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 A355069 a(n) is the number of solutions to x^y == y^x (mod p) where 0 < x,y <= p^2 - p and p is the n-th prime. 3
 2, 14, 104, 366, 1550, 3048, 6272, 9774, 14894, 34664, 48750, 84456, 108320, 128814, 128846, 209768, 255374, 424680, 479886, 563150, 700704, 782574, 712334, 1068320, 1614336, 1649000, 1721454, 1527566, 2299752, 2328704, 3654126, 3428750, 3834656, 4201134, 4596584 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS Chai Wah Wu, Table of n, a(n) for n = 1..558 Project Euler, Problem 801. x^y = y^x. FORMULA a(n) = Sum_{k=0..p-1} b(k)^2, where b(k) is the number of solutions to x^y == k (mod p) with p = prime(n), 0 < x <= p and 0 < y <= p-1. - Jinyuan Wang, Jun 19 2022 EXAMPLE For p=2: p x y x^y mod p y^x mod p - - - --------- --------- 2 1 1 1 1 2 2 2 0 0 Solutions: 2 . For p=3: p x y x^y mod p y^x mod p - - - --------- --------- 3 1 1 1 1 3 1 4 1 1 3 2 2 1 1 3 2 4 1 1 3 3 3 0 0 3 3 6 0 0 3 4 1 1 1 3 4 2 1 1 3 4 4 1 1 3 4 5 1 1 3 5 4 1 1 3 5 5 2 2 3 6 3 0 0 3 6 6 0 0 Solutions: 14 PROG (Python) from sympy import prime def f(n): S = 0 n2n=(n*n) - n for x in range(1, n2n + 1): for y in range(x + 1 , n2n + 1): if ((pow(x, y, n) == pow(y, x, n))): S += 2 return S + n2n def a(n): return f(prime(n)) (Python) from sympy import prime from sympy.ntheory.residue_ntheory import nthroot_mod def A355069(n): p = prime(n) return sum(sum(len(nthroot_mod(k, y, p, True)) for y in range(1, p))**2 for k in range(p)) # Chai Wah Wu, Aug 31 2022 (PARI) a(n) = my(p=prime(n), v=vector(p)); for(x=1, p, for(y=1, p-1, v[1+lift(Mod(x, p)^y)]++)); sum(i=1, p, v[i]^2); \\ Jinyuan Wang, Jun 19 2022 CROSSREFS Cf. A000040, A355419. Sequence in context: A123525 A286310 A295865 * A293044 A343818 A160780 Adjacent sequences: A355066 A355067 A355068 * A355070 A355071 A355072 KEYWORD nonn AUTHOR Darío Clavijo, Jun 17 2022 EXTENSIONS a(10)-a(26) from Michel Marcus, Jun 18 2022 More terms from Jinyuan Wang, Jun 19 2022 STATUS approved

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Last modified June 8 03:07 EDT 2023. Contains 363157 sequences. (Running on oeis4.)