A213859 a(n) = 2^n mod (n+2).

1, 2, 0, 3, 4, 4, 0, 2, 6, 6, 4, 7, 8, 2, 0, 9, 16, 10, 4, 2, 12, 12, 16, 8, 14, 20, 4, 15, 16, 16, 0, 2, 18, 22, 16, 19, 20, 2, 24, 21, 16, 22, 4, 38, 24, 24, 16, 32, 6, 2, 4, 27, 34, 52, 8, 2, 30, 30, 4, 31, 32, 2, 0, 8, 16, 34, 4, 2, 46, 36, 16, 37, 38, 17

Offset: 0

Author

Alex Ratushnyak, Jun 22 2012

Keywords

• nonn
• easy

Comments

Conjectures:

1. Indices of zeros: 2^(x+2)-2, x >= 0.

2. There are infinitely many n's such that a(n)=n.

3. Every integer k >= 0 appears in a(n) at least once.

4. Every k >= 0 appears in a(n) infinitely many times.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A015910, A066606, A062173, A106262.

Programs

Magma

[Modexp(2,n,n+2): n in [0..120]]; // G. C. Greubel, Jan 11 2023

Mathematica

Table[PowerMod[2, n, n+2], {n, 0, 100}] (* T. D. Noe, Jun 26 2012 *)

Python

print([2**n % (n+2) for n in range(222)])

SageMath

[power_mod(2,n,n+2) for n in range(121)] # G. C. Greubel, Jan 11 2023

Formula

a(n) = 2^n mod (n+2).

a(n) = A106262(2*n, n). - G. C. Greubel, Jan 11 2023