A100714 Number of runs in binary expansion of A000040(n) (the n-th prime number) for n > 0.

2, 1, 3, 1, 3, 3, 3, 3, 3, 3, 1, 5, 5, 5, 3, 5, 3, 3, 3, 3, 5, 3, 5, 5, 3, 5, 3, 5, 5, 3, 1, 3, 5, 5, 7, 5, 5, 5, 5, 7, 5, 7, 3, 3, 5, 3, 5, 3, 3, 5, 5, 3, 3, 3, 3, 3, 5, 3, 7, 5, 5, 7, 5, 5, 5, 5, 7, 7, 7, 7, 5, 5, 5, 7, 5, 3, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 3, 5, 5, 3, 5, 3, 3, 5, 3, 3, 3, 5, 5, 5, 5, 7, 5, 5, 5

Offset

1,1

Comments

Record values of a(n) = 2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, ... are set at the indices n = 1, 3, 12, 35, 121, 355, 1317, 4551, 15897, 56475, 197249, 737926, ... - R. J. Mathar, Mar 02 2007

Links

Table of n, a(n) for n=1..105.

Eric Weisstein's World of Mathematics, Run-Length Encoding.

Formula

a(n) = A005811(A000040(n)).

Example

a(5)=3 because A000040(5) = 11_10 = 1011_2, which splits into three runs ({1}, {0}, {1,1}).

Mathematica

Table[Length[Split[IntegerDigits[Prime[n], 2]]], {n, 1, 128}]

Prog

(PARI) a(n, p=prime(n))=hammingweight(bitxor(p, p>>1)) \\ Charles R Greathouse IV, Oct 19 2015
(Python)
from sympy import prime
def a(n): return ((p:=prime(n))^(p>>1)).bit_count()
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Feb 25 2023

Crossrefs

Cf. A005811, A000040.

Sequence in context: A145652 A368290 A111248 * A339364 A050123 A081386

Adjacent sequences: A100711 A100712 A100713 * A100715 A100716 A100717

Keyword

base,easy,nonn

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 11 2004

Status

approved