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a(n) is the number of total solutions (minus the n-th prime) to x^y == y^x (mod p) where 0 < x,y <= p and p is the n-th prime.
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%I #32 Sep 09 2022 17:14:05

%S 0,0,2,10,10,16,22,40,56,48,70,64,66,74,114,130,118,122,138,168,220,

%T 174,158,270,242,242,234,212,238,308,284,272,334,296,318,332,424,364,

%U 368,416,370,470,524,510,464,474,552,542,480,604,586,554,768,578,752,618,628,880,752,634,702,606,846

%N a(n) is the number of total solutions (minus the n-th prime) to x^y == y^x (mod p) where 0 < x,y <= p and p is the n-th prime.

%H Chai Wah Wu, <a href="/A355486/b355486.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A355419(n) - A000040(n).

%F a(n) = 2*(number of solutions to x^y == y^x (mod p) where 1 < x < y < p). - _Chai Wah Wu_, Aug 30 2022

%p f:= proc(n) local p,x,y,t;

%p p:= ithprime(n);

%p t:= 0;

%p for x from 2 to p-1 do

%p for y from x+1 to p-1 do

%p if x&^y - y&^x mod p = 0 then t:= t+1 fi

%p od od:

%p 2*t

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Aug 31 2022

%o (Python)

%o from sympy import prime

%o def f(n):

%o S = 0

%o for x in range(1, n + 1):

%o for y in range(x + 1 , n + 1):

%o if ((pow(x, y, n) == pow(y, x, n))):

%o S += 2

%o return S

%o def a(n): return f(prime(n))

%o (Python)

%o from sympy import prime

%o def A355486(n):

%o p = prime(n)

%o return sum(2 for x in range(2,p-1) for y in range(x+1,p) if pow(x,y,p)==pow(y,x,p)) # _Chai Wah Wu_, Aug 30 2022

%Y Cf. A000040, A355419.

%K nonn

%O 1,3

%A _DarĂ­o Clavijo_, Jul 04 2022

%E More terms from _Robert Israel_, Aug 31 2022