login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Fibonacci(p-J(p,5)) mod p^3, where p is the n-th prime and J is the Jacobi symbol.
1

%I #38 Feb 28 2022 19:34:32

%S 2,3,5,21,55,377,2584,2584,9867,754,27683,34706,55391,77486,2961,

%T 49237,178121,151768,269809,180340,137459,440741,304859,634125,3589,

%U 224018,925249,689508,276097,389850,1566164,488892,101791,731140,1838362,3406409,31557,2311014,3158805,4571698,2914836,3267050,1294789,6599056,7246251,159399

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

%C Very similar to A113650 but modulo p^3.

%t a[n_]:= Mod[Fibonacci[(n-JacobiSymbol[n, 5])], Power[n, 3]]; Table[a[Prime[n]], {n, 50}]

%o (Sage)

%o p = 1

%o while p < 200:

%o print(fibonacci(p-jacobi_symbol(p,5))%pow(p,3), end=', ')

%o p = next_prime(p)

%o (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

%o (Python)

%o from sympy import prime, fibonacci

%o from sympy.ntheory import jacobi_symbol

%o def A351989(n): return fibonacci((p := prime(n))-jacobi_symbol(p,5)) % p**3 # _Chai Wah Wu_, Feb 28 2022

%Y Cf. A113650.

%K nonn,easy

%O 1,1

%A _Javier Rivera Romeu_, Feb 27 2022