%I #54 Apr 22 2023 10:34:14

%S 0,0,2,0,2,2,6,0,2,2,6,2,6,6,14,0,2,2,6,2,6,6,14,2,6,6,14,6,14,14,30,

%T 0,2,2,6,2,6,6,14,2,6,6,14,6,14,14,30,2,6,6,14,6,14,14,30,6,14,14,30,

%U 14,30,30,62,0,2,2,6,2,6,6,14,2,6,6,14,6,14,14,30,2,6,6,14,6

%N Number of 0's in n-th row of triangle in A071038.

%C a(n) is also the number of pairs of consecutive entries in the n-th row of Pascal's triangle with opposite parity.

%C All terms appear to be of the form 2^k - 2 (checked for n <= 10000). - _Michael De Vlieger_, Mar 02 2015

%C This appears to be equal to the number of previous values k, from 1..n-1, such that k AND n = k, where 'AND' is binary AND, and where the sequence starts at 1. For example, 1 AND 2 = 0, so a(2) = 0, while 1 AND 3 = 1 and 2 AND 3 = 2, so a(3) = 2. It follows from this that if n = 2^m - 1 then a(n) = n - 1 = 2^m - 2, giving the right border values noted below. - _Scott R. Shannon_, Apr 19 2023

%D S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

%H Charles R Greathouse IV, <a href="/A071055/b071055.txt">Table of n, a(n) for n = 0..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Rule182.html">Rule 182</a>

%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>

%F a(n)=b(n+1), with b(0)=0, b(2n)=b(n), b(2n+1)=2b(n)+2-2[n==0] (conjectured). - _Ralf Stephan_, Mar 05 2004

%F a(n) = pext(n, n + 1) (conjectured) where pext is the "parallel bits extract" instruction of the x86 CPU; pext(x, mask) extracts bits from x at the bit locations specified by mask to contiguous low bits. - _Falk Hüffner_, Jul 26 2019

%e From _Omar E. Pol_, Mar 02 2015: (Start)

%e Also, written as an irregular triangle in which the row lengths are the powers of 2, the sequence begins:

%e 0;

%e 0,2;

%e 0,2,2,6;

%e 0,2,2,6,2,6,6,14;

%e 0,2,2,6,2,6,6,14,2,6,6,14,6,14,14,30;

%e 0,2,2,6,2,6,6,14,2,6,6,14,6,14,14,30,2,6,6,14,6,14,14,30,6,14,14,30,14,30,30,62;

%e ...

%e It appears that the right border gives the nonnegative terms of A000918.

%e It appears that the row sums give A056182.

%e (End)

%t Count[#, n_ /; n == 0] & /@

%t Flatten[CellularAutomaton[182, {{1}, 0}, {{#}}] & /@ Range[0, 100],

%t 1] (* _Michael De Vlieger_, Mar 02 2015 *)

%o (PARI) A011371(n)=my(s); while(n>>=1, s+=n); s

%o a(n)=my(t=A011371(n)); sum(k=1,n,(A011371(k)+A011371(n-k)==t)!=(A011371(k-1)+A011371(n-k+1)==t)) \\ _Charles R Greathouse IV_, Mar 02 2015

%Y Cf. A071042.

%K nonn

%O 0,3

%A _Hans Havermann_, May 26 2002