A285097 a(n) = difference between the positions of two least significant 1-bits in base-2 representation of n, or 0 if there are less than two 1-bits in n (when n is either zero or a power of 2).

0, 0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 1, 1, 2, 1, 1, 0, 4, 3, 1, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 5, 4, 1, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 1, 4, 3, 1, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 6, 5, 1, 4, 2, 1, 1, 3, 3, 2, 1, 1, 2, 1, 1, 2, 4, 3, 1, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 4, 1, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 1, 4, 3, 1, 2, 2, 1, 1, 1

Offset: 0

Author

Antti Karttunen, Apr 20 2017

Keywords

• nonn
• base
• easy

Comments

a(1+n) is the length of the least significant run of 0-bits in n, or 0 if n is one of terms of A000225. - Antti Karttunen, Oct 14 2023

Examples

			For n = 3, "11" in binary, the second least significant 1-bit (the second 1-bit from the right) is at position 1 and the rightmost 1-bit is at position 0, thus a(3) = 1-0 = 1.
For n = 4, "100" in binary, there is just one 1-bit present, thus a(4) = 0.
For n = 5, "101" in binary, the second 1-bit from the right is at position 2, and the least significant 1 is at position 0, thus a(5) = 2-0 = 2.
For n = 26, "11010" in binary, the second 1-bit from the right is at position 3, and the least significant 1 is at position 1, thus a(26) = 3-1 = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000120, A000225, A000265, A007814, A119387, A129760, A285099, A366370.

Programs

Mathematica

a007814[n_]:=IntegerExponent[n, 2]; a285099[n_]:=If[DigitCount[n, 2, 1]<2, 0, a007814[BitAnd[n, n - 1]]]; a[n_]:=If[DigitCount[n, 2, 1]<2, 0,a285099[n] - a007814[n]]; Table[a[n], {n, 0, 150}] (* Indranil Ghosh, Apr 20 2017 *)

PARI

A285097(n) = if(!n || !bitand(n,n-1), 0, valuation((n>>valuation(n,2))-1, 2)); \\ Antti Karttunen, Oct 14 2023

Python

import math
def a007814(n): return int(math.log(n - (n & n - 1), 2))
def a285099(n): return 0 if bin(n)[2:].count("1") < 2 else a007814(n & (n - 1))
def a(n): return 0 if bin(n)[2:].count("1")<2 else a285099(n) - a007814(n) # Indranil Ghosh, Apr 20 2017

Scheme

(define (A285097 n) (if (<= (A000120 n) 1) 0 (- (A285099 n) (A007814 n))))

Formula

If A000120(n) < 2, a(n) = 0, otherwise a(n) = A285099(n) - A007814(n) = A007814(A129760(n)) - A007814(n).

a(n) = 0 if n is 0 or of the form 2^k, (k>=0), otherwise a(n) = v_2(A000265(n)-1), where v_2(i) = A007814(i). - Ridouane Oudra, Oct 20 2019