A351989 Fibonacci(p-J(p,5)) mod p^3, where p is the n-th prime and J is the Jacobi symbol.

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.

Links

Table of n, a(n) for n=1..46.

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

Adjacent sequences: A351986 A351987 A351988 * A351990 A351991 A351992

Keyword

nonn,easy

Author

Javier Rivera Romeu, Feb 27 2022

Status

approved