%I
%S 0,3,6,9,12,15,18,24,27,30,33,36,39,45,48,51,54,57,60,63,66,72,75,78,
%T 90,96,99,102,105,108,111,114,120,123,126,129,132,135,141,144,147,150,
%U 153,156,159,165,177,180,183,189,192,195,198,201,204,207,210,216,219
%N Numbers n such that representation in base 4 has same number of 1's and 2's.
%C Numbers such that sum (1)^k*b(k) = 0 where b(k)=kth binary digit of n.  _Benoit Cloitre_, Nov 18 2003
%C Conjecture: a(C(2n,n)1) = 4^n  1. (A000984 is C(2n,n).  _Gerald McGarvey_, Nov 18 2007
%C From _Russell Jay Hendel_, Jun 23 2015: (Start)
%C We prove the McGarvey conjecture (A) a(e(n,n)1) = 4^n1, with e(n,m) = a034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{ndigit, base4 numbers with nk more 1digits than 2digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of ndigit, base4 numbers with nk more 1digits than 2digits.
%C We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1digits and 2digits. The biggest ndigit, base4 number is 333...3 (n copies of 3). Since 333...33 has zero 1digits and zero 2digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n1 is the (e(n,n)1)st member of A039004, proving the McGarvey conjecture. (End)
%H Alois P. Heinz, <a href="/A039004/b039004.txt">Table of n, a(n) for n = 1..10000</a>
%F Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n.  _Benoit Cloitre_, Nov 24 2002
%F That conjecture is false. The number of members of the sequence from 0 to 4^d1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction.  _Robert Israel_, Jun 24 2015
%p N:= 1000: # to get all terms up to N, which should be divisible by 4
%p B:= Array(0..N1):
%p d:= ceil(log[4](N));
%p S:= Array(0..N1,[seq(op([0,1,1,0]),i=1..N/4)]):
%p for i from 1 to d do
%p B:= B + S;
%p S:= Array(0..N1,i> S[floor(i/4)]);
%p od:
%p select(t > B[t]=0, [$0..N1]); # _Robert Israel_, Jun 24 2015
%o (PARI) for(n=0,219,if(sum(i=1,length(binary(n)),(1)^i*component(binary(n),i))==0,print1(n,",")))
%o See link in A139351 for Fortran program.
%Y Cf. A139370, A139371, A139372, A139373.
%Y Cf. A139351, A139352, A139353, A139354, A139355.
%K nonn,base,easy
%O 1,2
%A _Olivier GĂ©rard_
