A276561 For n-th odd prime prime(n) in binary form, a(n) is the decimal value of the bits in between the most significant and least significant bits which are both 1. Since there are no middle bits for odd_prime(1) = 3 = (11)_2, a(1) = 0.

0, 0, 1, 1, 2, 0, 1, 3, 6, 7, 2, 4, 5, 7, 10, 13, 14, 1, 3, 4, 7, 9, 12, 16, 18, 19, 21, 22, 24, 31, 1, 4, 5, 10, 11, 14, 17, 19, 22, 25, 26, 31, 32, 34, 35, 41, 47, 49, 50, 52, 55, 56, 61, 0, 3, 6, 7, 10, 12, 13, 18, 25, 27, 28, 30, 37, 40, 45, 46, 48, 51, 55, 58, 61, 63, 66, 70, 72, 76, 81, 82, 87, 88, 91, 93, 96, 100, 102, 103, 105, 111, 115, 117, 121, 123, 126, 4, 5, 14, 17, 22, 25, 28, 29, 32, 37, 40, 43, 44, 47, 50, 52, 53

Offset

1,5

Comments

Sequence of prime numbers can be used as a pseudo random bit sequence. But, since for every odd prime the most and least significant bits are 1, this introduces redundancy into this sequence. Removing the first and last bits therefore makes sense. Zeros in a(n) are primes of the form 2^k + 1, for which k must be a power of two, hence zeros of a(n) corresponds to Fermat primes.

Links

Table of n, a(n) for n=1..113.

Formula

a(n) = A164089(prime(n+1)). - Michel Marcus, May 29 2019

Example

Since the 9th odd prime is 29 = (11101)_2, a(9) = (110)_2 = 6.

Prog

(SageMath)
a = []
for p in range(3, 1000):
    if is_prime(p):
        t = p - (1<<int(floor(log(float(p))/log(2.0))))
        a.append(t>>1)
print(a)
(PARI) a(n) = my(b = binary(prime(n+1))); fromdigits(vector(#b-2, k, b[k+1]), 2); \\ Michel Marcus, May 29 2019

Crossrefs

Cf. A000215, A019434, A164089.

Sequence in context: A257783 A226874 A267901 * A325746 A325842 A153506

Adjacent sequences: A276558 A276559 A276560 * A276562 A276563 A276564

Keyword

nonn,base

Author

Adnan Baysal, Apr 10 2017

Status

approved