1 and b^a-b == 0 mod a+1 then b^c+b+c is a term for c = ab(b^(a-1)-1)/(a+1), y = c/a, x = b^a.

If b > 1 and b <> 2 mod 3, then b^(2b(b-1)/3)+b(2b+1)/3 is a term.

If b > 2, then b^((b-1)(b^(b-2)-1)) + b + (b-1)(b^(b-2)-1) is a term. (End)

From Chai Wah Wu, May 18 2021: (Start)

Either c>=3 or y>=3. If c=y=2, we get b^2+b+2=x^2+x-2, i.e. (x-b)(x+b+1) = 4. Since x>1 and b>1, x+b+1>4, a contradiction.

This allows for a faster search algorithm by assuming c>=3 and y>=3. The cases c=2 and y>=3 can be dealt with by picking y>=3 and solving for b in the quadratic equation b^2+b+2=x^y+x-y. Similarly for c>=3 and y=2. This approach was used to confirm a(9). (End)

For n >= 3, we have max(c,y) >= 5. First note that c == y (mod 2). Case (c,y) = (3,3) implies (x-b)|6 and leads to quadratic equations with no integer roots. Case (c,y) = (4,2) corresponds to a quartic curve and has the only solution (b,x) = (3,9) giving a(2)=88, while case (c,y) = (2,4) has the only solution (b,x) = (3,2) giving a(1)=14. Finally, case (c,y) = (4,4) implies (x-b)|8 and leads to cubic equations with no integer roots. - Max Alekseyev, Feb 10 2025

Since a number can't be expressed simultaneously as 2*t^2 and 4*u^4, any number in this sequence must have a representation with A >= 5.

Up to 10^52, the only two terms not of the form 2*k^2 are 3*41841412812^3 = 4*86093442^4 = 64*3^64 and 3*54043195528445952^3 = 4*3298534883328^4 = 81*4^81.

From Charlie Neder, Jul 21 2018: (Start)

For each prime p and each value of A, the p-adic valuation of n must be congruent to the p-adic valuation of A modulo A. As a consequence, if two numbers k and m have greatest common divisor g and at least one of (k/g)^(1/g) or (m/g)^(1/g) is not an integer then no number n can have both A = k and A = m since this would lead to an unsolvable system of modular congruences.

If a number n is in this sequence with corresponding A-values {a,b,c}, then n*k^lcm(a,b,c) is also in this sequence for all k. Of the first 1198 terms, 949 of these are of the form 344373768*k^24, and 164 more are of the form 30233088000000*k^30. As values of n get larger, the proportion of primitive values rapidly decreases. (End)