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A355069
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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.
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3
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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
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OFFSET
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1,1
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LINKS
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FORMULA
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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
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EXAMPLE
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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
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PROG
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(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
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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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