A175930 Concatenation of run lengths in binary expansion of n, written in base 2, then converted to base 10.

1, 3, 2, 6, 7, 5, 3, 7, 13, 15, 14, 10, 11, 7, 4, 12, 15, 27, 26, 30, 31, 29, 15, 11, 21, 23, 22, 14, 15, 9, 5, 13, 25, 31, 30, 54, 55, 53, 27, 31, 61, 63, 62, 58, 59, 31, 28, 20, 23, 43, 42, 46, 47, 45, 23, 15, 29, 31, 30, 18, 19, 11, 6, 14, 27, 51, 50, 62, 63, 61, 31, 55, 109, 111

Offset: 1

Author

Dylan Hamilton, Oct 23 2010

Keywords

• base
• easy
• look
• nonn

Examples

			6 = 110, two runs, lengths 2 and 1, so we write down 101 and convert it to base 10, getting 5. So a(6) = 5.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..8192

Crossrefs

Cf. A004755, A101211, A321226.

Programs

Mathematica

f[n_] := FromDigits[Flatten[IntegerDigits[Length /@ Split[IntegerDigits[n, \ 2]], 2]], 2]
Array[f, NUMBER OF TERMS]

PARI

a(n) = my (b=[]); while (n, my (x=valuation(n+(n%2), 2)); b = concat(binary(x), b); n \= 2^x); fromdigits(b, 2) \\ Rémy Sigrist, Jul 02 2019

Python

from itertools import groupby
def a(n):
    c = "".join(bin(len(list(g)))[2:] for k, g in groupby(bin(n)[2:]))
    return int(c, 2)
print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Oct 02 2021

Formula

From Rémy Sigrist, Jul 02 2019: (Start)

a(2^k-1) = k for any k > 0.

a(2^k) = A004755(k) for any k > 0. (End)

Extensions

Edited by N. J. A. Sloane, Oct 23 2010