OFFSET
1,1
COMMENTS
Very similar to A113650 but modulo p^3.
MATHEMATICA
a[n_]:= Mod[Fibonacci[(n-JacobiSymbol[n, 5])], Power[n, 3]]; Table[a[Prime[n]], {n, 50}]
PROG
(Sage)
p = 1
while p < 200:
print(fibonacci(p-jacobi_symbol(p, 5))%pow(p, 3), end=', ')
p = next_prime(p)
(PARI) a(n) = my(p=prime(n)); lift(Mod([1, 1; 1, 0]^(p-kronecker(p, 5)), p^3)[1, 2]); \\ Michel Marcus, Feb 28 2022
(Python)
from sympy import prime, fibonacci
from sympy.ntheory import jacobi_symbol
def A351989(n): return fibonacci((p := prime(n))-jacobi_symbol(p, 5)) % p**3 # Chai Wah Wu, Feb 28 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Javier Rivera Romeu, Feb 27 2022
STATUS
approved