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A351989
Fibonacci(p-J(p,5)) mod p^3, where p is the n-th prime and J is the Jacobi symbol.
1
2, 3, 5, 21, 55, 377, 2584, 2584, 9867, 754, 27683, 34706, 55391, 77486, 2961, 49237, 178121, 151768, 269809, 180340, 137459, 440741, 304859, 634125, 3589, 224018, 925249, 689508, 276097, 389850, 1566164, 488892, 101791, 731140, 1838362, 3406409, 31557, 2311014, 3158805, 4571698, 2914836, 3267050, 1294789, 6599056, 7246251, 159399
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
Cf. A113650.
Sequence in context: A113650 A259376 A060321 * A093522 A352124 A093499
KEYWORD
nonn,easy
AUTHOR
Javier Rivera Romeu, Feb 27 2022
STATUS
approved