login
A100714
Number of runs in binary expansion of A000040(n) (the n-th prime number) for n > 0.
6
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
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
Sequence in context: A373745 A368290 A111248 * A339364 A050123 A081386
KEYWORD
base,easy,nonn
AUTHOR
Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 11 2004
STATUS
approved